https://karnatakaeducation.org.in/KOER/en/api.php?action=feedcontributions&user=Anilkumar&feedformat=atomKarnataka Open Educational Resources - User contributions [en]2024-03-29T00:02:41ZUser contributionsMediaWiki 1.35.6https://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=17928Circles Tangents Problems2015-02-15T16:59:06Z<p>Anilkumar: </p>
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<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
a)<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br><br />
b)<br><br />
Let ∠QBP=x˚<br><br />
<br />
∴∠QPB=x˚ (∵PQ=BQ)<br><br />
Now Let ∠PAQ=y˚<br><br />
∠QPA=y˚ (∵ PQ=AQ)<br><br />
∴In △PAB<br><br />
<br />
∠PAB+∠PBA+∠APB=180˚<br><br />
<br />
y˚+x˚+(x˚+y˚)=180˚<br><br />
2x˚+2y˚=180˚<br><br />
2(x˚+y˚)=180˚<br><br />
x˚+y˚=90˚<br><br />
∴ ∠APB=90˚<br />
<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[Image:problem 3 on circle.png|300px]]<br />
==Concepts used==<br />
#The radii of a circle are equal<br />
#Properties of isosceles triangle.<br />
#SAS postulate<br />
#Properties of congruent triangles.<br />
<br />
==Prerequisite knowledge==<br />
#The radii of a circle are equal.<br />
# In an isosceles triangle angles opposite to equal sides are equal.<br />
#All the elements of congruent triangles are equal.<br />
<br />
==Algoritham==<br />
In ∆AOB <br><br />
AO=BO [Radii of a same circle]<br><br />
∴ ∠OAB = ∠OBA --------------I [∆AOB is an isosceles ∆}<br><br />
Then,<br> <br />
In ∆AOP and ∆BOP,<br><br />
AO = BO [Radii of a same circle]<br><br />
OP=OP [common side]<br><br />
∠OAP = ∠OBP [ from I]<br><br />
∴ ∆AOP ≅ ∆BOP [SAS postulate] <br><br />
∴ AP = BP [corresponding sides of congruent triangles ]<br />
<br />
=problem-4=<br />
'''In the given Quadrilateral ABCD , BC=38cm , QB=27cm , DC=25cm and AD⊥DC find the radius of the circle.(Ex:15.2. A-6)'''<br><br />
[[File:fig3.png|200px]]<br />
==Interpretation of problem==<br />
# In the quadrilateral ABCD sides BC , DC & QB are given . <br />
#AD⊥DC.<br />
# Asked to find the radius OS or OP<br />
<br />
==Concepts used==<br />
#Tangents drawn from an external point to a circle are equal.<br />
#In a quadrilateral, if all angles are equal and a pair of adjacent sides are equal then it is a square<br />
#In a circle, the radius drawn at the point of contact is perpendicular to the tangent<br />
==Algorithm== <br />
In the fig BC=38 cm and BQ=27 cm <br><br />
BQ=BR=27 cm (∵ by concept 1) <br><br />
∴CR=BC-BR=38-27=11 cm <br><br />
CR=SC=11 cm (∵ by concept 1) <br><br />
DC=25 cm <br><br />
∴ DS=DC-SC=25-11=14 cm <br><br />
DS=DP=14 cm (∵ by concept 1) <br><br />
Also AD⊥DC, OP ⊥ AD and OS ⊥ DC <br><br />
∠D=∠S=∠P=90˚ <br><br />
⇒ ∠O=90˚ <br><br />
∴ DSOP is a Square <br><br />
SO=OP=14 cm <br> hence Radius of given circle is 14 cm<br />
<br />
=Problem 5 [Ex-15.2-B8]=<br />
A circle is touching the side BC of △ABC at P. AB and AC when produced are touching the circle at Q and R respectively. Prove that AQ =<math>\frac{1}{2}</math> [perimeter of △ABC].<br />
[[File:123.png|300px]]<br />
<br />
==Algorithm==<br />
In the figure AQ , AR and BC are tangents to the circle with center O.<br><br />
BP=BQ and PC=CR (Tangents drawn from external point are equal) ---------- (1)<br><br />
<br />
Perimeter of △ABC=AB+BC+CA<br><br />
=AB+(BP+PC)+CA<br><br />
=AB+BQ+CR+CA ------ (From eq-1)<br><br />
=(AB+BQ)+(CR+CA)<br><br />
=AQ+AR ----- (From fig)<br><br />
=AQ+AQ -- --- (∵AQ=AR)<br><br />
=2AQ<br><br />
<br />
∴ AQ = <math>\frac{1}{2}</math> [perimeter of △ABC]<br />
<br />
=Problem-6 [Ex-15.4-B3]=<br />
In circle with center O , diameter AB and a chord AD are drawn. Another circle drawn with OA as diameter to cut AD at C. Prove that BD=2OC.[[File:15.4B3.png|300px]]<br />
==Algorithm==<br />
In figure, AB is the diameter of circle <math>C_{1}</math> and AO is the diameter of the circle <math>C_{2}</math><br>in △ADB and △ACO<br><br />
∠ADB=90° and ∠ACO=90° [∵angles in the semi circles]<br>∠DAB=∠CAO [∵common angles]<br>∴△ADB∼△ACO [∵equiangular triangles are similar]<br>∴<math>\frac{AB}{OA}</math>=<math>\frac{BD}{OC}</math>=<math>\frac{AD}{AC}</math> [∵corresonding sides of a similar triangles are proportional]<br>But AB=2OA----1 (∵diameter is twice the radius of a cicle)<br><math>\frac{AB}{OA}</math>=<math>\frac{BD}{OC}</math><br>from (1)<br><math>\frac{2OA}{OA}</math>=<math>\frac{BD}{OC}</math><br>∴BD=2OC<br />
<br />
<br />
<br />
=Problem-7 [Ex-15.4-A3]=<br />
<br />
In the figure AB=10cm,AC=6cm and the radius of the smaller circle is xcm. find x.<br />
<br />
[[File:circle.png|400px]]<br />
<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch internally.<br />
#OB and OF are the radii of the semicircle with center "O".<br />
#PC and PF are the radii of the circle with center "P".</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=17927Circles Tangents Problems2015-02-15T16:57:17Z<p>Anilkumar: /* == Problem-7 [Ex-15.4-A3] == */</p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
a)<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br><br />
b)<br><br />
Let ∠QBP=x˚<br><br />
<br />
∴∠QPB=x˚ (∵PQ=BQ)<br><br />
Now Let ∠PAQ=y˚<br><br />
∠QPA=y˚ (∵ PQ=AQ)<br><br />
∴In △PAB<br><br />
<br />
∠PAB+∠PBA+∠APB=180˚<br><br />
<br />
y˚+x˚+(x˚+y˚)=180˚<br><br />
2x˚+2y˚=180˚<br><br />
2(x˚+y˚)=180˚<br><br />
x˚+y˚=90˚<br><br />
∴ ∠APB=90˚<br />
<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[Image:problem 3 on circle.png|300px]]<br />
==Concepts used==<br />
#The radii of a circle are equal<br />
#Properties of isosceles triangle.<br />
#SAS postulate<br />
#Properties of congruent triangles.<br />
<br />
==Prerequisite knowledge==<br />
#The radii of a circle are equal.<br />
# In an isosceles triangle angles opposite to equal sides are equal.<br />
#All the elements of congruent triangles are equal.<br />
<br />
==Algoritham==<br />
In ∆AOB <br><br />
AO=BO [Radii of a same circle]<br><br />
∴ ∠OAB = ∠OBA --------------I [∆AOB is an isosceles ∆}<br><br />
Then,<br> <br />
In ∆AOP and ∆BOP,<br><br />
AO = BO [Radii of a same circle]<br><br />
OP=OP [common side]<br><br />
∠OAP = ∠OBP [ from I]<br><br />
∴ ∆AOP ≅ ∆BOP [SAS postulate] <br><br />
∴ AP = BP [corresponding sides of congruent triangles ]<br />
<br />
=problem-4=<br />
'''In the given Quadrilateral ABCD , BC=38cm , QB=27cm , DC=25cm and AD⊥DC find the radius of the circle.(Ex:15.2. A-6)'''<br><br />
[[File:fig3.png|200px]]<br />
==Interpretation of problem==<br />
# In the quadrilateral ABCD sides BC , DC & QB are given . <br />
#AD⊥DC.<br />
# Asked to find the radius OS or OP<br />
<br />
==Concepts used==<br />
#Tangents drawn from an external point to a circle are equal.<br />
#In a quadrilateral, if all angles are equal and a pair of adjacent sides are equal then it is a square<br />
#In a circle, the radius drawn at the point of contact is perpendicular to the tangent<br />
==Algorithm== <br />
In the fig BC=38 cm and BQ=27 cm <br><br />
BQ=BR=27 cm (∵ by concept 1) <br><br />
∴CR=BC-BR=38-27=11 cm <br><br />
CR=SC=11 cm (∵ by concept 1) <br><br />
DC=25 cm <br><br />
∴ DS=DC-SC=25-11=14 cm <br><br />
DS=DP=14 cm (∵ by concept 1) <br><br />
Also AD⊥DC, OP ⊥ AD and OS ⊥ DC <br><br />
∠D=∠S=∠P=90˚ <br><br />
⇒ ∠O=90˚ <br><br />
∴ DSOP is a Square <br><br />
SO=OP=14 cm <br> hence Radius of given circle is 14 cm<br />
<br />
=Problem 5 [Ex-15.2-B8]=<br />
A circle is touching the side BC of △ABC at P. AB and AC when produced are touching the circle at Q and R respectively. Prove that AQ =<math>\frac{1}{2}</math> [perimeter of △ABC].<br />
[[File:123.png|300px]]<br />
<br />
==Algorithm==<br />
In the figure AQ , AR and BC are tangents to the circle with center O.<br><br />
BP=BQ and PC=CR (Tangents drawn from external point are equal) ---------- (1)<br><br />
<br />
Perimeter of △ABC=AB+BC+CA<br><br />
=AB+(BP+PC)+CA<br><br />
=AB+BQ+CR+CA ------ (From eq-1)<br><br />
=(AB+BQ)+(CR+CA)<br><br />
=AQ+AR ----- (From fig)<br><br />
=AQ+AQ -- --- (∵AQ=AR)<br><br />
=2AQ<br><br />
<br />
∴ AQ = <math>\frac{1}{2}</math> [perimeter of △ABC]<br />
<br />
=Problem-6 [Ex-15.4-B3]=<br />
In circle with center O , diameter AB and a chord AD are drawn. Another circle drawn with OA as diameter to cut AD at C. Prove that BD=2OC.[[File:15.4B3.png|300px]]<br />
==Algorithm==<br />
In figure, AB is the diameter of circle <math>C_{1}</math> and AO is the diameter of the circle <math>C_{2}</math><br>in △ADB and △ACO<br><br />
∠ADB=90° and ∠ACO=90° [∵angles in the semi circles]<br>∠DAB=∠CAO [∵common angles]<br>∴△ADB∼△ACO [∵equiangular triangles are similar]<br>∴<math>\frac{AB}{OA}</math>=<math>\frac{BD}{OC}</math>=<math>\frac{AD}{AC}</math> [∵corresonding sides of a similar triangles are proportional]<br>But AB=2OA----1 (∵diameter is twice the radius of a cicle)<br><math>\frac{AB}{OA}</math>=<math>\frac{BD}{OC}</math><br>from (1)<br><math>\frac{2OA}{OA}</math>=<math>\frac{BD}{OC}</math><br>∴BD=2OC<br />
<br />
<br />
<br />
== Problem-7 [Ex-15.4-A3] == <br />
<br />
In the figure AB=10cm,AC=6cm and the radius of the smaller circle is xcm. find x.<br />
<br />
[[File:circle.png|400px]]<br />
<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch internally.<br />
#OB and OF are the radii of the semicircle with center "O".<br />
#PC and PF are the radii of the circle with center "P".</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=17926Circles Tangents Problems2015-02-15T16:54:11Z<p>Anilkumar: </p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
a)<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br><br />
b)<br><br />
Let ∠QBP=x˚<br><br />
<br />
∴∠QPB=x˚ (∵PQ=BQ)<br><br />
Now Let ∠PAQ=y˚<br><br />
∠QPA=y˚ (∵ PQ=AQ)<br><br />
∴In △PAB<br><br />
<br />
∠PAB+∠PBA+∠APB=180˚<br><br />
<br />
y˚+x˚+(x˚+y˚)=180˚<br><br />
2x˚+2y˚=180˚<br><br />
2(x˚+y˚)=180˚<br><br />
x˚+y˚=90˚<br><br />
∴ ∠APB=90˚<br />
<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[Image:problem 3 on circle.png|300px]]<br />
==Concepts used==<br />
#The radii of a circle are equal<br />
#Properties of isosceles triangle.<br />
#SAS postulate<br />
#Properties of congruent triangles.<br />
<br />
==Prerequisite knowledge==<br />
#The radii of a circle are equal.<br />
# In an isosceles triangle angles opposite to equal sides are equal.<br />
#All the elements of congruent triangles are equal.<br />
<br />
==Algoritham==<br />
In ∆AOB <br><br />
AO=BO [Radii of a same circle]<br><br />
∴ ∠OAB = ∠OBA --------------I [∆AOB is an isosceles ∆}<br><br />
Then,<br> <br />
In ∆AOP and ∆BOP,<br><br />
AO = BO [Radii of a same circle]<br><br />
OP=OP [common side]<br><br />
∠OAP = ∠OBP [ from I]<br><br />
∴ ∆AOP ≅ ∆BOP [SAS postulate] <br><br />
∴ AP = BP [corresponding sides of congruent triangles ]<br />
<br />
=problem-4=<br />
'''In the given Quadrilateral ABCD , BC=38cm , QB=27cm , DC=25cm and AD⊥DC find the radius of the circle.(Ex:15.2. A-6)'''<br><br />
[[File:fig3.png|200px]]<br />
==Interpretation of problem==<br />
# In the quadrilateral ABCD sides BC , DC & QB are given . <br />
#AD⊥DC.<br />
# Asked to find the radius OS or OP<br />
<br />
==Concepts used==<br />
#Tangents drawn from an external point to a circle are equal.<br />
#In a quadrilateral, if all angles are equal and a pair of adjacent sides are equal then it is a square<br />
#In a circle, the radius drawn at the point of contact is perpendicular to the tangent<br />
==Algorithm== <br />
In the fig BC=38 cm and BQ=27 cm <br><br />
BQ=BR=27 cm (∵ by concept 1) <br><br />
∴CR=BC-BR=38-27=11 cm <br><br />
CR=SC=11 cm (∵ by concept 1) <br><br />
DC=25 cm <br><br />
∴ DS=DC-SC=25-11=14 cm <br><br />
DS=DP=14 cm (∵ by concept 1) <br><br />
Also AD⊥DC, OP ⊥ AD and OS ⊥ DC <br><br />
∠D=∠S=∠P=90˚ <br><br />
⇒ ∠O=90˚ <br><br />
∴ DSOP is a Square <br><br />
SO=OP=14 cm <br> hence Radius of given circle is 14 cm<br />
<br />
=Problem 5 [Ex-15.2-B8]=<br />
A circle is touching the side BC of △ABC at P. AB and AC when produced are touching the circle at Q and R respectively. Prove that AQ =<math>\frac{1}{2}</math> [perimeter of △ABC].<br />
[[File:123.png|300px]]<br />
<br />
==Algorithm==<br />
In the figure AQ , AR and BC are tangents to the circle with center O.<br><br />
BP=BQ and PC=CR (Tangents drawn from external point are equal) ---------- (1)<br><br />
<br />
Perimeter of △ABC=AB+BC+CA<br><br />
=AB+(BP+PC)+CA<br><br />
=AB+BQ+CR+CA ------ (From eq-1)<br><br />
=(AB+BQ)+(CR+CA)<br><br />
=AQ+AR ----- (From fig)<br><br />
=AQ+AQ -- --- (∵AQ=AR)<br><br />
=2AQ<br><br />
<br />
∴ AQ = <math>\frac{1}{2}</math> [perimeter of △ABC]<br />
<br />
=Problem-6 [Ex-15.4-B3]=<br />
In circle with center O , diameter AB and a chord AD are drawn. Another circle drawn with OA as diameter to cut AD at C. Prove that BD=2OC.[[File:15.4B3.png|300px]]<br />
