Rs Aggarwal 2015 for Class 10 Math Chapter 12 - Circles

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Rs Aggarwal 2015 Solutions for Class 10 Math Chapter 12 Circles are provided here with simple step-by-step explanations. These solutions for Circles are extremely popular among Class 10 students for Math Circles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2015 Book of Class 10 Math Chapter 12 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2015 Solutions. All Rs Aggarwal 2015 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 529:

Question 1:

Find the length of the tangent drawn to a circle with radius 8 cm from a point 17 cm away from the centre of the circle.

Answer:

$Let O be the centre of the given circle . Let P be a point, such that OP = 17 cm . Let OT be the radius, where OT = 5 cm Join TP,where TP is a tangent . Now,tangent drawn from an external point is perpendicular to the radius at the point of contact . ∴ OT ⊥ PT In the right △ OTP, we have: {OP}^{2} = {OT}^{2} + {TP}^{2} [By Pythagoras' theorem:] TP = \sqrt{{OP}^{2} - O T^{2}} = \sqrt{17^{2} - 8^{2}} = \sqrt{289 - 64} = \sqrt{225} =15 cm . ∴ The length of the tangent is 15 cm .$

Page No 529:

Question 2:

A point P is 25 cm away from the centre of a circle and the length of tangent drawn from P to the circle is 24 cm. Find the radius of the circle.

Answer:

$Draw a circle and let P be a point such that OP = 25 cm . Let TP be the tangent, so that TP = 24 cm Join OT,where OT is radius . Now, tangent drawn from an external point is perpendicular to the radius at the point of contact . ∴ OT ⊥ PT In the right △ OTP, we have: {OP}^{2} = {OT}^{2} + {TP}^{2} [By Pythagoras' theorem:] {OT}^{2} = \sqrt{{OP}^{2} - {TP}^{2}} = \sqrt{25^{2} - 24^{2}} = \sqrt{625 - 576} = \sqrt{49} =7 cm ∴ The length of the radius is 7 cm .$

Page No 529:

Question 3:

In the given figure, PA and PB are the tangent segments to a circle with centre O. Show that the points A, O, B and P are concyclic.

Answer:

$Given, PA and PB are tangents to the circle with centre O and OA is a radius . Now, tangents drawn from an external point are perpendicular to the radius at the point of contact . ∴ O A ⊥ P A and O B ⊥ P B . ∴ ∠ OAP = 90^{0} and ∠ OBP = 90^{0} ∴ ∠ OAP + ∠ OBP = 90^{0} + 90^{0} = 180^{0} Thus, the sum of a pair of opposite angles of quadrilateral AOBP {is 180}^{0} . Now, ∠ A O B + ∠ A P B + ∠ P A O + ∠ P B O = 360^{0} (∵ {Sum of all the angles of a quadrilateral is 360}^{0}) = > ∠ A O B + ∠ A P B = 180^{0} (∵ ∠ OAP + ∠ OBP= 180^{0}) ∴ AOBP is a cyclic quadrilateral .$

Page No 529:

Question 4:

From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at a point E and PA = 14 cm, find the perimeter of ΔPCD.

Answer:

$Given, PA and PB are the tangents to a circle with centre O and CD is a tangent at E and PA = 14 cm . Tangents drawn from an external point are equal. ∴ PA = PB,CA = CE and DB = DE Perimeter of △ PCD = PC + CD + PD = (PA - CA) + (CE + DE) + (PB - DB) = (PA - C E) + (CE + DE) + (PB - DE) = (PA + PB) =2PA (∵ PA=PB) = (2 \times 14) cm =28 cm ∴ Perimeter of △ PCD = 28 cm .$

Page No 529:

Question 5:

A circle is inscribed in ΔABC, touching AB, BC and AC at P, Q and R, respectively. If AB = 10 cm, AR = 7 cm and CR = 5 cm, find the length of BC.

Answer:

$Given, a circle inscribed in triangle ABC, such that the circle touches the sides of the triangle at P,Q,R . Tangents drawn to a circle from an external point are equal . ∴ AP = AR = 7 cm, CQ = CR = 5 cm . Now, BP = (AB - AP) = (10 - 7) = 3 cm ∴ BP = BQ = 3 cm ∴ BC = (BQ + QC) = > B C = 3 + 5 = > B C = 8 ∴ The length of BC is 8 cm .$

Page No 529:

Question 6:

In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose three sides are AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD.