==Algorithm==<br />
In figure, AB is the diameter of circle <math>C_{1}</math> and AO is the diameter of the circle <math>C_{2}</math><br>in △ADB and △ACO<br><br />
∠ADB=90° and ∠ACO=90° [∵angles in the semi circles]<br>∠DAB=∠CAO [∵common angles]<br>∴△ADB∼△ACO [∵equiangular triangles are similar]<br>∴<math>\frac{AB}{OA}</math>=<math>\frac{BD}{OC}</math>=<math>\frac{AD}{AC}</math> [∵corresonding sides of a similar triangles are proportional]<br>But AB=2OA----1 (∵diameter is twice the radius of a cicle)<br><math>\frac{AB}{OA}</math>=<math>\frac{BD}{OC}</math><br>from (1)<br><math>\frac{2OA}{OA}</math>=<math>\frac{BD}{OC}</math><br>∴BD=2OC<br />
<br />
<br />
<br />
== == Problem-7 [Ex-15.4-A3] == ==<br />
<br />
In the figure AB=10cm,AC=6cm and the radius of the smaller circle is xcm. find x.<br />
<br />
[[File:circle.png|400px]]<br />
<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch internally.<br />
#OB and OF are the radii of the semicircle with center "O".<br />
#PC and PF are the radii of the circle with center "P".</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=17925Circles Tangents Problems2015-02-15T16:42:34Z<p>Anilkumar: </p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
a)<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br><br />
b)<br><br />
Let ∠QBP=x˚<br><br />
<br />
∴∠QPB=x˚ (∵PQ=BQ)<br><br />
Now Let ∠PAQ=y˚<br><br />
∠QPA=y˚ (∵ PQ=AQ)<br><br />
∴In △PAB<br><br />
<br />
∠PAB+∠PBA+∠APB=180˚<br><br />
<br />
y˚+x˚+(x˚+y˚)=180˚<br><br />
2x˚+2y˚=180˚<br><br />
2(x˚+y˚)=180˚<br><br />
x˚+y˚=90˚<br><br />
∴ ∠APB=90˚<br />
<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[Image:problem 3 on circle.png|300px]]<br />
==Concepts used==<br />
#The radii of a circle are equal<br />
#Properties of isosceles triangle.<br />
#SAS postulate<br />
#Properties of congruent triangles.<br />
<br />
==Prerequisite knowledge==<br />
#The radii of a circle are equal.<br />
# In an isosceles triangle angles opposite to equal sides are equal.<br />
#All the elements of congruent triangles are equal.<br />
<br />
==Algoritham==<br />
In ∆AOB <br><br />
AO=BO [Radii of a same circle]<br><br />
∴ ∠OAB = ∠OBA --------------I [∆AOB is an isosceles ∆}<br><br />
Then,<br> <br />
In ∆AOP and ∆BOP,<br><br />
AO = BO [Radii of a same circle]<br><br />
OP=OP [common side]<br><br />
∠OAP = ∠OBP [ from I]<br><br />
∴ ∆AOP ≅ ∆BOP [SAS postulate] <br><br />
∴ AP = BP [corresponding sides of congruent triangles ]<br />
<br />
=problem-4=<br />
'''In the given Quadrilateral ABCD , BC=38cm , QB=27cm , DC=25cm and AD⊥DC find the radius of the circle.(Ex:15.2. A-6)'''<br><br />
[[File:fig3.png|200px]]<br />
==Interpretation of problem==<br />
# In the quadrilateral ABCD sides BC , DC & QB are given . <br />
#AD⊥DC.<br />
# Asked to find the radius OS or OP<br />
<br />
==Concepts used==<br />
#Tangents drawn from an external point to a circle are equal.<br />
#In a quadrilateral, if all angles are equal and a pair of adjacent sides are equal then it is a square<br />
#In a circle, the radius drawn at the point of contact is perpendicular to the tangent<br />
==Algorithm== <br />
In the fig BC=38 cm and BQ=27 cm <br><br />
BQ=BR=27 cm (∵ by concept 1) <br><br />
∴CR=BC-BR=38-27=11 cm <br><br />
CR=SC=11 cm (∵ by concept 1) <br><br />
DC=25 cm <br><br />
∴ DS=DC-SC=25-11=14 cm <br><br />
DS=DP=14 cm (∵ by concept 1) <br><br />
Also AD⊥DC, OP ⊥ AD and OS ⊥ DC <br><br />
∠D=∠S=∠P=90˚ <br><br />
⇒ ∠O=90˚ <br><br />
∴ DSOP is a Square <br><br />
SO=OP=14 cm <br> hence Radius of given circle is 14 cm<br />
<br />
=Problem 5 [Ex-15.2-B8]=<br />
A circle is touching the side BC of △ABC at P. AB and AC when produced are touching the circle at Q and R respectively. Prove that AQ =<math>\frac{1}{2}</math> [perimeter of △ABC].<br />
[[File:123.png|300px]]<br />
<br />
==Algorithm==<br />
In the figure AQ , AR and BC are tangents to the circle with center O.<br><br />
BP=BQ and PC=CR (Tangents drawn from external point are equal) ---------- (1)<br><br />
<br />
Perimeter of △ABC=AB+BC+CA<br><br />
=AB+(BP+PC)+CA<br><br />
=AB+BQ+CR+CA ------ (From eq-1)<br><br />
=(AB+BQ)+(CR+CA)<br><br />
=AQ+AR ----- (From fig)<br><br />
=AQ+AQ -- --- (∵AQ=AR)<br><br />
=2AQ<br><br />
<br />
∴ AQ = <math>\frac{1}{2}</math> [perimeter of △ABC]<br />
<br />
=Problem-6 [Ex-15.4-B3]=<br />
In circle with center O , diameter AB and a chord AD are drawn. Another circle drawn with OA as diameter to cut AD at C. Prove that BD=2OC.[[File:15.4B3.png|300px]]<br />
==Algorithm==<br />
In figure, AB is the diameter of circle <math>C_{1}</math> and AO is the diameter of the circle <math>C_{2}</math><br>in △ADB and △ACO<br><br />
∠ADB=90° and ∠ACO=90° [∵angles in the semi circles]<br>∠DAB=∠CAO [∵common angles]<br>∴△ADB∼△ACO [∵equiangular triangles are similar]<br>∴<math>\frac{AB}{OA}</math>=<math>\frac{BD}{OC}</math>=<math>\frac{AD}{AC}</math> [∵corresonding sides of a similar triangles are proportional]<br>But AB=2OA----1 (∵diameter is twice the radius of a cicle)<br><math>\frac{AB}{OA}</math>=<math>\frac{BD}{OC}</math><br>from (1)<br><math>\frac{2OA}{OA}</math>=<math>\frac{BD}{OC}</math><br>∴BD=2OC<br />
<br />
<br />
<br />
== == Problem-7 [Ex-15.4-A3] == ==<br />
<br />
In the figure AB=10cm,AC=6cm and the radius of the smaller circle is xcm. find x.<br />
<br />
[[File:circle.png|400px]]</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=17924Circles Tangents Problems2015-02-15T16:37:26Z<p>Anilkumar: /* == Problem-7 [Ex-15.4-A3] == */</p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
a)<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br><br />
b)<br><br />
Let ∠QBP=x˚<br><br />
<br />
∴∠QPB=x˚ (∵PQ=BQ)<br><br />
Now Let ∠PAQ=y˚<br><br />
∠QPA=y˚ (∵ PQ=AQ)<br><br />
∴In △PAB<br><br />
<br />
∠PAB+∠PBA+∠APB=180˚<br><br />
<br />
y˚+x˚+(x˚+y˚)=180˚<br><br />
2x˚+2y˚=180˚<br><br />
2(x˚+y˚)=180˚<br><br />
x˚+y˚=90˚<br><br />
∴ ∠APB=90˚<br />
<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[Image:problem 3 on circle.png|300px]]<br />
==Concepts used==<br />
#The radii of a circle are equal<br />
#Properties of isosceles triangle.<br />
#SAS postulate<br />
#Properties of congruent triangles.<br />
<br />
==Prerequisite knowledge==<br />
#The radii of a circle are equal.<br />
# In an isosceles triangle angles opposite to equal sides are equal.<br />
#All the elements of congruent triangles are equal.<br />
<br />
==Algoritham==<br />
In ∆AOB <br><br />
AO=BO [Radii of a same circle]<br><br />
∴ ∠OAB = ∠OBA --------------I [∆AOB is an isosceles ∆}<br><br />
Then,<br> <br />
In ∆AOP and ∆BOP,<br><br />
AO = BO [Radii of a same circle]<br><br />
OP=OP [common side]<br><br />
∠OAP = ∠OBP [ from I]<br><br />
∴ ∆AOP ≅ ∆BOP [SAS postulate] <br><br />
∴ AP = BP [corresponding sides of congruent triangles ]<br />
<br />
=problem-4=<br />
'''In the given Quadrilateral ABCD , BC=38cm , QB=27cm , DC=25cm and AD⊥DC find the radius of the circle.(Ex:15.2. A-6)'''<br><br />
[[File:fig3.png|200px]]<br />
==Interpretation of problem==<br />
# In the quadrilateral ABCD sides BC , DC & QB are given . <br />
#AD⊥DC.<br />
# Asked to find the radius OS or OP<br />
<br />
==Concepts used==<br />
#Tangents drawn from an external point to a circle are equal.<br />
#In a quadrilateral, if all angles are equal and a pair of adjacent sides are equal then it is a square<br />
#In a circle, the radius drawn at the point of contact is perpendicular to the tangent<br />
==Algorithm== <br />
In the fig BC=38 cm and BQ=27 cm <br><br />
BQ=BR=27 cm (∵ by concept 1) <br><br />
∴CR=BC-BR=38-27=11 cm <br><br />
CR=SC=11 cm (∵ by concept 1) <br><br />
DC=25 cm <br><br />
∴ DS=DC-SC=25-11=14 cm <br><br />
DS=DP=14 cm (∵ by concept 1) <br><br />
Also AD⊥DC, OP ⊥ AD and OS ⊥ DC <br><br />
∠D=∠S=∠P=90˚ <br><br />
⇒ ∠O=90˚ <br><br />
∴ DSOP is a Square <br><br />
SO=OP=14 cm <br> hence Radius of given circle is 14 cm<br />
<br />
=Problem 5 [Ex-15.2-B8]=<br />
A circle is touching the side BC of △ABC at P. AB and AC when produced are touching the circle at Q and R respectively. Prove that AQ =<math>\frac{1}{2}</math> [perimeter of △ABC].<br />
[[File:123.png|300px]]<br />
<br />
==Algorithm==<br />
In the figure AQ , AR and BC are tangents to the circle with center O.<br><br />
BP=BQ and PC=CR (Tangents drawn from external point are equal) ---------- (1)<br><br />
<br />
Perimeter of △ABC=AB+BC+CA<br><br />
=AB+(BP+PC)+CA<br><br />
=AB+BQ+CR+CA ------ (From eq-1)<br><br />
=(AB+BQ)+(CR+CA)<br><br />
=AQ+AR ----- (From fig)<br><br />
=AQ+AQ -- --- (∵AQ=AR)<br><br />
=2AQ<br><br />
<br />
∴ AQ = <math>\frac{1}{2}</math> [perimeter of △ABC]<br />
<br />
=Problem-6 [Ex-15.4-B3]=<br />
In circle with center O , diameter AB and a chord AD are drawn. Another circle drawn with OA as diameter to cut AD at C. Prove that BD=2OC.[[File:15.4B3.png|300px]]<br />
==Algorithm==<br />
In figure, AB is the diameter of circle <math>C_{1}</math> and AO is the diameter of the circle <math>C_{2}</math><br>in △ADB and △ACO<br><br />
∠ADB=90° and ∠ACO=90° [∵angles in the semi circles]<br>∠DAB=∠CAO [∵common angles]<br>∴△ADB∼△ACO [∵equiangular triangles are similar]<br>∴<math>\frac{AB}{OA}</math>=<math>\frac{BD}{OC}</math>=<math>\frac{AD}{AC}</math> [∵corresonding sides of a similar triangles are proportional]<br>But AB=2OA----1 (∵diameter is twice the radius of a cicle)<br><math>\frac{AB}{OA}</math>=<math>\frac{BD}{OC}</math><br>from (1)<br><math>\frac{2OA}{OA}</math>=<math>\frac{BD}{OC}</math><br>∴BD=2OC<br />
<br />
<br />
<br />
== == Problem-7 [Ex-15.4-A3] == ==<br />
<br />
In the figure AB=10cm,AC=6cm and the radius of the smaller circle is xcm. find x.<br />
<br />
<br />
<br />
<br />
<br />
[[File:[[File:circle.png|400px]]Example.jpg]]</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=File:Circle.png&diff=17923File:Circle.png2015-02-15T16:36:31Z<p>Anilkumar: MsUpload</p>
<hr />
<div>MsUpload</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=17922Circles Tangents Problems2015-02-15T16:31:35Z<p>Anilkumar: </p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
a)<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br><br />
b)<br><br />
Let ∠QBP=x˚<br><br />
<br />
∴∠QPB=x˚ (∵PQ=BQ)<br><br />
Now Let ∠PAQ=y˚<br><br />
∠QPA=y˚ (∵ PQ=AQ)<br><br />
∴In △PAB<br><br />
<br />
∠PAB+∠PBA+∠APB=180˚<br><br />
<br />
y˚+x˚+(x˚+y˚)=180˚<br><br />
2x˚+2y˚=180˚<br><br />
2(x˚+y˚)=180˚<br><br />
x˚+y˚=90˚<br><br />
∴ ∠APB=90˚<br />
<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[Image:problem 3 on circle.png|300px]]<br />
==Concepts used==<br />
#The radii of a circle are equal<br />
#Properties of isosceles triangle.<br />
#SAS postulate<br />
#Properties of congruent triangles.<br />
<br />
==Prerequisite knowledge==<br />
#The radii of a circle are equal.<br />
# In an isosceles triangle angles opposite to equal sides are equal.<br />
#All the elements of congruent triangles are equal.<br />
<br />
==Algoritham==<br />
In ∆AOB <br><br />
AO=BO [Radii of a same circle]<br><br />
∴ ∠OAB = ∠OBA --------------I [∆AOB is an isosceles ∆}<br><br />
Then,<br> <br />
In ∆AOP and ∆BOP,<br><br />
AO = BO [Radii of a same circle]<br><br />
OP=OP [common side]<br><br />
∠OAP = ∠OBP [ from I]<br><br />
∴ ∆AOP ≅ ∆BOP [SAS postulate] <br><br />
∴ AP = BP [corresponding sides of congruent triangles ]<br />
<br />
=problem-4=<br />
'''In the given Quadrilateral ABCD , BC=38cm , QB=27cm , DC=25cm and AD⊥DC find the radius of the circle.(Ex:15.2. A-6)'''<br><br />
[[File:fig3.png|200px]]<br />
==Interpretation of problem==<br />
# In the quadrilateral ABCD sides BC , DC & QB are given . <br />
#AD⊥DC.<br />
# Asked to find the radius OS or OP<br />
<br />
==Concepts used==<br />
#Tangents drawn from an external point to a circle are equal.<br />
#In a quadrilateral, if all angles are equal and a pair of adjacent sides are equal then it is a square<br />
#In a circle, the radius drawn at the point of contact is perpendicular to the tangent<br />
==Algorithm== <br />
In the fig BC=38 cm and BQ=27 cm <br><br />
BQ=BR=27 cm (∵ by concept 1) <br><br />
∴CR=BC-BR=38-27=11 cm <br><br />
CR=SC=11 cm (∵ by concept 1) <br><br />
DC=25 cm <br><br />
∴ DS=DC-SC=25-11=14 cm <br><br />
DS=DP=14 cm (∵ by concept 1) <br><br />
Also AD⊥DC, OP ⊥ AD and OS ⊥ DC <br><br />
∠D=∠S=∠P=90˚ <br><br />
⇒ ∠O=90˚ <br><br />
∴ DSOP is a Square <br><br />
SO=OP=14 cm <br> hence Radius of given circle is 14 cm<br />
<br />
=Problem 5 [Ex-15.2-B8]=<br />
A circle is touching the side BC of △ABC at P. AB and AC when produced are touching the circle at Q and R respectively. Prove that AQ =<math>\frac{1}{2}</math> [perimeter of △ABC].<br />
[[File:123.png|300px]]<br />
<br />
==Algorithm==<br />