Answer:

$Let the circle touch the sides of the quadrilateral AB,BC,CD and DA at P,Q,R and S respectively . Given, A B = 6 cm, B C = 7 cm and C D = 4 cm . Tangents drawn from an external point are equal. ∴ A P = A S, B P = B Q, C R = C Q and D R = D S \begin{array}{l} Now, A B + C D = (A P + B P) + (C R + D R) \\ = > A B + C D = (A S + B Q) + (C Q + D S) \\ = > A B + C D = (A S + D S) + (B Q + C Q) \\ = > A B + C D = A D + B C \\ = > A D = (A B + C D) - B C \\ = > A D = (6 + 4) - 7 \\ = > A D = 3 cm . \end{array} ∴ The length of A D is 3 cm .$

Page No 529:

Question 7:

Two tangent segments BC and BD are drawn to a circle with centre O, such that ∠CBD = 120°. Prove that OB = 2BC.

Answer:

$Given, B C and B D are the two tangents drawn to the circle with centre O, such that ∠ C B D = 120^{0} . They are equally inclined to the line segment joining the centre to that point. ∴ ∠ O B C = ∠ O B D = \frac{1}{2} ∠ C B D = 60^{0} Now, tangent drawn from an external point is perpendicular to the radius at the point of contact . ∴ ∠ O C B = ∠ O D B = 90^{0} \begin{array}{l} ∴ From Δ O B C, ∠ B O C = 180^{0} - (∠ O C B + ∠ O B C) \\ \Rightarrow ∠ B O C = 180^{0} - (90^{0} + 60^{0}) \\ \Rightarrow ∠ B O C = 30^{0} . \end{array} \begin{array}{l} Now from right-angled △ O B C, \frac{B C}{O B} = \sin 30^{0} \\ \Rightarrow \frac{B C}{O B} = \frac{1}{2} \\ \Rightarrow O B = 2 B C . \end{array} Hence proved .$

Page No 530:

Question 8:

In the given figure, O is the centre of two concentric circles of radii 4 cm and 6 cm respectively. PA and PB are tangents to the outer and inner circles, respectively. If PA = 10 cm, find the length of PB up to one decimal place.

Answer:

$Given, O is the centre of two concentric circles of radii O A = 6 cm and O B = 4 cm . P A and P B are the two tangents to the outer and inner circles respectively and P A = 10 cm . Now, tangent drawn from an external point is perpendicular to the radius at the point of contact . ∴ ∠ O A P = ∠ O B P = 90^{0} ∴ From right-angled △ O A P, O P^{2} = O A^{2} + P A^{2} \begin{array}{l} = > O P = \sqrt{O A^{2} + P A^{2}} \\ = > O P = \sqrt{6^{2} + 10^{2}} \\ = > O P = \sqrt{136} cm . \end{array} ∴ From right-angled △ O A P, O P^{2} = O B^{2} + P B^{2} \begin{array}{l} = > P B = \sqrt{O P^{2} - O B^{2}} \\ = > P B = \sqrt{136 - 16} \\ = > P B = \sqrt{120} cm \\ => P B = 10.9 cm . \end{array} ∴ The length of P B is 10.9 cm .$

Page No 530:

Question 9:

In the given figure, ABCD is a quadrilateral in which ∠D = 90°. A circle C(O, r) touches the sides AB, BC, CD and DA at P, Q, R, S, respectively. If BC = 38 cm, CD = 25 cm and BP = 27 cm, find the value of r.

Answer:

$\begin{array}{l} Given, A B C D is a quadrilateral in which ∠ D = 90^{0} . A circle C (O, r) touches the sides \\ A B, B C, C D and D A at P, Q, R, S respectively and B C = 38 cm, C D = 25 cm and B P = 27 cm . \end{array} Now, O R = O S (Radii of the same circle) Tangents drawn from an external point are perpendicular to the radius at the point of contact . ∴ O R ⊥ D R and O S ⊥ D S . ∴ O R D S is a square . Tangents drawn from an external point are equal. ∴ B P = B Q, C Q = C R and D R = D S ∴ B Q = B P = 27 cm \begin{array}{l} => B C - C Q = 27 \\ = > 38 - C Q = 27 \\ = > C Q = 38 - 27 \\ = > C Q = 11 cm \end{array} ∴ C R = 11 cm => C D - D R = 11 cm =>25 - D R = 11 => D R = 25 - 11 = 14 cm Now, D R = D S = O R = O S (∵ Sides of a square are equal) ∴ O R = O S = 14 cm ∴ r = 14 cm ∴ The radius is 14 cm .$

Page No 531:

Question 1:

In the given figure, PT is a tangent to the circle with centre O. If OT = 6 cm and OP = 10 cm, then the length of tangent PT is

(a) 8 cm
(b) 10 cm
(c) 12 cm
(d) 16 cm

Answer:

(a) 8 cm

$P T is a tangent to the circle with center O . Given that O T = 6 cm and O P = 10 cm . T he tangent at any point of a circle is perpendicular to the radius at the point of contact . ∴ O T ⊥ P T In right - angled triangle P T O, P T^{2} + O T^{2} = O P^{2} [U s i n g P y t h a g o r a s' t h e o r e m] \Rightarrow P T^{2} = O P^{2} - O T^{2} \Rightarrow P T^{2} = 100 - 36 \Rightarrow P T^{2} = 64 \Rightarrow P T = \sqrt{64} \Rightarrow P T = 8 cm$

Page No 531:

Question 2:

In a circle of radius 7 cm, tangent PT is drawn from a point P, such that PT = 24 cm. If O is the centre of the circle, then OP = ?

(a) 30 cm
(b) 28 cm
(c) 25 cm
(d) 18 cm

Answer:

(c) 25 cm
The tangent at any point of a circle is perpendicular to the radius at the point of contact.
$∴ O T ⊥ P T From right - angled triangle P T O, ∴ O P^{2} = O T^{2} + P T^{2} [U s i n g P y t h a g o r a s' t h e o r e m] \Rightarrow O P^{2} = {(7)}^{2} + {(24)}^{2} \Rightarrow O P^{2} = 49 + 576 \Rightarrow O P^{2} = 625 \Rightarrow O P = \sqrt{625} \Rightarrow O P = 25 cm$

Page No 532:

Question 3:

A point P is 26 cm away from the centre of a circle and the length of the tangent drawn from P to the circle is 24 cm. The radius of the circle is

(a) 8 cm
(b) 10 cm
(c) 12 cm
(d) 14 cm

Answer:

(b) 10 cm
Given, point P is 26 cm away from the centre of a circle and the length of the tangent drawn from P to the circle is 24 cm. The centre of the circle is O.
The tangent at any point of a circle is perpendicular to the radius at the point of contact.
$∴ O T ⊥ P T ∴ O T^{2} = O P^{2} - P T^{2} \Rightarrow O T^{2} = {(26)}^{2} - {(24)}^{2} \Rightarrow O T^{2} = 676 - 576 \Rightarrow O T^{2} = 100 \Rightarrow O T = \sqrt{100} \Rightarrow O T = 10 cm ∴ Radius of the circle is 10 cm .$

Page No 532:

Question 4:

If two tangents inclined at an angle of 60° are drawn to a circle of radius 3 cm, then the length of each tangent is

(a) 3 cm
(b) 6 cm
(c) $\frac{3 \sqrt{3}}{2} cm$
(d) $3 \sqrt{3} cm$

Answer:

(d) $3 \sqrt{3} cm$

$Given, P A and P B are tangents to circle with cent re O and radius 3 cm and ∠ A P B = 60^{0} . T angents drawn from an external point are equal; so, P A = P B . A nd O P is the bisector of ∠ A P B, which gives ∠ O P B = ∠ O P A = 30^{0} . O A ⊥ P A . S o, from right - angled ∆ O P A, we have : \frac{O A}{A P} = \tan 30^{0} \Rightarrow \frac{O A}{A P} = \frac{1}{\sqrt{3}} \Rightarrow \frac{3}{A P} = \frac{1}{\sqrt{3}} \Rightarrow A P = 3 \sqrt{3} cm$

Page No 532:

Question 5:

If PA and PB are two tangents to a circle with centre O, such that ∠AOB = 110°, find ∠APB.

(a) 90°
(b) 60°
(c) 70°
(d) 55°

Answer:

(c) 70°
$Given, P A and P B are tangents to a circle with centre O, with ∠ A O B = 110^{0} . Now, we know that tangents drawn from an external point are perpendicular to the radius at the point of contact . So, ∠ O A P = 90^{0} a n d ∠ O B P = 90^{0} . \Rightarrow ∠ O A P + ∠ O B P = 90^{0} + 90^{0} = 180^{0}, which shows that O A B P is a cyclic quadrilateral . ∴ ∠ A O B + ∠ A P B = 180^{0} \Rightarrow 110^{0} + ∠ A P B = 180^{0} \Rightarrow ∠ A P B = 180^{0} - 110^{0} \Rightarrow ∠ A P B = 70^{0}$

Page No 532:

Question 6:

If PA and PB are two tangents to a circle with centre O, such that ∠APB = 80°, then ∠AOP = ?