In the figure AQ , AR and BC are tangents to the circle with center O.<br><br />
BP=BQ and PC=CR (Tangents drawn from external point are equal) ---------- (1)<br><br />
<br />
Perimeter of △ABC=AB+BC+CA<br><br />
=AB+(BP+PC)+CA<br><br />
=AB+BQ+CR+CA ------ (From eq-1)<br><br />
=(AB+BQ)+(CR+CA)<br><br />
=AQ+AR ----- (From fig)<br><br />
=AQ+AQ -- --- (∵AQ=AR)<br><br />
=2AQ<br><br />
<br />
∴ AQ = <math>\frac{1}{2}</math> [perimeter of △ABC]<br />
<br />
=Problem-6 [Ex-15.4-B3]=<br />
In circle with center O , diameter AB and a chord AD are drawn. Another circle drawn with OA as diameter to cut AD at C. Prove that BD=2OC.[[File:15.4B3.png|300px]]<br />
==Algorithm==<br />
In figure, AB is the diameter of circle <math>C_{1}</math> and AO is the diameter of the circle <math>C_{2}</math><br>in △ADB and △ACO<br><br />
∠ADB=90° and ∠ACO=90° [∵angles in the semi circles]<br>∠DAB=∠CAO [∵common angles]<br>∴△ADB∼△ACO [∵equiangular triangles are similar]<br>∴<math>\frac{AB}{OA}</math>=<math>\frac{BD}{OC}</math>=<math>\frac{AD}{AC}</math> [∵corresonding sides of a similar triangles are proportional]<br>But AB=2OA----1 (∵diameter is twice the radius of a cicle)<br><math>\frac{AB}{OA}</math>=<math>\frac{BD}{OC}</math><br>from (1)<br><math>\frac{2OA}{OA}</math>=<math>\frac{BD}{OC}</math><br>∴BD=2OC<br />
<br />
<br />
<br />
== == Problem-7 [Ex-15.4-A3] == ==<br />
<br />
In the figure AB=10cm,AC=6cm and the radius of the smaller circle is xcm. find x.</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=15205Circles Tangents Problems2014-08-18T15:38:16Z<p>Anilkumar: /* Problem 5 */</p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
a)<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br><br />
b)<br><br />
Let ∠QBP=x˚<br><br />
<br />
∴∠QPB=x˚ (∵PQ=BQ)<br><br />
Now Let ∠PAQ=y˚<br><br />
∠QPA=y˚ (∵ PQ=AQ)<br><br />
∴In △PAB<br><br />
<br />
∠PAB+∠PBA+∠APB=180˚<br><br />
<br />
y˚+x˚+(x˚+y˚)=180˚<br><br />
2x˚+2y˚=180˚<br><br />
2(x˚+y˚)=180˚<br><br />
x˚+y˚=90˚<br><br />
∴ ∠APB=90˚<br />
<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[Image:problem 3 on circle.png|300px]]<br />
==Concepts used==<br />
#The radii of a circle are equal<br />
#Properties of isosceles triangle.<br />
#SAS postulate<br />
#Properties of congruent triangles.<br />
<br />
==Prerequisite knowledge==<br />
#The radii of a circle are equal.<br />
# In an isosceles triangle angles opposite to equal sides are equal.<br />
#All the elements of congruent triangles are equal.<br />
<br />
==Algoritham==<br />
In ∆AOB <br><br />
AO=BO [Radii of a same circle]<br><br />
∴ ∠OAB = ∠OBA --------------I [∆AOB is an isosceles ∆}<br><br />
Then,<br> <br />
In ∆AOP and ∆BOP,<br><br />
AO = BO [Radii of a same circle]<br><br />
OP=OP [common side]<br><br />
∠OAP = ∠OBP [ from I]<br><br />
∴ ∆AOP ≅ ∆BOP [SAS postulate] <br><br />
∴ AP = BP [corresponding sides of congruent triangles ]<br />
<br />
=problem-4=<br />
'''In the given Quadrilateral ABCD , BC=38cm , QB=27cm , DC=25cm and AD⊥DC find the radius of the circle.(Ex:15.2. A-6)'''<br><br />
[[File:fig3.png|200px]]<br />
==Interpretation of problem==<br />
# In the quadrilateral ABCD sides BC , DC & QB are given . <br />
#AD⊥DC.<br />
# Asked to find the radius OS or OP<br />
<br />
==Concepts used==<br />
#Tangents drawn from an external point to a circle are equal.<br />
#In a quadrilateral, if all angles are equal and a pair of adjacent sides are equal then it is a square<br />
#In a circle, the radius drawn at the point of contact is perpendicular to the tangent<br />
==Algorithm== <br />
In the fig BC=38 cm and BQ=27 cm <br><br />
BQ=BR=27 cm (∵ by concept 1) <br><br />
∴CR=BC-BR=38-27=11 cm <br><br />
CR=SC=11 cm (∵ by concept 1) <br><br />
DC=25 cm <br><br />
∴ DS=DC-SC=25-11=14 cm <br><br />
DS=DP=14 cm (∵ by concept 1) <br><br />
Also AD⊥DC, OP ⊥ AD and OS ⊥ DC <br><br />
∠D=∠S=∠P=90˚ <br><br />
⇒ ∠O=90˚ <br><br />
∴ DSOP is a Square <br><br />
SO=OP=14 cm <br> hence Radius of given circle is 14 cm<br />
<br />
=Problem 5 [Ex-15.2-B8]=<br />
A circle is touching the side BC of △ABC at P. AB and AC when produced are touching the circle at Q and R respectively. Prove that AQ =<math>\frac{1}{2}</math> [perimeter of △ABC].<br />
[[File:123.png|300px]]<br />
<br />
==Algorithm==<br />
In the figure AQ , AR and BC are tangents to the circle with center O.<br><br />
BP=BQ and PC=CR (Tangents drawn from external point are equal) ---------- (1)<br><br />
<br />
Perimeter of △ABC=AB+BC+CA<br><br />
=AB+(BP+PC)+CA<br><br />
=AB+BQ+CR+CA ------ (From eq-1)<br><br />
=(AB+BQ)+(CR+CA)<br><br />
=AQ+AR ----- (From fig)<br><br />
=AQ+AQ -- --- (∵AQ=AR)<br><br />
=2AQ<br><br />
<br />
∴ AQ = <math>\frac{1}{2}</math> [perimeter of △ABC]<br />
<br />
=Problem-6 [Ex-15.4-B3]=<br />
In circle with center O , diameter AB and a chord AD are drawn. Another circle drawn with OA as diameter to cut AD at C. Prove that BD=2OC.[[File:15.4B3.png|300px]]<br />
==Algorithm==<br />
In figure, AB is the diameter of circle <math>C_{1}</math> and AO is the diameter of the circle <math>C_{2}</math><br>in △ADB and △ACO<br><br />
∠ADB=90° and ∠ACO=90° [∵angles in the semi circles]<br>∠DAB=∠CAO [∵common angles]<br>∴△ADB∼△ACO [∵equiangular triangles are similar]<br>∴<math>\frac{AB}{OA}</math>=<math>\frac{BD}{OC}</math>=<math>\frac{AD}{AC}</math> [∵corresonding sides of a similar triangles are proportional]<br>But AB=2OA----1 (∵diameter is twice the radius of a cicle)<br><math>\frac{AB}{OA}</math>=<math>\frac{BD}{OC}</math><br>from (1)<br><math>\frac{2OA}{OA}</math>=<math>\frac{BD}{OC}</math><br>∴BD=2OC</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=14996Circles Tangents Problems2014-08-14T07:51:00Z<p>Anilkumar: /* Algorithm */</p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
a)<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br><br />
b)<br><br />
Let ∠QBP=x˚<br><br />
<br />
∴∠QPB=x˚ (∵PQ=BQ)<br><br />
Now Let ∠PAQ=y˚<br><br />
∠QPA=y˚ (∵ PQ=AQ)<br><br />
∴In △PAB<br><br />
<br />
∠PAB+∠PBA+∠APB=180˚<br><br />
<br />
y˚+x˚+(x˚+y˚)=180˚<br><br />
2x˚+2y˚=180˚<br><br />
2(x˚+y˚)=180˚<br><br />
x˚+y˚=90˚<br><br />
∴ ∠APB=90˚<br />
<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[Image:problem 3 on circle.png|300px]]<br />
==Concepts used==<br />
#The radii of a circle are equal<br />
#Properties of isosceles triangle.<br />
#SAS postulate<br />
#Properties of congruent triangles.<br />
<br />
==Prerequisite knowledge==<br />
#The radii of a circle are equal.<br />
# In an isosceles triangle angles opposite to equal sides are equal.<br />
#All the elements of congruent triangles are equal.<br />
<br />
==Algoritham==<br />
In ∆AOB <br><br />
AO=BO [Radii of a same circle]<br><br />
∴ ∠OAB = ∠OBA --------------I [∆AOB is an isosceles ∆}<br><br />
Then,<br> <br />
In ∆AOP and ∆BOP,<br><br />
AO = BO [Radii of a same circle]<br><br />
OP=OP [common side]<br><br />
∠OAP = ∠OBP [ from I]<br><br />
∴ ∆AOP ≅ ∆BOP [SAS postulate] <br><br />
∴ AP = BP [corresponding sides of congruent triangles ]<br />
<br />
=problem-4=<br />
'''In the given Quadrilateral ABCD , BC=38cm , QB=27cm , DC=25cm and AD⊥DC find the radius of the circle.(Ex:15.2. A-6)'''<br><br />
[[File:fig3.png|200px]]<br />
==Interpretation of problem==<br />
# In the quadrilateral ABCD sides BC , DC & QB are given . <br />
#AD⊥DC.<br />
# Asked to find the radius OS or OP<br />
<br />
==Concepts used==<br />
#Tangents drawn from an external point to a circle are equal.<br />
#In a quadrilateral, if all angles are equal and a pair of adjacent sides are equal then it is a square<br />
#In a circle, the radius drawn at the point of contact is perpendicular to the tangent<br />
==Algorithm== <br />
In the fig BC=38 cm and BQ=27 cm <br><br />
BQ=BR=27 cm (∵ by concept 1) <br><br />
∴CR=BC-BR=38-27=11 cm <br><br />
CR=SC=11 cm (∵ by concept 1) <br><br />
DC=25 cm <br><br />
∴ DS=DC-SC=25-11=14 cm <br><br />
DS=DP=14 cm (∵ by concept 1) <br><br />
Also AD⊥DC, OP ⊥ AD and OS ⊥ DC <br><br />
∠D=∠S=∠P=90˚ <br><br />
⇒ ∠O=90˚ <br><br />
∴ DSOP is a Square <br><br />
SO=OP=14 cm <br> hence Radius of given circle is 14 cm<br />
<br />
=Problem 5=<br />
A circle is touching the side BC of △ABC at P. AB and AC when produced are touching the circle at Q and R respectively. Prove that AQ =<math>\frac{1}{2}</math> [perimeter of △ABC].<br />
[[File:123.png|300px]]<br />
<br />
==Algorithm==<br />
In the figure AQ , AR and BC are tangents to the circle with center O.<br><br />
BP=BQ and PC=CR (Tangents drawn from external point are equal) ---------- (1)<br><br />
<br />
Perimeter of △ABC=AB+BC+CA<br><br />
=AB+(BP+PC)+CA<br><br />
=AB+BQ+CR+CA ------ (From eq-1)<br><br />
=(AB+BQ)+(CR+CA)<br><br />
=AQ+AR ----- (From fig)<br><br />
=AQ+AQ -- --- (∵AQ=AR)<br><br />
=2AQ<br><br />
<br />
∴ AQ = <math>\frac{1}{2}</math> [perimeter of △ABC]<br />
=Problem-6 [Ex-15.4-B3]=<br />
In circle with center O , diameter AB and a chord AD are drawn. Another circle drawn with OA as diameter to cut AD at C. Prove that BD=2OC.[[File:15.4B3.png|300px]]<br />
==Algorithm==<br />
In figure, AB is the diameter of circle <math>C_{1}</math> and AO is the diameter of the circle <math>C_{2}</math><br>in △ADB and △ACO<br><br />
∠ADB=90° and ∠ACO=90° [∵angles in the semi circles]<br>∠DAB=∠CAO [∵common angles]<br>∴△ADB∼△ACO [∵equiangular triangles are similar]<br>∴<math>\frac{AB}{OA}</math>=<math>\frac{BD}{OC}</math>=<math>\frac{AD}{AC}</math> [∵corresonding sides of a similar triangles are proportional]<br>But AB=2OA----1 (∵diameter is twice the radius of a cicle)<br><math>\frac{AB}{OA}</math>=<math>\frac{BD}{OC}</math><br>from (1)<br><math>\frac{2OA}{OA}</math>=<math>\frac{BD}{OC}</math><br>∴BD=2OC</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=14990Circles Tangents Problems2014-08-14T07:46:25Z<p>Anilkumar: /* Algorithm */</p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
a)<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br><br />
b)<br><br />
Let ∠QBP=x˚<br><br />
<br />
∴∠QPB=x˚ (∵PQ=BQ)<br><br />
Now Let ∠PAQ=y˚<br><br />
∠QPA=y˚ (∵ PQ=AQ)<br><br />
∴In △PAB<br><br />
<br />
∠PAB+∠PBA+∠APB=180˚<br><br />
<br />
y˚+x˚+(x˚+y˚)=180˚<br><br />
2x˚+2y˚=180˚<br><br />
2(x˚+y˚)=180˚<br><br />
x˚+y˚=90˚<br><br />
∴ ∠APB=90˚<br />
<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[Image:problem 3 on circle.png|300px]]<br />
==Concepts used==<br />
#The radii of a circle are equal<br />
#Properties of isosceles triangle.<br />
#SAS postulate<br />
#Properties of congruent triangles.<br />
<br />
==Prerequisite knowledge==<br />
#The radii of a circle are equal.<br />
# In an isosceles triangle angles opposite to equal sides are equal.<br />
#All the elements of congruent triangles are equal.<br />
<br />
==Algoritham==<br />
In ∆AOB <br><br />
AO=BO [Radii of a same circle]<br><br />
∴ ∠OAB = ∠OBA --------------I [∆AOB is an isosceles ∆}<br><br />
Then,<br> <br />
In ∆AOP and ∆BOP,<br><br />
AO = BO [Radii of a same circle]<br><br />
OP=OP [common side]<br><br />
∠OAP = ∠OBP [ from I]<br><br />
∴ ∆AOP ≅ ∆BOP [SAS postulate] <br><br />
∴ AP = BP [corresponding sides of congruent triangles ]<br />
<br />
=problem-4=<br />
'''In the given Quadrilateral ABCD , BC=38cm , QB=27cm , DC=25cm and AD⊥DC find the radius of the circle.(Ex:15.2. A-6)'''<br><br />
[[File:fig3.png|200px]]<br />
==Interpretation of problem==<br />
# In the quadrilateral ABCD sides BC , DC & QB are given . <br />
#AD⊥DC.<br />
# Asked to find the radius OS or OP<br />
<br />
==Concepts used==<br />
#Tangents drawn from an external point to a circle are equal.<br />
#In a quadrilateral, if all angles are equal and a pair of adjacent sides are equal then it is a square<br />
#In a circle, the radius drawn at the point of contact is perpendicular to the tangent<br />
==Algorithm== <br />
In the fig BC=38 cm and BQ=27 cm <br><br />
BQ=BR=27 cm (∵ by concept 1) <br><br />
∴CR=BC-BR=38-27=11 cm <br><br />
CR=SC=11 cm (∵ by concept 1) <br><br />
DC=25 cm <br><br />
∴ DS=DC-SC=25-11=14 cm <br><br />
DS=DP=14 cm (∵ by concept 1) <br><br />
Also AD⊥DC, OP ⊥ AD and OS ⊥ DC <br><br />
∠D=∠S=∠P=90˚ <br><br />
⇒ ∠O=90˚ <br><br />
∴ DSOP is a Square <br><br />
SO=OP=14 cm <br> hence Radius of given circle is 14 cm<br />
<br />
=Problem 5=<br />
A circle is touching the side BC of △ABC at P. AB and AC when produced are touching the circle at Q and R respectively. Prove that AQ =<math>\frac{1}{2}</math> [perimeter of △ABC].<br />
[[File:123.png|300px]]<br />
<br />
==Algorithm==<br />
In the figure AQ , AR and BC are tangents to the circle with center O.<br><br />
BP=BQ and PC=CR (Tangents drawn from external point are equal) ---------- (1)<br><br />
<br />
Perimeter of △ABC=AB+BC+CA<br><br />
=AB+(BP+PC)+CA<br><br />