(a) 40°
(b) 50°
(c) 60°
(d) 70°

Answer:

(b) 50°
$Given, P A and P B are two tangents to a circle with centre O and ∠ A P B = 80^{0} . ∴ ∠ A P O = \frac{1}{2} ∠ A P B = 40^{0} [S ince they are equally inclined to the line segment joining the centre to that point] and ∠ O A P = 90^{0} [S ince tangents drawn from an external point are perpendicular to the radius at the point of contact] Now, in triangle A O P : ∠ A O P + ∠ O A P + ∠ A P O = 180^{0} \Rightarrow ∠ A O P + 90^{0} + 40^{0} = 180^{0} \Rightarrow ∠ A O P = 180^{0} - 130^{0} \Rightarrow ∠ A O P = 50^{0}$

Page No 532:

Question 7:

In the given figure, PQR is a tangent to the circle at Q, whose centre is O and AB is a chord parallel to PR, such that ∠BQR = 70°. Then, ∠AQB = ?

(a) 20°
(b) 35°
(c) 40°
(d) 45°

Answer:

(c) 40°

$S i n c e, A B ∥ P R, B Q is transversal . ∠ B Q R = ∠ A B Q = 70^{0} [A lternate angles] O Q ⊥ P Q R (T angents drawn from an external point are perpendicular to the radius at the point of contact) a n d A B ∥ P Q R ∴ Q L ⊥ A B; so, O L ⊥ A B ∴ O L bisects chord A B [P erpendicular drawn from the centre bisects the chord] From ∆ Q L A a n d ∆ Q L B : ∠ Q L A = ∠ Q L B = 90^{0} L A = L B (O L bisects chord A B) Q L is the common side . ∴ ∆ Q L A ≅ ∆ Q L B [By SAS congruency] ∴ ∠ Q A L = ∠ Q B L \Rightarrow ∠ Q A B = ∠ Q B A ∴ ∆ A Q B is isosceles . ∴ ∠ L Q A = ∠ L Q R ∠ L Q P = ∠ L Q R = 90^{0} ∠ L Q B = (90^{0} - 70^{0}) = 20^{0} ∴ ∠ L Q A = ∠ L Q B = 20^{0} \Rightarrow ∠ A Q B = ∠ L Q A + ∠ L Q B = 40^{0}$

Page No 532:

Question 8:

In the given figure, O is the centre of a circle and PT is the tangent to the circle. If PQ is a chord, such that ∠QPT = 50°, then ∠PQT = ?

(a) 100°
(b) 90°
(c) 80°
(d) 75°

Answer:

$(a) 100 ° Given, ∠ Q P T = 50^{0} and ∠ O P T = 90^{0} (T angents drawn from an external point are perpendicular to the radius at the point of contact) ∴ ∠ O P Q = (∠ O P T - ∠ Q P T) = (90^{0} - 50^{0}) = 40^{0} O P = O Q (R adius of the same circle) \Rightarrow ∠ O Q P = ∠ O P Q = 40^{0} In ∆ P O Q, ∠ P O Q + ∠ O Q P + ∠ O P Q = 180^{0} ∴ ∠ P O Q = 180^{0} - (40^{0} + 40^{0}) = 100^{0}$

Page No 532:

Question 9:

In the given figure, PQ is a chord of a circle with centre O and PT is a tangent at P, such that ∠QPT = 60°. Then, ∠PRQ = ?

(a) 135°
(b) 150°
(c) 120°
(d) 110°

Answer:

$(c) 120 ° A ngle between a chord and a tangent is equal to the angle formed in the corresponding alternate segment . ∴ ∠ Q P A = ∠ P R Q ∠ Q P A + ∠ Q P T = 180^{0} (linear pair) B u t ∠ Q P A = (180^{0} - 60^{0}) = 120^{0} ∴ ∠ P R Q = 120^{0}$

Page No 533:

Question 10:

If the angle between two radii of a circle is 130°, the angle between the tangents at the ends of the radii is

(a) 65°
(b) 50°
(c) 40°
(d) 70°

Answer:

(b) 50°

$O A a n d O B are the two radii of a circle with centre O . Also A T and B T are the tangents to the circle . Then, ∠ T A O = ∠ T B O = 90^{0} \Rightarrow ∠ T A O + ∠ T B O = 180^{0} ∴ ∠ A O B + ∠ A T B = 180^{0} \Rightarrow 130^{0} + ∠ A T B = 180^{0} \Rightarrow ∠ A T B = (180^{0} - 130^{0}) \Rightarrow ∠ A T B = 50^{0}$

Page No 533:

Question 11:

PA and PB are tangents to the circle with centre O, such that ∠APB = 50°. Then, ∠OAB = ?