=AB+BQ+CR+CA ------ (From eq-1)<br><br />
=(AB+BQ)+(CR+CA)<br><br />
=AQ+AR ----- (From fig)<br><br />
=AQ+AQ -- --- (∵AQ=AR)<br><br />
=2AQ<br><br />
<br />
∴ AQ = <math>\frac{1}{2}</math> [perimeter of △ABC]<br />
=Problem-6 [Ex-15.4-B3]=<br />
In circle with center O , diameter AB and a chord AD are drawn. Another circle drawn with OA as diameter to cut AD at C. Prove that BD=2OC.[[File:15.4B3.png|300px]]<br />
==Algorithm==<br />
In figure, AB is the diameter of circle C_{1} and AO is the diameter of the circle C_{2}<br>in △ADB and △ACO<br><br />
∠ADB=90° and ∠ACO=90° [∵angles in the semi circles]<br>∠DAB=∠CAO [∵common angles]<br>∴△ADB∼△ACO [∵equiangular triangles are similar]<br>∴<math>\frac{AB}{AO}</math>=<math>\frac{BD}{OC}</math>=<math>\frac{AD}{AC}</math> [∵corresonding sides of a similar triangles are proportional]<br>But AB=2OA----1 (∵diameter is twice the radius of a cicle)<br><math>\frac{AB}{AO}</math>=<math>\frac{BD}{OC}</math><br>from (1)<br><math>\frac{2OA}{AO}</math>=<math>\frac{BD}{OC}</math><br>∴BD=2OC</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=14984Circles Tangents Problems2014-08-14T07:42:25Z<p>Anilkumar: /* Algorithm */</p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
a)<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br><br />
b)<br><br />
Let ∠QBP=x˚<br><br />
<br />
∴∠QPB=x˚ (∵PQ=BQ)<br><br />
Now Let ∠PAQ=y˚<br><br />
∠QPA=y˚ (∵ PQ=AQ)<br><br />
∴In △PAB<br><br />
<br />
∠PAB+∠PBA+∠APB=180˚<br><br />
<br />
y˚+x˚+(x˚+y˚)=180˚<br><br />
2x˚+2y˚=180˚<br><br />
2(x˚+y˚)=180˚<br><br />
x˚+y˚=90˚<br><br />
∴ ∠APB=90˚<br />
<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[Image:problem 3 on circle.png|300px]]<br />
==Concepts used==<br />
#The radii of a circle are equal<br />
#Properties of isosceles triangle.<br />
#SAS postulate<br />
#Properties of congruent triangles.<br />
<br />
==Prerequisite knowledge==<br />
#The radii of a circle are equal.<br />
# In an isosceles triangle angles opposite to equal sides are equal.<br />
#All the elements of congruent triangles are equal.<br />
<br />
==Algoritham==<br />
In ∆AOB <br><br />
AO=BO [Radii of a same circle]<br><br />
∴ ∠OAB = ∠OBA --------------I [∆AOB is an isosceles ∆}<br><br />
Then,<br> <br />
In ∆AOP and ∆BOP,<br><br />
AO = BO [Radii of a same circle]<br><br />
OP=OP [common side]<br><br />
∠OAP = ∠OBP [ from I]<br><br />
∴ ∆AOP ≅ ∆BOP [SAS postulate] <br><br />
∴ AP = BP [corresponding sides of congruent triangles ]<br />
<br />
=problem-4=<br />
'''In the given Quadrilateral ABCD , BC=38cm , QB=27cm , DC=25cm and AD⊥DC find the radius of the circle.(Ex:15.2. A-6)'''<br><br />
[[File:fig3.png|200px]]<br />
==Interpretation of problem==<br />
# In the quadrilateral ABCD sides BC , DC & QB are given . <br />
#AD⊥DC.<br />
# Asked to find the radius OS or OP<br />
<br />
==Concepts used==<br />
#Tangents drawn from an external point to a circle are equal.<br />
#In a quadrilateral, if all angles are equal and a pair of adjacent sides are equal then it is a square<br />
#In a circle, the radius drawn at the point of contact is perpendicular to the tangent<br />
==Algorithm== <br />
In the fig BC=38 cm and BQ=27 cm <br><br />
BQ=BR=27 cm (∵ by concept 1) <br><br />
∴CR=BC-BR=38-27=11 cm <br><br />
CR=SC=11 cm (∵ by concept 1) <br><br />
DC=25 cm <br><br />
∴ DS=DC-SC=25-11=14 cm <br><br />
DS=DP=14 cm (∵ by concept 1) <br><br />
Also AD⊥DC, OP ⊥ AD and OS ⊥ DC <br><br />
∠D=∠S=∠P=90˚ <br><br />
⇒ ∠O=90˚ <br><br />
∴ DSOP is a Square <br><br />
SO=OP=14 cm <br> hence Radius of given circle is 14 cm<br />
<br />
=Problem 5=<br />
A circle is touching the side BC of △ABC at P. AB and AC when produced are touching the circle at Q and R respectively. Prove that AQ =<math>\frac{1}{2}</math> [perimeter of △ABC].<br />
[[File:123.png|300px]]<br />
<br />
==Algorithm==<br />
In the figure AQ , AR and BC are tangents to the circle with center O.<br><br />
BP=BQ and PC=CR (Tangents drawn from external point are equal) ---------- (1)<br><br />
<br />
Perimeter of △ABC=AB+BC+CA<br><br />
=AB+(BP+PC)+CA<br><br />
=AB+BQ+CR+CA ------ (From eq-1)<br><br />
=(AB+BQ)+(CR+CA)<br><br />
=AQ+AR ----- (From fig)<br><br />
=AQ+AQ -- --- (∵AQ=AR)<br><br />
=2AQ<br><br />
<br />
∴ AQ = <math>\frac{1}{2}</math> [perimeter of △ABC]<br />
=Problem-6 [Ex-15.4-B3]=<br />
In circle with center O , diameter AB and a chord AD are drawn. Another circle drawn with OA as diameter to cut AD at C. Prove that BD=2OC.[[File:15.4B3.png|300px]]<br />
==Algorithm==<br />
In figure, AB is the diameter of circle C_{1} and AO is the diameter of the circle C_{2}<br>in △ADB and △ACO<br><br />
∠ADB=90° and ∠ACO=90° [∵angles in the semi circles]<br>∠DAB=∠CAO [∵common angles]<br>△ADB∼△ACO [equiangular triangles are similar]<br><math>\frac{AB}{AO}</math>=<math>\frac{BD}{OC}</math>=<math>\frac{AD}{AC}</math> [corresonding sides of a similar triangles are proportional]<br>But AB=2OA----1 (diameter is twice the radius of a cicle)<br><math>\frac{AB}{AO}</math>=<math>\frac{BD}{OC}</math><br>from (1)<br><math>\frac{2OA}{AO}</math>=<math>\frac{BD}{OC}</math><br>BD=2OC</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=14943Circles Tangents Problems2014-08-14T06:32:50Z<p>Anilkumar: /* Problem-6 [Ex-15.4-B3] */</p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
a)<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br><br />
b)<br><br />
Let ∠QBP=x˚<br><br />
<br />
∴∠QPB=x˚ (∵PQ=BQ)<br><br />
Now Let ∠PAQ=y˚<br><br />
∠QPA=y˚ (∵ PQ=AQ)<br><br />
∴In △PAB<br><br />
<br />
∠PAB+∠PBA+∠APB=180˚<br><br />
<br />
y˚+x˚+(x˚+y˚)=180˚<br><br />
2x˚+2y˚=180˚<br><br />
2(x˚+y˚)=180˚<br><br />
x˚+y˚=90˚<br><br />
∴ ∠APB=90˚<br />
<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[Image:problem 3 on circle.png|300px]]<br />
==Concepts used==<br />
#The radii of a circle are equal<br />
#Properties of isosceles triangle.<br />
#SAS postulate<br />
#Properties of congruent triangles.<br />
<br />
==Prerequisite knowledge==<br />
#The radii of a circle are equal.<br />
# In an isosceles triangle angles opposite to equal sides are equal.<br />
#All the elements of congruent triangles are equal.<br />
<br />
==Algoritham==<br />
In ∆AOB <br><br />
AO=BO [Radii of a same circle]<br><br />
∴ ∠OAB = ∠OBA --------------I [∆AOB is an isosceles ∆}<br><br />
Then,<br> <br />
In ∆AOP and ∆BOP,<br><br />
AO = BO [Radii of a same circle]<br><br />
OP=OP [common side]<br><br />
∠OAP = ∠OBP [ from I]<br><br />
∴ ∆AOP ≅ ∆BOP [SAS postulate] <br><br />
∴ AP = BP [corresponding sides of congruent triangles ]<br />
<br />
=problem-4=<br />
'''In the given Quadrilateral ABCD , BC=38cm , QB=27cm , DC=25cm and AD⊥DC find the radius of the circle.(Ex:15.2. A-6)'''<br><br />
[[File:fig3.png|200px]]<br />
==Interpretation of problem==<br />
# In the quadrilateral ABCD sides BC , DC & QB are given . <br />
#AD⊥DC.<br />
# Asked to find the radius OS or OP<br />
<br />
==Concepts used==<br />
#Tangents drawn from an external point to a circle are equal.<br />
#In a quadrilateral, if all angles are equal and a pair of adjacent sides are equal then it is a square<br />
#In a circle, the radius drawn at the point of contact is perpendicular to the tangent<br />
==Algorithm== <br />
In the fig BC=38 cm and BQ=27 cm <br><br />
BQ=BR=27 cm (∵ by concept 1) <br><br />
∴CR=BC-BR=38-27=11 cm <br><br />
CR=SC=11 cm (∵ by concept 1) <br><br />
DC=25 cm <br><br />
∴ DS=DC-SC=25-11=14 cm <br><br />
DS=DP=14 cm (∵ by concept 1) <br><br />
Also AD⊥DC, OP ⊥ AD and OS ⊥ DC <br><br />
∠D=∠S=∠P=90˚ <br><br />
⇒ ∠O=90˚ <br><br />
∴ DSOP is a Square <br><br />
SO=OP=14 cm <br> hence Radius of given circle is 14 cm<br />
<br />
=Problem 5=<br />
A circle is touching the side BC of △ABC at P. AB and AC when produced are touching the circle at Q and R respectively. Prove that AQ =<math>\frac{1}{2}</math> [perimeter of △ABC].<br />
[[File:123.png|300px]]<br />
<br />
==Algorithm==<br />
In the figure AQ , AR and BC are tangents to the circle with center O.<br><br />
BP=BQ and PC=CR (Tangents drawn from external point are equal) ---------- (1)<br><br />
<br />
Perimeter of △ABC=AB+BC+CA<br><br />
=AB+(BP+PC)+CA<br><br />
=AB+BQ+CR+CA ------ (From eq-1)<br><br />
=(AB+BQ)+(CR+CA)<br><br />
=AQ+AR ----- (From fig)<br><br />
=AQ+AQ -- --- (∵AQ=AR)<br><br />
=2AQ<br><br />
<br />
∴ AQ = <math>\frac{1}{2}</math> [perimeter of △ABC]<br />
=Problem-6 [Ex-15.4-B3]=<br />
In circle with center O , diameter AB and a chord AD are drawn. Another circle drawn with OA as diameter to cut AD at C. Prove that BD=2OC.[[File:15.4B3.png|300px]]<br />
==Algorithm==<br />
In figure, AB is the diameter of circle C_{1} and AO is the diameter of the circle C_{2}<br>in △ADB and △ACO<br><br />
∠ADB=90° and ∠ACO=90° [∵angles in the semi circles]<br>∠DAB=∠CAO [∵common angles]<br>△ADB∼△ACO [equiangular triangles are similar]<br><math>\frac{AB}{AO}</math>=<math>\frac{PD}{OC}</math>=<math>\frac{AD}{AC}</math> [corresonding sides of a similar triangles are proportional]<br></div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=14921Circles Tangents Problems2014-08-14T06:14:56Z<p>Anilkumar: /* Problem-6 [Ex-15.4-B3] */</p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
a)<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br><br />
b)<br><br />
Let ∠QBP=x˚<br><br />
<br />
∴∠QPB=x˚ (∵PQ=BQ)<br><br />
Now Let ∠PAQ=y˚<br><br />
∠QPA=y˚ (∵ PQ=AQ)<br><br />
∴In △PAB<br><br />
<br />
∠PAB+∠PBA+∠APB=180˚<br><br />
<br />
y˚+x˚+(x˚+y˚)=180˚<br><br />
2x˚+2y˚=180˚<br><br />
2(x˚+y˚)=180˚<br><br />
x˚+y˚=90˚<br><br />
∴ ∠APB=90˚<br />
<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[Image:problem 3 on circle.png|300px]]<br />
==Concepts used==<br />
#The radii of a circle are equal<br />
#Properties of isosceles triangle.<br />
#SAS postulate<br />
#Properties of congruent triangles.<br />
<br />
==Prerequisite knowledge==<br />
#The radii of a circle are equal.<br />
# In an isosceles triangle angles opposite to equal sides are equal.<br />
#All the elements of congruent triangles are equal.<br />
<br />
==Algoritham==<br />
In ∆AOB <br><br />
AO=BO [Radii of a same circle]<br><br />
∴ ∠OAB = ∠OBA --------------I [∆AOB is an isosceles ∆}<br><br />
Then,<br> <br />
In ∆AOP and ∆BOP,<br><br />
AO = BO [Radii of a same circle]<br><br />
OP=OP [common side]<br><br />
∠OAP = ∠OBP [ from I]<br><br />
∴ ∆AOP ≅ ∆BOP [SAS postulate] <br><br />
∴ AP = BP [corresponding sides of congruent triangles ]<br />
<br />
=problem-4=<br />
'''In the given Quadrilateral ABCD , BC=38cm , QB=27cm , DC=25cm and AD⊥DC find the radius of the circle.(Ex:15.2. A-6)'''<br><br />
[[File:fig3.png|200px]]<br />
==Interpretation of problem==<br />
# In the quadrilateral ABCD sides BC , DC & QB are given . <br />
#AD⊥DC.<br />
# Asked to find the radius OS or OP<br />
<br />
==Concepts used==<br />
#Tangents drawn from an external point to a circle are equal.<br />
#In a quadrilateral, if all angles are equal and a pair of adjacent sides are equal then it is a square<br />
#In a circle, the radius drawn at the point of contact is perpendicular to the tangent<br />
==Algorithm== <br />
In the fig BC=38 cm and BQ=27 cm <br><br />
BQ=BR=27 cm (∵ by concept 1) <br><br />
∴CR=BC-BR=38-27=11 cm <br><br />
CR=SC=11 cm (∵ by concept 1) <br><br />
DC=25 cm <br><br />
∴ DS=DC-SC=25-11=14 cm <br><br />
DS=DP=14 cm (∵ by concept 1) <br><br />
Also AD⊥DC, OP ⊥ AD and OS ⊥ DC <br><br />
∠D=∠S=∠P=90˚ <br><br />
⇒ ∠O=90˚ <br><br />
∴ DSOP is a Square <br><br />
SO=OP=14 cm <br> hence Radius of given circle is 14 cm<br />
<br />
=Problem 5=<br />
A circle is touching the side BC of △ABC at P. AB and AC when produced are touching the circle at Q and R respectively. Prove that AQ =<math>\frac{1}{2}</math> [perimeter of △ABC].<br />
[[File:123.png|300px]]<br />
<br />
==Algorithm==<br />
In the figure AQ , AR and BC are tangents to the circle with center O.<br><br />
BP=BQ and PC=CR (Tangents drawn from external point are equal) ---------- (1)<br><br />
<br />
Perimeter of △ABC=AB+BC+CA<br><br />
=AB+(BP+PC)+CA<br><br />
=AB+BQ+CR+CA ------ (From eq-1)<br><br />
=(AB+BQ)+(CR+CA)<br><br />
=AQ+AR ----- (From fig)<br><br />
=AQ+AQ -- --- (∵AQ=AR)<br><br />
=2AQ<br><br />
<br />
∴ AQ = <math>\frac{1}{2}</math> [perimeter of △ABC]<br />
=Problem-6 [Ex-15.4-B3]=<br />
In circle with center O , diameter AB and a chord AD are drawn. Another circle drawn with OA as diameter to cut AD at C. Prove that BD=2OC.[[File:15.4B3.png|300px]]<br />
==Algorithm==<br />
In figure, AB is the diameter of circle C_{1} and AO is the diameter of the circle C_{2}<br>in △ADB and △ACO<br><br />