(a) 25°
(b) 30°
(c) 40°
(d) 50°

Answer:

$(a) 25 ° I n q u a drilateral O A P B, w e h a v e : ∠ O A P = ∠ O B P = 90^{0} (T angents drawn from an external point are perpendicular to the radius at the point of contact) \Rightarrow ∠ O A P + ∠ O B P = 180^{0} ∴ ∠ A O B + ∠ A P B = 180^{0} (S um of the angles of a quadrilateral is 360^{0}) \Rightarrow ∠ A O B + 50^{0} = 180^{0} \Rightarrow ∠ A O B = (180^{0} - 50^{0}) \Rightarrow ∠ A O B = 130^{0} In ∆ A O B, ∠ O A B + ∠ O B A = 180^{0} - ∠ A O B = (180^{0} - 130^{0}) = 50^{0} N o w O A = O B (R adi i of the same circle) \Rightarrow ∠ O A B = ∠ O B A = 25^{0} H e n c e, ∠ O A B = 25^{0}$

Page No 533:

Question 12:

O is the centre of a circle of radius 5 cm. At a distance of 13 cm from O, a point P is taken. From this point, two tangents PQ and PR are drawn to the circle. Then, the area of quadrilateral PQOR is

(a) 60 cm²
(b) 32.5 cm²
(c) 65 cm²
(d) 30 cm²

Answer:

(a) 60 cm²

$G i v e n, O Q = O R = 5 cm, O P = 13 cm . ∠ O Q P = ∠ O R P = 90^{0} (T angents drawn from an external point are perpendicular to the radius at the point of contact) From right - angled ∆ P O Q : P Q^{2} = (O P^{2} - O Q^{2}) \Rightarrow P Q^{2} = 13^{2} - 5^{2} \Rightarrow P Q^{2} = 169 - 25 \Rightarrow P Q^{2} = 144 \Rightarrow P Q = \sqrt{144} \Rightarrow P Q = 12 cm ∴ a r (△ O Q P) = \frac{1}{2} \times P Q \times O Q \Rightarrow a r (△ O Q P) = (\frac{1}{2} \times 12 \times 5) {cm}^{2} \Rightarrow a r (△ O Q P) = 30 {cm}^{2} S i m i l a r l y, a r (△ O R P) = 30 {cm}^{2} ∴ a r (q u a d . P Q O R) = (30 + 30) {cm}^{2} = 60 {cm}^{2}$

Page No 533:

Question 13:

In the given figure, O is the centre of a circle. AOC is its diameter, such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT = ?

(a) 40°
(b) 50°
(c) 60°
(d) 65°

Answer:

(b) 50°
$∠ A B C = 90^{0} (A ngle in a semicircle) I n △ A B C, w e h a v e : ∠ A C B + ∠ C A B + ∠ A B C = 180^{0} \Rightarrow 50^{0} + ∠ C A B + 90^{0} = 180^{0} \Rightarrow ∠ C A B = (180^{0} - 140^{0}) \Rightarrow ∠ C A B = 40^{0} Now, ∠ C A T = 90^{0} (T angents drawn from an external point are perpendicular to the radius at the point of contact) ∴ ∠ C A B + ∠ B A T = 90^{0} \Rightarrow 40^{0} + ∠ B A T = 90^{0} \Rightarrow ∠ B A T = (90^{0} - 40^{0}) \Rightarrow ∠ B A T = 50^{0}$

Page No 533:

Question 14:

In the given figure, AT is a tangent to the circle with centre O, such that OT = 4 cm and ∠OTA = 30°. Then, AT = ?

(a) 4 cm
(b) 2 cm
(c) $2 \sqrt{3} cm$
(d) $4 \sqrt{3} cm$

Answer:

(c) $2 \sqrt{3} cm$

$O A ⊥ A T So, \frac{A T}{O T} = \cos 30^{0} \Rightarrow \frac{A T}{4} = \frac{\sqrt{3}}{2} \Rightarrow A T = (\frac{\sqrt{3}}{2} \times 4) \Rightarrow A T = 2 \sqrt{3}$

Page No 533:

Question 15:

In the given figure, O is the centre of a circle. BOA is its diameter and the tangent at the point P meets BA extended at T. If ∠PBO = 30°, then ∠PTA = ?