∠ADB=90° and ∠ACO=90° [∵angles in the semi circles]<br>∠DAB=∠CAO [∵common angles]</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=14875Circles Tangents Problems2014-08-14T05:44:18Z<p>Anilkumar: /* Probiem-6 [Ex-15.4-B3] */</p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
a)<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br><br />
b)<br><br />
Let ∠QBP=x˚<br><br />
<br />
∴∠QPB=x˚ (∵PQ=BQ)<br><br />
Now Let ∠PAQ=y˚<br><br />
∠QPA=y˚ (∵ PQ=AQ)<br><br />
∴In △PAB<br><br />
<br />
∠PAB+∠PBA+∠APB=180˚<br><br />
<br />
y˚+x˚+(x˚+y˚)=180˚<br><br />
2x˚+2y˚=180˚<br><br />
2(x˚+y˚)=180˚<br><br />
x˚+y˚=90˚<br><br />
∴ ∠APB=90˚<br />
<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[Image:problem 3 on circle.png|300px]]<br />
==Concepts used==<br />
#The radii of a circle are equal<br />
#Properties of isosceles triangle.<br />
#SAS postulate<br />
#Properties of congruent triangles.<br />
<br />
==Prerequisite knowledge==<br />
#The radii of a circle are equal.<br />
# In an isosceles triangle angles opposite to equal sides are equal.<br />
#All the elements of congruent triangles are equal.<br />
<br />
==Algoritham==<br />
In ∆AOB <br><br />
AO=BO [Radii of a same circle]<br><br />
∴ ∠OAB = ∠OBA --------------I [∆AOB is an isosceles ∆}<br><br />
Then,<br> <br />
In ∆AOP and ∆BOP,<br><br />
AO = BO [Radii of a same circle]<br><br />
OP=OP [common side]<br><br />
∠OAP = ∠OBP [ from I]<br><br />
∴ ∆AOP ≅ ∆BOP [SAS postulate] <br><br />
∴ AP = BP [corresponding sides of congruent triangles ]<br />
<br />
=problem-4=<br />
'''In the given Quadrilateral ABCD , BC=38cm , QB=27cm , DC=25cm and AD⊥DC find the radius of the circle.(Ex:15.2. A-6)'''<br><br />
[[File:fig3.png|200px]]<br />
==Interpretation of problem==<br />
# In the quadrilateral ABCD sides BC , DC & QB are given . <br />
#AD⊥DC.<br />
# Asked to find the radius OS or OP<br />
<br />
==Concepts used==<br />
#Tangents drawn from an external point to a circle are equal.<br />
#In a quadrilateral, if all angles are equal and a pair of adjacent sides are equal then it is a square<br />
#In a circle, the radius drawn at the point of contact is perpendicular to the tangent<br />
==Algorithm== <br />
In the fig BC=38 cm and BQ=27 cm <br><br />
BQ=BR=27 cm (∵ by concept 1) <br><br />
∴CR=BC-BR=38-27=11 cm <br><br />
CR=SC=11 cm (∵ by concept 1) <br><br />
DC=25 cm <br><br />
∴ DS=DC-SC=25-11=14 cm <br><br />
DS=DP=14 cm (∵ by concept 1) <br><br />
Also AD⊥DC, OP ⊥ AD and OS ⊥ DC <br><br />
∠D=∠S=∠P=90˚ <br><br />
⇒ ∠O=90˚ <br><br />
∴ DSOP is a Square <br><br />
SO=OP=14 cm <br> hence Radius of given circle is 14 cm<br />
<br />
=Problem 5=<br />
A circle is touching the side BC of △ABC at P. AB and AC when produced are touching the circle at Q and R respectively. Prove that AQ =<math>\frac{1}{2}</math> [perimeter of △ABC].<br />
[[File:123.png|300px]]<br />
<br />
==Algorithm==<br />
In the figure AQ , AR and BC are tangents to the circle with center O.<br><br />
BP=BQ and PC=CR (Tangents drawn from external point are equal) ---------- (1)<br><br />
<br />
Perimeter of △ABC=AB+BC+CA<br><br />
=AB+(BP+PC)+CA<br><br />
=AB+BQ+CR+CA ------ (From eq-1)<br><br />
=(AB+BQ)+(CR+CA)<br><br />
=AQ+AR ----- (From fig)<br><br />
=AQ+AQ -- --- (∵AQ=AR)<br><br />
=2AQ<br><br />
<br />
∴ AQ = <math>\frac{1}{2}</math> [perimeter of △ABC]<br />
=Problem-6 [Ex-15.4-B3]=<br />
In circle with center O , diameter AB and a chord AD are drawn. Another circle drawn with OA as diameter to cut AD at C. Prove that BD=2OC.[[File:15.4B3.png|300px]]</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=File:15.4B3.png&diff=14869File:15.4B3.png2014-08-14T05:42:40Z<p>Anilkumar: MsUpload</p>
<hr />
<div>MsUpload</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=14862Circles Tangents Problems2014-08-14T05:39:28Z<p>Anilkumar: </p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
a)<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br><br />
b)<br><br />
Let ∠QBP=x˚<br><br />
<br />
∴∠QPB=x˚ (∵PQ=BQ)<br><br />
Now Let ∠PAQ=y˚<br><br />
∠QPA=y˚ (∵ PQ=AQ)<br><br />
∴In △PAB<br><br />
<br />
∠PAB+∠PBA+∠APB=180˚<br><br />
<br />
y˚+x˚+(x˚+y˚)=180˚<br><br />
2x˚+2y˚=180˚<br><br />
2(x˚+y˚)=180˚<br><br />
x˚+y˚=90˚<br><br />
∴ ∠APB=90˚<br />
<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[Image:problem 3 on circle.png|300px]]<br />
==Concepts used==<br />
#The radii of a circle are equal<br />
#Properties of isosceles triangle.<br />
#SAS postulate<br />
#Properties of congruent triangles.<br />
<br />
==Prerequisite knowledge==<br />
#The radii of a circle are equal.<br />
# In an isosceles triangle angles opposite to equal sides are equal.<br />
#All the elements of congruent triangles are equal.<br />
<br />
==Algoritham==<br />
In ∆AOB <br><br />
AO=BO [Radii of a same circle]<br><br />
∴ ∠OAB = ∠OBA --------------I [∆AOB is an isosceles ∆}<br><br />
Then,<br> <br />
In ∆AOP and ∆BOP,<br><br />
AO = BO [Radii of a same circle]<br><br />
OP=OP [common side]<br><br />
∠OAP = ∠OBP [ from I]<br><br />
∴ ∆AOP ≅ ∆BOP [SAS postulate] <br><br />
∴ AP = BP [corresponding sides of congruent triangles ]<br />
<br />
=problem-4=<br />
'''In the given Quadrilateral ABCD , BC=38cm , QB=27cm , DC=25cm and AD⊥DC find the radius of the circle.(Ex:15.2. A-6)'''<br><br />
[[File:fig3.png|200px]]<br />
==Interpretation of problem==<br />
# In the quadrilateral ABCD sides BC , DC & QB are given . <br />
#AD⊥DC.<br />
# Asked to find the radius OS or OP<br />
<br />
==Concepts used==<br />
#Tangents drawn from an external point to a circle are equal.<br />
#In a quadrilateral, if all angles are equal and a pair of adjacent sides are equal then it is a square<br />
#In a circle, the radius drawn at the point of contact is perpendicular to the tangent<br />
==Algorithm== <br />
In the fig BC=38 cm and BQ=27 cm <br><br />
BQ=BR=27 cm (because by concept 1) <br><br />
∴CR=BC-BR=38-27=11 cm <br><br />
CR=SC=11 cm (because by concept 1) <br><br />
DC=25 cm <br><br />
∴ DS=DC-SC=25-11=14 cm <br><br />
DS=DP=14 cm (because by concept 1) <br><br />
Also {AD ortho DC} , {OP ortho AD} and {OS ortho DC} <br><br />
∠D=∠S=∠P=90˚ <br><br />
⇒ ∠O=90˚ <br><br />
therefore DSOP is a Square <br><br />
SO=OP=14 cm <br> hence Radius of given circle is 14 cm<br />
<br />
=Problem 5=<br />
A circle is touching the side BC of △ABC at P. AB and AC when produced are touching the circle at Q and R respectively. Prove that AQ =<math>\frac{1}{2}</math> [perimeter of △ABC].<br />
[[File:123.png|300px]]<br />
<br />
==Algorithm==<br />
In the figure AQ , AR and BC are tangents to the circle with center O.<br><br />
BP=BQ and PC=CR (Tangents drawn from external point are equal) ---------- (1)<br><br />
<br />
Perimeter of △ABC=AB+BC+CA<br><br />
=AB+(BP+PC)+CA<br><br />
=AB+BQ+CR+CA ------ (From eq-1)<br><br />
=(AB+BQ)+(CR+CA)<br><br />
=AQ+AR ----- (From fig)<br><br />
=AQ+AQ -- --- (∵AQ=AR)<br><br />
=2AQ<br><br />
<br />
∴ AQ = <math>\frac{1}{2}</math> [perimeter of △ABC]<br />
=Probiem-6 [Ex-15.4-B3]=</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=14843Circles Tangents Problems2014-08-14T05:28:25Z<p>Anilkumar: /* Problem 5 */</p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
a)<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br><br />
b)<br><br />
Let ∠QBP=x˚<br><br />
<br />
∴∠QPB=x˚ (∵PQ=BQ)<br><br />
Now Let ∠PAQ=y˚<br><br />
∠QPA=y˚ (∵ PQ=AQ)<br><br />
∴In △PAB<br><br />
<br />
∠PAB+∠PBA+∠APB=180˚<br><br />
<br />
y˚+x˚+(x˚+y˚)=180˚<br />
2x˚+2y˚=180˚<br />
2(x˚+y˚)=180˚<br />
x˚+y˚=90˚<br />
∴ ∠APB=90˚<br />
<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[Image:problem 3 on circle.png|300px]]<br />
==Concepts used==<br />
#The radii of a circle are equal<br />
#Properties of isosceles triangle.<br />
#SAS postulate<br />
#Properties of congruent triangles.<br />
<br />
==Prerequisite knowledge==<br />
#The radii of a circle are equal.<br />
# In an isosceles triangle angles opposite to equal sides are equal.<br />
#All the elements of congruent triangles are equal.<br />
<br />
==Algoritham==<br />
In ∆AOB <br><br />
AO=BO [Radii of a same circle]<br><br />
∴ ∠OAB = ∠OBA --------------I [∆AOB is an isosceles ∆}<br><br />
Then,<br> <br />
In ∆AOP and ∆BOP,<br><br />
AO = BO [Radii of a same circle]<br><br />
OP=OP [common side]<br><br />
∠OAP = ∠OBP [ from I]<br><br />
∴ ∆AOP ≅ ∆BOP [SAS postulate] <br><br />
∴ AP = BP [corresponding sides of congruent triangles ]<br />
<br />
=problem-4=<br />
'''In the given Quadrilateral ABCD , BC=38cm , QB=27cm , DC=25cm and AD⊥DC find the radius of the circle.(Ex:15.2. A-6)'''<br><br />
[[File:fig3.png|200px]]<br />
==Interpretation of problem==<br />
=Problem 5=<br />
A circle is touching the side BC of △ABC at P. AB and AC when produced are touching the circle at Q and R respectively. Prove that AQ =<math>\frac{1}{2}</math> [perimeter of △ABC].<br />
[[File:123.png|300px]]<br />
<br />
==Algorithm==<br />
In the figure AQ , AR and BC are tangents to the circle with center O.<br><br />
BP=BQ and PC=CR (Tangents drawn from external point are equal) ---------- (1)<br><br />
<br />
Perimeter of △ABC=AB+BC+CA<br><br />
=AB+(BP+PC)+CA<br><br />
=AB+BQ+CR+CA ------ (From eq-1)<br><br />
=(AB+BQ)+(CR+CA)<br><br />
=AQ+AR ----- (From fig)<br><br />
=AQ+AQ -- --- (∵AQ=AR)<br><br />
=2AQ<br><br />
<br />
∴ AQ = <math>\frac{1}{2}</math> [perimeter of △ABC]</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=14836Circles Tangents Problems2014-08-14T05:24:30Z<p>Anilkumar: /* Problem 5 */</p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
a)<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br><br />
b)<br><br />
Let ∠QBP=x˚<br><br />
<br />
∴∠QPB=x˚ (∵PQ=BQ)<br><br />
Now Let ∠PAQ=x˚<br><br />
<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[Image:problem 3 on circle.png|300px]]<br />
==Concepts used==<br />
#The radii of a circle are equal<br />
#Properties of isosceles triangle.<br />
#SAS postulate<br />
#Properties of congruent triangles.<br />
<br />
==Prerequisite knowledge==<br />
#The radii of a circle are equal.<br />
# In an isosceles triangle angles opposite to equal sides are equal.<br />
#All the elements of congruent triangles are equal.<br />
<br />
==Algoritham==<br />
In ∆AOB <br><br />
AO=BO [Radii of a same circle]<br><br />
∴ ∠OAB = ∠OBA --------------I [∆AOB is an isosceles ∆}<br><br />
Then,<br> <br />
In ∆AOP and ∆BOP,<br><br />
AO = BO [Radii of a same circle]<br><br />
OP=OP [common side]<br><br />
∠OAP = ∠OBP [ from I]<br><br />
∴ ∆AOP ≅ ∆BOP [SAS postulate] <br><br />
∴ AP = BP [corresponding sides of congruent triangles ]<br />
<br />
=problem-4=<br />
'''In the given Quadrilateral ABCD , BC=38cm , QB=27cm , DC=25cm and AD⊥DC find the radius of the circle.(Ex:15.2. A-6)'''<br><br />
[[File:fig3.png|200px]]<br />
==Interpretation of problem==<br />
=Problem 5=<br />
A circle is touching the side BC of △ABC at P. AB and AC when produced are touching the circle at Q and R respectively. Prove that AQ =<math>\frac{1}{2}</math> [perimeter of △ABC].<br />
[[File:123.png|300px]]<br />
<br />
==Algorithm==<br />
In the figure AQ , AR and BC are tangents to the circle with center O.<br><br />
BP=BQ and PC=CR (Tangents drawn from external point are equal) ---------- (1)<br><br />
<br />
Perimeter of △ABC=AB+BC+CA<br><br />
=AB+(BP+PC)+CA<br><br />
=AB+BQ+CR+CA ------ (From eq-1)<br><br />
=(AB+BQ)+(CR+CA)<br><br />
=AQ+AR ----- (From fig)<br><br />
=AQ+AQ -- --- (∵∵∵∵∵∵∵∵∵∵∵∵∵∵∵∵∵∵AQ=AR)<br><br />
=2AQ<br><br />
<br />
∴ AQ = <math>\frac{1}{2}</math> [perimeter of △ABC]</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=14815Circles Tangents Problems2014-08-14T05:06:06Z<p>Anilkumar: /* Problem 5 */</p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
a)<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br><br />
b)<br><br />
<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[Image:problem 3 on circle.png|300px]]<br />
==Concepts used==<br />
#The radii of a circle are equal<br />
#Properties of isosceles triangle.<br />
#SAS postulate<br />
#Properties of congruent triangles.<br />
<br />
==Prerequisite knowledge==<br />
#The radii of a circle are equal.<br />
# In an isosceles triangle angles opposite to equal sides are equal.<br />
#All the elements of congruent triangles are equal.<br />
<br />
==Algoritham==<br />
In ∆AOB <br><br />
AO=BO [Radii of a same circle]<br><br />
∴ ∠OAB = ∠OBA --------------I [∆AOB is an isosceles ∆}<br><br />
Then,<br> <br />
In ∆AOP and ∆BOP,<br><br />
AO = BO [Radii of a same circle]<br><br />
OP=OP [common side]<br><br />
∠OAP = ∠OBP [ from I]<br><br />
∴ ∆AOP ≅ ∆BOP [SAS postulate] <br><br />
∴ AP = BP [corresponding sides of congruent triangles ]<br />
<br />
=problem-4=<br />
'''In the given Quadrilateral ABCD , BC=38cm , QB=27cm , DC=25cm and AD⊥DC find the radius of the circle.(Ex:15.2. A-6)'''<br><br />
[[File:fig3.png|200px]]<br />
==Interpretation of problem==<br />
=Problem 5=<br />
A circle is touching the side BC of △ABC at P. AB and AC when produced are touching the circle at Q and R respectively. Prove that AQ =<math>\frac{1}{2}</math> [perimeter of △ABC].<br />
[[File:123.png|300px]]</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=File:123.png&diff=14814File:123.png2014-08-14T05:03:58Z<p>Anilkumar: Anilkumar uploaded a new version of &quot;File:123.png&quot;</p>
<hr />
<div></div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=File:Screenshot_from_2014-08-13_13-08-29.png&diff=14812File:Screenshot from 2014-08-13 13-08-29.png2014-08-14T04:59:42Z<p>Anilkumar: MsUpload</p>
<hr />
<div>MsUpload</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=14805Circles Tangents Problems2014-08-14T04:50:01Z<p>Anilkumar: </p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
'''Step-1''''''Bold text'''<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br />
<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[Image:problem 3 on circle.png|300px]]<br />
==Concepts used==<br />
#The radii of a circle are equal<br />
#Properties of isosceles triangle.<br />
#SAS postulate<br />
#Properties of congruent triangles.<br />
<br />
==Prerequisite knowledge==<br />
#The radii of a circle are equal.<br />
# In an isosceles triangle angles opposite to equal sides are equal.<br />
#All the elements of congruent triangles are equal.<br />
<br />
==Algoritham==<br />
In ∆AOB <br><br />
AO=BO [Radii of a same circle]<br><br />
∴ ∠OAB = ∠OBA --------------I [∆AOB is an isosceles ∆}<br><br />
Then,<br> <br />
In ∆AOP and ∆BOP,<br><br />
AO = BO [Radii of a same circle]<br><br />
OP=OP [common side]<br><br />
∠OAP = ∠OBP [ from I]<br><br />
∴ ∆AOP ≅ ∆BOP [SAS postulate] <br><br />
∴ AP = BP [corresponding sides of congruent triangles ]<br />
<br />
=problem-4=<br />
'''In the given Quadrilateral ABCD , BC=38cm , QB=27cm , DC=25cm and AD⊥DC find the radius of the circle.(Ex:15.2. A-6)'''<br><br />
[[File:fig3.png|200px]]<br />
==Interpretation of problem==<br />
=Problem 5=</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=14663Circles Tangents Problems2014-08-13T06:43:53Z<p>Anilkumar: </p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
'''Step-1''''''Bold text'''<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[Image:problem 3 on circle.png|300px]]<br />
==Concepts used==<br />
#The radii of a circle are equal<br />
#Properties of isosceles triangle.<br />
#SAS postulate<br />
#Properties of congruent triangles.<br />
<br />
==Prerequisite knowledge==<br />
#The radii of a circle are equal.<br />
# In an isosceles triangle angles opposite to equal sides are equal.<br />
#All the elements of congruent triangles are equal.<br />
<br />
==Algoritham==<br />
In ∆AOB <br><br />
AO=BO [Radii of a same circle]<br><br />
∴ ∠OAB = ∠OBA --------------I [∆AOB is an isosceles ∆}<br><br />
Then,<br> <br />
In ∆AOP and ∆BOP,<br><br />
AO = BO [Radii of a same circle]<br><br />
OP=OP [common side]<br><br />
∠OAP = ∠OBP [ from I]<br><br />
∴ ∆AOP ≅ ∆BOP [SAS postulate] <br><br />
∴ AP = BP [corresponding sides of congruent triangles ]<br />
=problem 4 [Ex-15.2 B.8 ]</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=14658Circles Tangents Problems2014-08-13T06:37:47Z<p>Anilkumar: /* Algoritham */</p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
'''Step-1''''''Bold text'''<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[Image:problem 3 on circle.png|300px]]<br />
==Concepts used==<br />
#The radii of a circle are equal<br />
#Properties of isosceles triangle.<br />
#SAS postulate<br />
#Properties of congruent triangles.<br />
<br />
==Prerequisite knowledge==<br />
#The radii of a circle are equal.<br />
# In an isosceles triangle angles opposite to equal sides are equal.<br />
#All the elements of congruent triangles are equal.<br />
<br />
==Algoritham==<br />
In ∆AOB <br><br />
AO=BO [Radii of a same circle]<br><br />
∴ ∠OAB = ∠OBA --------------I [∆AOB is an isosceles ∆}<br><br />
Then,<br> <br />
In ∆AOP and ∆BOP,<br><br />
AO = BO [Radii of a same circle]<br><br />
OP=OP [common side]<br><br />
∠OAP = ∠OBP [ from I]<br><br />
∴ ∆AOP ≅ ∆BOP [SAS postulate] <br><br />
∴ AP = BP [corresponding sides of congruent triangles ]</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=14655Circles Tangents Problems2014-08-13T06:37:01Z<p>Anilkumar: /* Algoritham */</p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
'''Step-1''''''Bold text'''<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[Image:problem 3 on circle.png|300px]]<br />
==Concepts used==<br />
#The radii of a circle are equal<br />
#Properties of isosceles triangle.<br />
#SAS postulate<br />
#Properties of congruent triangles.<br />
<br />
==Prerequisite knowledge==<br />
#The radii of a circle are equal.<br />
# In an isosceles triangle angles opposite to equal sides are equal.<br />
#All the elements of congruent triangles are equal.<br />
<br />
==Algoritham==<br />
In ∆AOB <br><br />
AO=BO [Radii of a same circle]<br><br />
∴ ∠OAB = ∠OBA --------------I [∆AOB is an isosceles ∆}<br><br />
Then, <br />
In ∆AOP and ∆BOP,<br><br />
AO = BO [Radii of a same circle]<br><br />
OP=OP [common side]<br><br />
∠OAP = ∠OBP [ from I]<br><br />
∴ ∆AOP ≅ ∆BOP [SAS postulate] <br><br />
∴ AP = BP [corresponding sides of congruent triangles ]</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=14647Circles Tangents Problems2014-08-13T06:34:21Z<p>Anilkumar: /* Prerequisite knowledge */</p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
'''Step-1''''''Bold text'''<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[Image:problem 3 on circle.png|300px]]<br />
==Concepts used==<br />
#The radii of a circle are equal<br />
#Properties of isosceles triangle.<br />
#SAS postulate<br />
#Properties of congruent triangles.<br />
<br />
==Prerequisite knowledge==<br />
#The radii of a circle are equal.<br />
# In an isosceles triangle angles opposite to equal sides are equal.<br />
#All the elements of congruent triangles are equal.<br />
<br />
==Algoritham==<br />
In ∆AOB <br />
AO=BO [Radii of a same circle]<br />
∴ ∠OAB = ∠OBA --------------I [∆AOB is an isosceles ∆}<br />
Then, <br />
In ∆AOP and ∆BOP,<br />
AO = BO [Radii of a same circle]<br />
OP=OP [common side]<br />
∠OAP = ∠OBP [ from I]<br />
∴ ∆AOP ≅ ∆BOP [SAS postulate] <br />
∴ AP = BP [corresponding sides of congruent triangles ]</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=14645Circles Tangents Problems2014-08-13T06:33:16Z<p>Anilkumar: /* Concepts used */</p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
'''Step-1''''''Bold text'''<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[Image:problem 3 on circle.png|300px]]<br />
==Concepts used==<br />
#The radii of a circle are equal<br />
#Properties of isosceles triangle.<br />
#SAS postulate<br />
#Properties of congruent triangles.<br />
<br />
==Prerequisite knowledge==<br />
1. The radii of a circle are equal.<br />
2. In an isosceles triangle angles opposite to equal sides are equal.<br />
3.All the elements of congruent triangles are equal.<br />
==Algoritham==<br />
In ∆AOB <br />
AO=BO [Radii of a same circle]<br />
∴ ∠OAB = ∠OBA --------------I [∆AOB is an isosceles ∆}<br />
Then, <br />
In ∆AOP and ∆BOP,<br />
AO = BO [Radii of a same circle]<br />
OP=OP [common side]<br />
∠OAP = ∠OBP [ from I]<br />
∴ ∆AOP ≅ ∆BOP [SAS postulate] <br />
∴ AP = BP [corresponding sides of congruent triangles ]</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=14635Circles Tangents Problems2014-08-13T06:15:51Z<p>Anilkumar: /* problem 3 [Ex-15.2 B.7] */</p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
'''Step-1''''''Bold text'''<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[Image:problem 3 on circle.png|300px]]<br />
==Concepts used==<br />
1. The radii of a circle are equal<br />
2.Properties of isosceles triangle.<br />
3.SAS postulate<br />
4.Properties of congruent triangles. <br />
==Prerequisite knowledge==<br />
1. The radii of a circle are equal.<br />
2. In an isosceles triangle angles opposite to equal sides are equal.<br />
3.All the elements of congruent triangles are equal.<br />
==Algoritham==<br />
In ∆AOB <br />
AO=BO [Radii of a same circle]<br />
∴ ∠OAB = ∠OBA --------------I [∆AOB is an isosceles ∆}<br />
Then, <br />
In ∆AOP and ∆BOP,<br />
AO = BO [Radii of a same circle]<br />
OP=OP [common side]<br />
∠OAP = ∠OBP [ from I]<br />
∴ ∆AOP ≅ ∆BOP [SAS postulate] <br />
∴ AP = BP [corresponding sides of congruent triangles ]</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=14617Circles Tangents Problems2014-08-13T05:57:17Z<p>Anilkumar: /* problem 3 [Ex-15.2 B.7] */</p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
'''Step-1''''''Bold text'''<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[Image:problem 3 on circle.png|300px]]<br />
Concepts used<br />
1. The radii of a circle are equal<br />
2.Properties of isosceles triangle.<br />
3.SAS postulate<br />
4.Properties of congruent triangles. <br />
Prerequisite knowledge<br />
1. The radii of a circle are equal.<br />
2. In an isosceles triangle angles opposite to equal sides are equal.<br />
3.All the elements of congruent triangles are equal.</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=File:Problem_3_on_circle.png&diff=14611File:Problem 3 on circle.png2014-08-13T05:52:15Z<p>Anilkumar: </p>
<hr />
<div></div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=14610Circles Tangents Problems2014-08-13T05:51:43Z<p>Anilkumar: /* problem 3 [Ex-15.2 B.7] */</p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
'''Step-1''''''Bold text'''<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[Image:problem 3 on circle.png|300px]]</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=14609Circles Tangents Problems2014-08-13T05:51:13Z<p>Anilkumar: /* problem 3 [Ex-15.2 B.7] */</p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
'''Step-1''''''Bold text'''<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br><br />
[[]Image:problem 3 on circle.png|300px]]</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=File:Screenshot_from_2014-08-12_15-29-42.png&diff=14599File:Screenshot from 2014-08-12 15-29-42.png2014-08-13T05:44:36Z<p>Anilkumar: Anilkumar uploaded a new version of &quot;File:Screenshot from 2014-08-12 15-29-42.png&quot;</p>
<hr />
<div>MsUpload</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=14596Circles Tangents Problems2014-08-13T05:41:52Z<p>Anilkumar: /* problem 3 [Ex-15.2 B.7] */</p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
'''Step-1''''''Bold text'''<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br />
[[File:Screenshot from 2014-08-12 15:29:42.png|400px]]</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=File:Screenshot_from_2014-08-12_15-29-42.png&diff=14595File:Screenshot from 2014-08-12 15-29-42.png2014-08-13T05:41:30Z<p>Anilkumar: MsUpload</p>
<hr />
<div>MsUpload</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=14589Circles Tangents Problems2014-08-13T05:37:36Z<p>Anilkumar: /* problem 3 [Ex-15.2 B.7] */</p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
'''Step-1''''''Bold text'''<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br />
=problem 3 [Ex-15.2 B.7]=<br />
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Circles_Tangents_Problems&diff=14568Circles Tangents Problems2014-08-13T05:02:44Z<p>Anilkumar: </p>
<hr />
<div>=Problem 1=<br />
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ <br><br />
[[File:image_circle_with_tangents.png|300px]]<br />
==Interpretation of the problem==<br />
#O is the centre of the circle and tangents AP and AQ are drawn from an external point A.<br />
#OP and OQ are the radii.<br />
#The students have to prove thne angle PAQ=twise the angle OPQ.<br />
===Geogebra file===<br />
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<br />
==Concepts used==<br />
#The radii of a circle are equal.<br />
#In any circle the radius drawn at the point of contact is perpendicular to the tangent.<br />
#The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.<br />
#Properties of isoscles triangle.<br />
#Properties of quadrillateral ( sum of all angles) is 360 degrees<br />
#Sum of three angles of triangle is 180 degrees.<br />
==Algorithm==<br />
OP=OQ ---- radii of the same circle<br />
OA is joined.<br><br />
In quadrillateral APOQ ,<br><br />
∠APO=∠AQO=<math>90^{0}</math> [radius drawn at the point of contact is perpendicular to the tangent]<br><br />
∠PAQ+∠POQ=<math>180^{0}</math><br> <br />
Or, ∠PAQ+∠POQ=<math>180^{0}</math><br><br />
∠PAQ = <math>180^{0}</math>-∠POQ ----------1<br><br />