(a) 60°
(b) 30°
(c) 15°
(d) 45°

Answer:

(b) 30°
$∠ B P A = 90^{0} [∵ angle in a semicircle] In △ P B A, ∠ B P A + ∠ P B A + ∠ B A P = 180^{0} \Rightarrow 90^{0} + 30^{0} + ∠ B A P = 180^{0} \Rightarrow ∠ B A P = (180^{0} - 120^{0}) \Rightarrow ∠ B A P = 60^{0} B A T is a straight angle . ∴ ∠ B A P + ∠ P A T = 180^{0} \Rightarrow 60^{0} + ∠ P A T = 180^{0} \Rightarrow ∠ P A T = (180^{0} - 60^{0}) \Rightarrow ∠ P A T = 120^{0} O A = O P (R adius of the same circle) \Rightarrow ∠ O A P = ∠ O P A = 60^{0} (S ince ∠ B A P = 60^{0}) N o w, ∠ O P T = 90^{0} (T angents drawn from an external point are perpendicular to the radius at the point of contact) \Rightarrow ∠ O P A + ∠ A P T = 90^{0} \Rightarrow 60^{0} + ∠ A P T = 90^{0} \Rightarrow ∠ A P T = (90^{0} - 60^{0}) \Rightarrow ∠ A P T = 30^{0} I n △ P A T, we have : ∠ P A T + ∠ A P T + ∠ P T A = 180^{0} \Rightarrow 120^{0} + 30^{0} + ∠ P T A = 180^{0} \Rightarrow ∠ P T A = (180^{0} - 150^{0}) \Rightarrow ∠ P T A = 30^{0}$

Page No 534:

Question 16:

In the given figure, O is the centre of a circle; PQL and PRM are the tangents at the points Q and R respectively, and S is a point on the circle, such that ∠SQL = 50° and ∠SRM = 60°. Find ∠QSR.

(a) 40°
(b) 50°
(c) 60°
(d) 70°

Answer:

$(d) 70 ° P Q L i s a tangent O Q is the radius; s o, ∠ O Q L = 90^{0} . ∴ ∠ O Q S = (90^{0} - 50^{0}) = 40^{0} N o w, O Q = O S (Ra d i u s o f t h e s a m e c i r c l e) \Rightarrow ∠ O S Q = ∠ O Q S = 40^{0} S i m i l a r l y, ∠ O R S = (90^{0} - 60^{0}) = 30^{0}, A n d, O R = O S (R a d i u s o f t h e s a m e c i r c l e) \Rightarrow ∠ O S R = ∠ O R S = 30^{0} ∴ ∠ Q S R = ∠ O S Q + ∠ O S R \Rightarrow ∠ Q S R = (40^{0} + 30^{0}) \Rightarrow ∠ Q S R = 70^{0}$

Page No 534:

Question 17:

In the given figure, O is the centre of two concentric circles of radii 3 cm and 5 cm. AB is a chord of the outer circle, which touches the inner circle. The length of AB is

(a) 4 cm
(b) 7 cm
(c) 8 cm
(d) $\sqrt{34} cm$

Answer:

(c) 8 cm

$Given, O P = 3 cm and O A = 5 cm N o w, ∠ O P A = ∠ O P B = 90^{0} (Ta n g e n t s d r a w n f r o m a n e x t e r n a l p o i n t a r e p e r p e n d i c u l a r t o t h e r a d i u s a t t h e p o i n t o f c o n t a c t) I n ∆ O P A, ∴ O A^{2} = {(O P}^{2} + A P^{2}) \Rightarrow O A^{2} - O P^{2} = A P^{2} \Rightarrow 5^{2} - 3^{2} = A P^{2} \Rightarrow A P^{2} = 16 \Rightarrow A P = \sqrt{16} \Rightarrow A P = 4 c m ∴ A B = (2 \times A P) c m = (2 \times 4) c m = 8 c m (Pe r p e n d i c u l a r d r a w n f r o m t h e c e n t r e t o a c h o r d b i s e c t s t h e c h o r d)$

Page No 534:

Question 18:

In the given figure, ΔABC is circumscribed, touching the circle at P, Q and R. AP = 4 cm, BP = 6 cm, AC = 12 cm and BC = x cm. Find x.

(a) 10 cm
(b) 14 cm
(c) 18 cm
(d) 12 cm

Answer:

(b) 14 cm
$T angents drawn from an external point to a circle are equal . So, AR = AP = 4 cm ∴ CR = (AC - AR) \Rightarrow CR = (12 - 4) cm \Rightarrow CR = 8 cm BQ = BP = 6 cm ∴ B C = (B Q + C Q) \Rightarrow B C = (6 + 8) c m [∵ C Q = C R = 8 cm] \Rightarrow B C = 14 c m$

Page No 534:

Question 19:

In the given figure, quadrilateral ABCD is circumscribed, touching the circle at P, Q, R and S. If AP = 5 cm, BC = 7 cm and CS = 3 cm, AB = ?

(a) 9 cm
(b) 10 cm
(c) 12 cm
(d) 8 cm

Answer:

(a) 9 cm
$T angents drawn from an external point to a circle are equal . So, A Q = A P = 5 c m C R = C S = 3 c m a n d B R = (B C - C R) \Rightarrow B R = (7 - 3) c m \Rightarrow B R = 4 c m B Q = B R = 4 c m ∴ A B = (A Q + B Q) \Rightarrow A B = (5 + 4) c m \Rightarrow A B = 9 c m$

Page No 534:

Question 20:

In the given figure, quadrilateral ABCD is circumscribed, touching the circle at P, Q, R and S. If AP = 6 cm, BP = 5 cm, CQ = 3 cm and DR = 4 cm, then the perimeter of quadrilateral ABCD is

(a) 18 cm
(b) 27 cm
(c) 36 cm
(d) 32 cm

Answer:

$(c) 36 cm Given, A P = 6 cm, B P = 5 cm, C Q = 3 cm and D R = 4 cm . T angents drawn from an external points to a circle are equal . So, A P = A S = 6 c m, B P = B Q = 5 c m, C Q = C R = 3 c m, D R = D S = 4 c m . ∴ A B = A P + P B = 6 + 5 = 11 c m B C = B Q + C Q = 5 + 3 = 8 c m C D = C R + D R = 3 + 4 = 7 c m A D = A S + D S = 6 + 4 = 10 c m ∴ Pe r i m e t e r o f q u a drilateral A BCD = AB + BC + CD + DA = (11 + 8 + 7 + 10) c m = 36 c m$

Page No 535:

Question 21:

In the given figure, quad. ABCD is circumscribed, touching the circle at P, Q, R and S such that ∠DAB = 90°, If CS = 27 cm and CB = 38 cm and radius of the circle is 10 cm, then AB = ?

(a) 28 cm
(b) 21 cm
(c) 19 cm
(d) 17 cm

Answer:

$(b) 21 cm Given, C S = 27 cm, C B = 38 cm and ∠ D A B = 90^{0} . T he length s of tangents drawn from an external point to a circle are equal . So, C R = C S = 27 c m ∴ B R = (B C - C R) \Rightarrow B R = (38 - 27) c m \Rightarrow B R = 11 c m ∴ B Q = B R = 11 c m J o i n O Q . ∠ A = ∠ Q = ∠ O = ∠ P = 90 ° and O P = O Q (radius) . So, P A Q O i s a square . ∴ A Q = P Q = 10 c m H e n c e, A B = (A Q + B Q) \Rightarrow A B = (10 + 11) c m \Rightarrow A B = 21 c m$

Page No 535:

Question 22:

In the given figure, ΔABC is right-angled at B, such that BC = 6 cm and AB = 8 cm. A circle with centre O has been inscribed in the triangle. OP ⊥ AB, OQ ⊥ BC and OR ⊥ AC.
If OP = OQ = OR = x cm, then x = ?