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP<br><br />
∠POQ+∠OPQ+∠OQP=<math>180^{0}</math><br><br />
Or ∠POQ+2∠OPQ=<math>180^{0}</math><br><br />
2∠OPQ=<math>180^{0}</math>- ∠POQ ------2<br><br />
From 1 and 2 <br><br />
∠PAQ=2∠OPQ<br />
=Problem-2=<br />
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that<br>a). Tangent at P bisects AB at Q<br> b). ∠APB=90° (Exescise-15.2, B.3)<br><br />
[[File:fig2.png|400px]]<br />
==Interpretation of the problem==<br />
#In the given figure two circles touch externally.<br />
#AB is the direct common tangent to these circles.<br />
#PQ is the transverse common tangent drawn to these circles at point P.<br />
#Using the tangent properties students have to show AQ=BQ and ∠APB=90°<br />
==Concepts used==<br />
#The tangent drawn from an external point to a circle <br> a) are equal <br> b] subtend equal angle at the center <br> c] are equally inclined to the line joining the center and external point.<br />
#Angle subtended by equal sides are equal.<br />
#Axiom-1:- "Things which are equal to same thing are equal"<br><br />
[[https://www.geogebratube.org/student/m86017 '''Click here for geogebra animation''']]<br />
<br />
==Algorithm==<br />
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.<br><br />
'''Step-1''''''Bold text'''<br><br />
AQ=QP and BQ=QP (Tangents drawn from external point are equal)<br><br />
By axiom-1, AQ=BQ<br><br />
∴tangent at P bisects AB at Q.<br />
=problem 3 [Ex-15.2 B.7]=</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Factorisation&diff=9714Factorisation2014-02-15T06:53:39Z<p>Anilkumar: /* hints for difficult problem */</p>
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= Concept Map =<br />
<mm>[[Factorisation of Polynomials.mm|Flash]]</mm><br />
<br />
<br />
<br />
__FORCETOC__<br />
<br />
= Textbook =<br />
To add textbook links, please follow these instructions to: <br />
([{{fullurl:{{FULLPAGENAME}}/textbook|action=edit}} Click to create the subpage])<br />
<br />
=Additional Information=<br />
==Useful websites==<br />
*Question Corner<br />
<br />
<br />
#[http://www.mathsisfun.com/algebra/polynomials.html Maths is Fun]. This website contains good worksheets for factorisation.<br />
#[http://reference.wolfram.com/mathematica/guide/PolynomialAlgebra.html Wolfram Mathworld]. This website contains good simulations for math identities.<br />
<br />
==Reference Books==<br />
===NCERT Books===<br />
#[http://www.ncert.nic.in/ncerts/textbook/textbook.htm?hemh1=9-16 Algebraic expressions and identities] <br />
#[http://www.ncert.nic.in/ncerts/textbook/textbook.htm?hemh1=14-16 Factorisation]<br />
<br />
= Teaching Outlines =<br />
<br />
==Concept #1 Monomial expressions==<br />
===Learning objectives===<br />
To introduce expressions and the need and method of splitting<br />
===Notes for teachers===<br />
===Activity No # ===<br />
{| style="height:10px; float:right; align:center;"<br />
|<div style="width:150px;border:none; border-radius:10px;box-shadow: 5px 5px 5px #888888; background:#f5f5f5; vertical-align:top; text-align:center; padding:5px;"><br />
''[http://karnatakaeducation.org.in/?q=node/305 Click to Comment]''</div><br />
|}<br />
*Estimated Time<br />
*Materials/ Resources needed<br />
*Prerequisites/Instructions, if any<br />
*Multimedia resources<br />
*Website interactives/ links/ / Geogebra Applets<br />
*Process/ Developmental Questions<br />
*Evaluation<br />
*Question Corner<br />
<br />
===Activity No #2 Demonstrate Binomial Cube===<br />
{| style="height:10px; float:right; align:center;"<br />
|<div style="width:150px;border:none; border-radius:10px;box-shadow: 5px 5px 5px #888888; background:#f5f5f5; vertical-align:top; text-align:center; padding:5px;"><br />
''[http://karnatakaeducation.org.in/?q=node/305 Click to Comment]''</div><br />
|}<br />
*Estimated Time <br />
*Materials/ Resources needed<br />
*Prerequisites/Instructions, if any<br />
*Multimedia resources<br />
http://www.infomontessori.com/sensorial/montessori_binomial_cube_1.jpg<br />
*Website interactives/ links/ / Geogebra Applets<br />
*Process/ Developmental Questions<br />
*Evaluation<br />
*Question Corner<br />
<br />
==Concept #==<br />
===Learning objectives===<br />
===Notes for teachers===<br />
===Activity No #1 Geogebra===<br />
{| style="height:10px; float:right; align:center;"<br />
|<div style="width:150px;border:none; border-radius:10px;box-shadow: 5px 5px 5px #888888; background:#f5f5f5; vertical-align:top; text-align:center; padding:5px;"><br />
''[http://karnatakaeducation.org.in/?q=node/305 Click to Comment]''</div><br />
|}<br />
*Estimated Time<br />
*Materials/ Resources needed<br />
*Prerequisites/Instructions, if any<br />
*Multimedia resources<br />
[[File:Geogebra_screenshot_for_identities.png|150px]]<br> This is a Geogebra screenshot for identity.<br><br />
{{#widget:YouTube|id=crjItXrt43Q}}<br><br />
This is a classroom demonstration of binomial cube. Show the children before you start the cubic identity.<br />
*Website interactives/ links/ / Geogebra Applets<br />
*Process/ Developmental Questions<br />
*Evaluation<br />
*Question Corner<br />
<br />
===Activity No # ===<br />
{| style="height:10px; float:right; align:center;"<br />
|<div style="width:150px;border:none; border-radius:10px;box-shadow: 5px 5px 5px #888888; background:#f5f5f5; vertical-align:top; text-align:center; padding:5px;"><br />
''[http://karnatakaeducation.org.in/?q=node/305 Click to Comment]''</div><br />
|}<br />
*Estimated Time<br />
*Materials/ Resources needed<br />
*Prerequisites/Instructions, if any<br />
*Multimedia resources<br />
*Website interactives/ links/ / Geogebra Applets<br />
*Process/ Developmental Questions<br />
*Evaluation<br />
*Question Corner<br />
<br />
<br />
<br />
# = Hints for difficult problems =<br />
Question : If x= and y= find <br />
<br />
<br />
Solution : <br />
<br />
Analysing the given condition<br />
<br />
Step 1 : = ()() => Using the formula : = ( )()<br />
<br />
step 2 : = ( + )(-) => substitute the value of x and y<br />
<br />
Step 3 : = x => take the L.C.M of the denominator , simplyfy using concept of addition and substaction of fraction <br />
<br />
Step 4: = x <br />
simply the above using basic concepts of addition and substraction<br />
<br />
Step 5 : = x <br />
<br />
Step 6 := x => take common term 2 ( H.C.F)<br />
<br />
Step 7 : = x <br />
<br />
Step 8 : =<br />
#<br />
=hints for difficult problem=<br />
<br />
If x-= 4 prove that <math>x^{3}+6x^{2}+\frac {6} {x^{2}}-\frac{1} {x^{3}}=184 </math><br />
<br />
=====If x+y=a and xy=b then prove that (1+)+(1+) = <br />
<br />
Steps for solution<br />
<br />
step 1: * Understanding the problem first.<br />
* Recalling the indentities <br />
<br />
step 2 : * consider the condition and squaring on both side<br />
* simplify to get the value<br />
<br />
step 3: * consider LHS<br />
* multiply the expression<br />
* substitute the value<br />
* simlpify the equqtion<br />
<br />
<br />
solution for the problem<br />
<br />
consider x+y=a<br />
=<br />
substitute x+y =a and xy=b<br />
then we get <br />
------->(1)<br />
consider xy=b squaring on both side<br />
then we get =------->(2)<br />
<br />
consider LHS=<br />
=(1+)+(1+)<br />
=1+<br />
= 1+ from eqn 1 & 2<br />
=<br />
= <br />
LHS = RHS=============<br />
<br />
= Project Ideas =<br />
<br />
= Math Fun =<br />
<br />
'''Usage''' <br />
<br />
Create a new page and type <nowiki>{{subst:Math-Content}}</nowiki> to use this template</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Factorisation&diff=9707Factorisation2014-02-15T06:43:45Z<p>Anilkumar: </p>
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[http://karnatakaeducation.org.in/KOER/en/index.php/Text_Books#Mathematics_-_Textbooks Textbooks]<br />
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|}<br />
While creating a resource page, please click here for a resource creation [http://karnatakaeducation.org.in/KOER/en/index.php/Resource_Creation_Checklist '''checklist'''].<br />
= Concept Map =<br />
<mm>[[Factorisation of Polynomials.mm|Flash]]</mm><br />
<br />
<br />
<br />
__FORCETOC__<br />
<br />
= Textbook =<br />
To add textbook links, please follow these instructions to: <br />
([{{fullurl:{{FULLPAGENAME}}/textbook|action=edit}} Click to create the subpage])<br />
<br />
=Additional Information=<br />
==Useful websites==<br />
*Question Corner<br />
<br />
<br />
#[http://www.mathsisfun.com/algebra/polynomials.html Maths is Fun]. This website contains good worksheets for factorisation.<br />
#[http://reference.wolfram.com/mathematica/guide/PolynomialAlgebra.html Wolfram Mathworld]. This website contains good simulations for math identities.<br />
<br />
==Reference Books==<br />
===NCERT Books===<br />
#[http://www.ncert.nic.in/ncerts/textbook/textbook.htm?hemh1=9-16 Algebraic expressions and identities] <br />
#[http://www.ncert.nic.in/ncerts/textbook/textbook.htm?hemh1=14-16 Factorisation]<br />
<br />
= Teaching Outlines =<br />
<br />
==Concept #1 Monomial expressions==<br />
===Learning objectives===<br />
To introduce expressions and the need and method of splitting<br />
===Notes for teachers===<br />
===Activity No # ===<br />
{| style="height:10px; float:right; align:center;"<br />
|<div style="width:150px;border:none; border-radius:10px;box-shadow: 5px 5px 5px #888888; background:#f5f5f5; vertical-align:top; text-align:center; padding:5px;"><br />
''[http://karnatakaeducation.org.in/?q=node/305 Click to Comment]''</div><br />
|}<br />
*Estimated Time<br />
*Materials/ Resources needed<br />
*Prerequisites/Instructions, if any<br />
*Multimedia resources<br />
*Website interactives/ links/ / Geogebra Applets<br />
*Process/ Developmental Questions<br />
*Evaluation<br />
*Question Corner<br />
<br />
===Activity No #2 Demonstrate Binomial Cube===<br />
{| style="height:10px; float:right; align:center;"<br />
|<div style="width:150px;border:none; border-radius:10px;box-shadow: 5px 5px 5px #888888; background:#f5f5f5; vertical-align:top; text-align:center; padding:5px;"><br />
''[http://karnatakaeducation.org.in/?q=node/305 Click to Comment]''</div><br />
|}<br />
*Estimated Time <br />
*Materials/ Resources needed<br />
*Prerequisites/Instructions, if any<br />
*Multimedia resources<br />
http://www.infomontessori.com/sensorial/montessori_binomial_cube_1.jpg<br />
*Website interactives/ links/ / Geogebra Applets<br />
*Process/ Developmental Questions<br />
*Evaluation<br />
*Question Corner<br />
<br />
==Concept #==<br />
===Learning objectives===<br />
===Notes for teachers===<br />
===Activity No #1 Geogebra===<br />
{| style="height:10px; float:right; align:center;"<br />
|<div style="width:150px;border:none; border-radius:10px;box-shadow: 5px 5px 5px #888888; background:#f5f5f5; vertical-align:top; text-align:center; padding:5px;"><br />
''[http://karnatakaeducation.org.in/?q=node/305 Click to Comment]''</div><br />
|}<br />
*Estimated Time<br />
*Materials/ Resources needed<br />
*Prerequisites/Instructions, if any<br />
*Multimedia resources<br />
[[File:Geogebra_screenshot_for_identities.png|150px]]<br> This is a Geogebra screenshot for identity.<br><br />
{{#widget:YouTube|id=crjItXrt43Q}}<br><br />
This is a classroom demonstration of binomial cube. Show the children before you start the cubic identity.<br />
*Website interactives/ links/ / Geogebra Applets<br />
*Process/ Developmental Questions<br />
*Evaluation<br />
*Question Corner<br />
<br />
===Activity No # ===<br />
{| style="height:10px; float:right; align:center;"<br />
|<div style="width:150px;border:none; border-radius:10px;box-shadow: 5px 5px 5px #888888; background:#f5f5f5; vertical-align:top; text-align:center; padding:5px;"><br />
''[http://karnatakaeducation.org.in/?q=node/305 Click to Comment]''</div><br />
|}<br />
*Estimated Time<br />
*Materials/ Resources needed<br />
*Prerequisites/Instructions, if any<br />
*Multimedia resources<br />
*Website interactives/ links/ / Geogebra Applets<br />
*Process/ Developmental Questions<br />
*Evaluation<br />
*Question Corner<br />
<br />
<br />
<br />
# = Hints for difficult problems =<br />
Question : If x= and y= find <br />
<br />
<br />
Solution : <br />
<br />
Analysing the given condition<br />
<br />
Step 1 : = ()() => Using the formula : = ( )()<br />
<br />
step 2 : = ( + )(-) => substitute the value of x and y<br />
<br />
Step 3 : = x => take the L.C.M of the denominator , simplyfy using concept of addition and substaction of fraction <br />
<br />
Step 4: = x <br />
simply the above using basic concepts of addition and substraction<br />
<br />
Step 5 : = x <br />
<br />
Step 6 := x => take common term 2 ( H.C.F)<br />
<br />
Step 7 : = x <br />
<br />
Step 8 : =<br />
#<br />
=hints for difficult problem=<br />
<br />
=====If x+y=a and xy=b then prove that (1+)+(1+) = <br />
<br />
Steps for solution<br />
<br />
step 1: * Understanding the problem first.<br />
* Recalling the indentities <br />
<br />
step 2 : * consider the condition and squaring on both side<br />
* simplify to get the value<br />
<br />
step 3: * consider LHS<br />
* multiply the expression<br />
* substitute the value<br />
* simlpify the equqtion<br />
<br />
<br />
solution for the problem<br />
<br />
consider x+y=a<br />
=<br />
substitute x+y =a and xy=b<br />
then we get <br />
------->(1)<br />
consider xy=b squaring on both side<br />
then we get =------->(2)<br />
<br />
consider LHS=<br />
=(1+)+(1+)<br />
=1+<br />
= 1+ from eqn 1 & 2<br />
=<br />
= <br />
LHS = RHS=============<br />
<br />
= Project Ideas =<br />
<br />
= Math Fun =<br />
<br />
'''Usage''' <br />
<br />
Create a new page and type <nowiki>{{subst:Math-Content}}</nowiki> to use this template</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Polynomials&diff=9699Polynomials2014-02-15T06:36:00Z<p>Anilkumar: </p>
<hr />
<div>'''Usage''' <br />
<br />
Create a new page and type <nowiki>{{subst:Math-Content}}</nowiki> to use this template<br />
<br />
Please click here for a resource creation [http://karnatakaeducation.org.in/KOER/en/index.php/Resource_Creation_Checklist '''checklist'''].<br />
<br />
= Concept Map =<br />
<mm>[[Multiplication.mm|Flash]]</mm><br />
<br />
= Textbook =<br />
To add textbook links, please follow these instructions to: <br />
([{{fullurl:{{FULLPAGENAME}}/textbook|action=edit}} Click to create the subpage])<br />
<br />
==NCERT Books==<br />
[http://karnatakaeducation.org.in/KOER/en/index.php/Text_Books_NCERT_Mathematics/ Class IX Chapter Polynomials]<br />
<br />
==Tamilnadu Text Books==<br />
[http://www.textbooksonline.tn.nic.in/Books/Std09/Std09-I-Maths-KM.pdf/ Class IX text book]<br />
<br />
=Additional Information=<br />
==Useful websites==<br />
[http://www.onlinemathlearning.com/multiplying-polynomials-techniques.html multipying polynomials with tabular method is explained in a very simple manner which could help you in classroom teching]<br />
<br />
[http://www.mathsisfun.com/algebra/polynomials.html Maths is fun website here you can understand polynomial in easy way]<br />
<br />
==Reference Books==<br />
<br />
= Teaching Outlines =<br />
<br />
==Concept #1==<br />
===Learning objectives===<br />
1. Can multiply Polynomial by monomial, binomial and Trinomial<br />
<br />
2. Can find the product of any two Polynomials by formal multiplication<br />
===Notes for teachers===<br />
<br />
===Activity No #1 ===<br />
'''Multiplying Polynomials Using Algebra Tiles''' <br />
*Estimated Time<br />
15 to 20 minutes<br />
*Materials/ Resources needed<br />
1. Algebra Tiles activity sheet , <br />
2. Algebra tiles <br />
*Prerequisites/Instructions, if any<br />
Knowledge Of the terms monomial, binomial, trinomial, polynomial, term, degree, base, exponent, coefficient <br />
*Multimedia resources<br />
{{#widget:YouTube|id=2FbEEgGq46o}}<br />
*Website interactives/ links/ / Geogebra Applets<br />
*Process/ Developmental Questions<br />
1. Demonstrate multiplying polynomials using algebra tiles, as shown in above video : <br />
<br />
(x x 1)(x + 6) = x2 + 7x + 6 <br />
<br />
<br />
2. Distribute algebra tiles and copies of the Multiplying Polynomials Using Algebra Tiles <br />
activity sheet. Instruct students to model each expression with the tiles, draw the model, <br />
simplify the expression, and write the simplified answer. <br />
*Evaluation<br />
o Draw a model of the multiplication of two binomials. Simplify your expression. <br />
<br />
o Explain why (2x)(3x) = 6x and not 6x. <br />
<br />
*Question Corner<br />
<br />
===Activity No #2 ===<br />
'''Multiplying Polynomials Using Tabular method'''<br />
*Estimated Time<br />
*Materials/ Resources needed<br />
<br />
Whiteboard, marker pens <br />
<br />
*Prerequisites/Instructions, if any<br />
Knowledge Of the terms monomial, binomial, trinomial, polynomial, term, degree, base, exponent, coefficient <br />
*Multimedia resources<br />
{{#widget:YouTube|id=XYP8oPVdrZw}}<br />
*Website interactives/ links/ / Geogebra Applets<br />
<br />
*Process/ Developmental Questions<br />
1. Demonstrate multiplying polynomials using tabular method<br />
<br />
(x x 1)(x + 6) = x2 + 7x + 6 <br />
<br />
<br />
2. Distribute copies of the Multiplying Polynomials Using Tabular method activity sheet. Instruct students to model each expression with the tabular method , draw the table, simplify the expression, and write the simplified answer. <br />
<br />
*Evaluation<br />
What is the product of (2x+3y)^2<br />
*Question Corner<br />
<br />
==Concept #2==<br />
===Learning objectives===<br />
1. Can find the product of binomials using special Identities<br />
2. Can prove some of the important Identities<br />
3. can expand binomial terms using the Identities<br />
4. Solve problems under Identities<br />
5. can prove some of the important conditional Identities<br />
===Notes for teachers===<br />
<br />
===Activity No #1 ===<br />
Verification the Identity (a+b)(a-b) = a2-b2<br />
{| style="height:10px; float:right; align:center;"<br />
|<div style="width:150px;border:none; border-radius:10px;box-shadow: 5px 5px 5px #888888; background:#f5f5f5; vertical-align:top; text-align:center; padding:5px;"><br />
''[http://karnatakaeducation.org.in/?q=node/305 Click to Comment]''</div><br />
|}<br />
*Estimated Time<br />
10 to 15 minutes<br />
*Materials/ Resources needed<br />
Laptop, Projector, Geogebra Software<br />
*Prerequisites/Instructions, if any<br />
Should know the multiplication of Integers <br />
*Multimedia resources<br />
<br />
*Website interactives/ links/ / Geogebra Applets<br />
<ggb_applet width="400" height="300" version="4.0" <br />
ggbBase64="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<ggb_applet width="1366" height="568" version="4.0" 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<br />
===Activity No # ===<br />
{| style="height:10px; float:right; align:center;"<br />
|<div style="width:150px;border:none; border-radius:10px;box-shadow: 5px 5px 5px #888888; background:#f5f5f5; vertical-align:top; text-align:center; padding:5px;"><br />
''[http://karnatakaeducation.org.in/?q=node/305 Click to Comment]''</div><br />
|}<br />
*Estimated Time<br />
*Materials/ Resources needed<br />
*Prerequisites/Instructions, if any<br />
*Multimedia resources<br />
*Website interactives/ links/ / Geogebra Applets<br />
*Process/ Developmental Questions<br />
*Evaluation<br />
*Question Corner<br />
<br />
*Question Corner<br />
<br />
<br />
=Hints for difficult problems=<br />
<br />
=Project Ideas=<br />
<br />
=Math Fun=</div>Anilkumarhttps://karnatakaeducation.org.in/KOER/en/index.php?title=Polynomials&diff=9696Polynomials2014-02-15T06:32:47Z<p>Anilkumar: </p>
<hr />
<div>'''Usage''' <br />
<br />
Create a new page and type <nowiki>{{subst:Math-Content}}</nowiki> to use this template<br />
<br />
Please click here for a resource creation [http://karnatakaeducation.org.in/KOER/en/index.php/Resource_Creation_Checklist '''checklist'''].<br />
<br />
= Concept Map =<br />
<mm>[[Multiplication.mm|Flash]]</mm><br />
<br />
= Textbook =<br />
To add textbook links, please follow these instructions to: <br />
([{{fullurl:{{FULLPAGENAME}}/textbook|action=edit}} Click to create the subpage])<br />
<br />
==NCERT Books==<br />
[http://karnatakaeducation.org.in/KOER/en/index.php/Text_Books_NCERT_Mathematics/ Class IX Chapter Polynomials]<br />
<br />
==Tamilnadu Text Books==<br />
[http://www.textbooksonline.tn.nic.in/Books/Std09/Std09-I-Maths-KM.pdf/ Class IX text book]<br />
<br />
=Additional Information=<br />
==Useful websites==<br />
[http://www.onlinemathlearning.com/multiplying-polynomials-techniques.html multipying polynomials with tabular method is explained in a very simple manner which could help you in classroom teching]<br />
<br />
[http://www.mathsisfun.com/algebra/polynomials.html Maths is fun website here you can understand polynomial in easy way]<br />
<br />
==Reference Books==<br />
<br />
= Teaching Outlines =<br />
<br />
==Concept #1==<br />
===Learning objectives===<br />
1. Can multiply Polynomial by monomial, binomial and Trinomial<br />
<br />
2. Can find the product of any two Polynomials by formal multiplication<br />
===Notes for teachers===<br />
<br />
===Activity No #1 ===<br />
'''Multiplying Polynomials Using Algebra Tiles''' <br />
*Estimated Time<br />
15 to 20 minutes<br />
*Materials/ Resources needed<br />
1. Algebra Tiles activity sheet , <br />
2. Algebra tiles <br />
*Prerequisites/Instructions, if any<br />
Knowledge Of the terms monomial, binomial, trinomial, polynomial, term, degree, base, exponent, coefficient <br />
*Multimedia resources<br />
{{#widget:YouTube|id=2FbEEgGq46o}}<br />
*Website interactives/ links/ / Geogebra Applets<br />
*Process/ Developmental Questions<br />
1. Demonstrate multiplying polynomials using algebra tiles, as shown in above video : <br />
<br />
(x x 1)(x + 6) = x2 + 7x + 6 <br />
<br />
<br />
2. Distribute algebra tiles and copies of the Multiplying Polynomials Using Algebra Tiles <br />
activity sheet. Instruct students to model each expression with the tiles, draw the model, <br />
simplify the expression, and write the simplified answer. <br />
*Evaluation<br />
o Draw a model of the multiplication of two binomials. Simplify your expression. <br />
<br />
o Explain why (2x)(3x) = 6x and not 6x. <br />
<br />
*Question Corner<br />
<br />
===Activity No #2 ===<br />
'''Multiplying Polynomials Using Tabular method'''<br />
*Estimated Time<br />
*Materials/ Resources needed<br />
<br />
Whiteboard, marker pens <br />
<br />
*Prerequisites/Instructions, if any<br />
Knowledge Of the terms monomial, binomial, trinomial, polynomial, term, degree, base, exponent, coefficient <br />
*Multimedia resources<br />
{{#widget:YouTube|id=XYP8oPVdrZw}}<br />
*Website interactives/ links/ / Geogebra Applets<br />
<br />
*Process/ Developmental Questions<br />
1. Demonstrate multiplying polynomials using tabular method<br />
<br />
(x x 1)(x + 6) = x2 + 7x + 6 <br />
<br />
<br />
2. Distribute copies of the Multiplying Polynomials Using Tabular method activity sheet. Instruct students to model each expression with the tabular method , draw the table, simplify the expression, and write the simplified answer. <br />
<br />
*Evaluation<br />
What is the product of (2x+3y)^2<br />
*Question Corner<br />
<br />
==Concept #2==<br />
===Learning objectives===<br />
1. Can find the product of binomials using special Identities<br />
2. Can prove some of the important Identities<br />
3. can expand binomial terms using the Identities<br />
4. Solve problems under Identities<br />
5. can prove some of the important conditional Identities<br />
===Notes for teachers===<br />
<br />
===Activity No #1 ===<br />
Verification the Identity (a+b)(a-b) = a2-b2<br />
{| style="height:10px; float:right; align:center;"<br />
|<div style="width:150px;border:none; border-radius:10px;box-shadow: 5px 5px 5px #888888; background:#f5f5f5; vertical-align:top; text-align:center; padding:5px;"><br />
''[http://karnatakaeducation.org.in/?q=node/305 Click to Comment]''</div><br />
|}<br />
*Estimated Time<br />
10 to 15 minutes<br />
*Materials/ Resources needed<br />
Laptop, Projector, Geogebra Software<br />
*Prerequisites/Instructions, if any<br />
Should know the multiplication of Integers <br />
*Multimedia resources<br />
<br />
*Website interactives/ links/ / Geogebra Applets<br />
<ggb_applet width="400" height="300" version="4.0" <br />
ggbBase64="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<ggb_applet width="1366" height="568" version="4.0" 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<br />
<br />
===Activity No # ===<br />
{| style="height:10px; float:right; align:center;"<br />
|<div style="width:150px;border:none; border-radius:10px;box-shadow: 5px 5px 5px #888888; background:#f5f5f5; vertical-align:top; text-align:center; padding:5px;"><br />
''[http://karnatakaeducation.org.in/?q=node/305 Click to Comment]''</div><br />
|}<br />
*Estimated Time<br />
*Materials/ Resources needed<br />
*Prerequisites/Instructions, if any<br />
*Multimedia resources<br />
*Website interactives/ links/ / Geogebra Applets<br />
*Process/ Developmental Questions<br />
*Evaluation<br />
*Question Corner<br />
<br />
*Question Corner<br />
= Hints for difficult problems =<br />
<br />
<br />
= Hints for difficult problems =<br />
<br />
= Project Ideas =<br />
<br />
= Math Fun =</div>Anilkumar