(a) 2 cm
(b) 2.5 cm
(c) 3 cm
(d) 3.5 cm

Answer:

(a) 2 cm
$Given, A B = 8 cm, B C = 6 cm Now, in ∆ A B C : A C^{2} = A B^{2} + B C^{2} \Rightarrow A C^{2} = (8^{2} + 6^{2}) \Rightarrow A C^{2} = (64 + 36) \Rightarrow A C^{2} = 100 \Rightarrow A C = \sqrt{100} \Rightarrow A C = 10 cm P B Q O is a square . C R = C Q (Since the length s of tangents drawn from an external point are equal) ∴ C Q = (B C - B Q) = (6 - x) cm Similarly, A R = A P = (A B - B P) = (8 - x) cm ∴ A C = (A R + C R) = [(8 - x) + (6 - x)] cm \Rightarrow 10 = (14 - 2 x) cm \Rightarrow 2 x = 4 \Rightarrow x = 2 cm ∴ The radius of the circle is 2 cm .$

Page No 535:

Question 23:

In the given figure, three circles with centres A, B, C, respectively, touch each other externally. If AB = 5 cm, BC = 7 cm and CA = 6 cm, the radius of the circle with centre A is

(a) 1.5 cm
(b) 2 cm
(c) 2.5 cm
(d) 3 cm

Answer:

(b) 2 cm

$Given, A B = 5 cm, B C = 7 cm a n d C A = 6 cm . Let, A R = A P = x cm . B Q = B P = y cm C R = C Q = z cm (Since the length of tangents drawn from an external point are equal) Then, A B = 5 cm \Rightarrow A P + P B = 5 cm \Rightarrow x + y = 5 \dots . . (i) Similarly, y + z = 7 \dots \dots \dots (i i) and z + x = 6 \dots . (i i i) Adding (i), (ii) and (iii), we get: (x + y) + (y + z) + (z + x) = 18 \Rightarrow 2 (x + y + z) = 18 \Rightarrow (x + y + z) = 9 \dots \dots . (i v) N o w, (i v) - (i i) : (x + y + z) - (y + z) = (9 - 7) \Rightarrow x = 2 ∴ T he radius of the circle with centre A is 2 cm .$

Page No 535:

Question 24:

In the given figure, O is the centre of two concentric circles of radii 5 cm and 3 cm. From an external point P, tangents PA and PB are drawn to these circles. If PA = 12, then PD = ?

(a) $5 \sqrt{2} cm$
(b) $3 \sqrt{5} cm$
(c) $4 \sqrt{10} cm$
(d) $5 \sqrt{10} cm$

Answer:

(c) $4 \sqrt{10} cm$

$Given, O P = 5 cm, P A = 12 cm Now, j oin O and B . Then, O B = 3 cm . Now, ∠ O A P = 90^{0} (T angents drawn from an external point are perpendicular to the radius at the point of contact) Now, in ∆ O A P : O P^{2} = O A^{2} + P A^{2} \Rightarrow O P^{2} = 5^{2} + 12^{2} \Rightarrow O P^{2} = 25 + 144 \Rightarrow O P^{2} = 169 \Rightarrow O P = \sqrt{169} \Rightarrow O P = 13 Now, in ∆ O B P : P B^{2} = O P^{2} - O B^{2} \Rightarrow P B^{2} = 13^{2} - 3^{2} \Rightarrow P B^{2} = 169 - 9 \Rightarrow P B^{2} = 160 \Rightarrow P B = \sqrt{160} \Rightarrow P B = 4 \sqrt{10} cm$

Page No 535:

Question 25:

In the given figure, PA and PB are tangents to the given circle, such that PA = 5 cm and ∠APB = 60°. The length of chord AB is

(a) $5 \sqrt{2} cm$
(b) 5 cm
(c) $5 \sqrt{3} cm$
(d) 7.5 cm

Answer:

(b) 5 cm
The lengths of tangents drawn from a point to a circle are equal.
$So, P A = P B and therefore, ∠ P A B = ∠ P B A = x (say) . Then, in ∆ P A B : ∠ P A B + ∠ P B A + ∠ A P B = 180^{0} \Rightarrow x + x + 60^{0} = 180^{0} \Rightarrow 2 x = 180^{0} - 60^{0} \Rightarrow 2 x = 120^{0} \Rightarrow x = 60^{0} ∴ Each angle of △ P A B i s 60^{0} and therefore, it is an equilateral triangle . ∴ A B = P A = P B = 5 cm ∴ T he length of the chord A B is 5 cm .$

Page No 535:

Question 26:

Which of the following statements is not true?
(a) If a point P lies inside a circle, no tangent can be drawn to the circle passing through P.
(b) If a point P lies on a circle, then one and only one tangent can be drawn to the circle at P.
(c) If a point P lies outside a circle, then only two tangents can be drawn to the circle from P.
(d) A circle can have more than two parallel tangents parallel to a given line.

Answer:

(d) A circle can have more than two parallel tangents, parallel to a given line.
This statement is false because there can only be two parallel tangents to the given line in a circle.

Page No 536:

Question 27:

Which of the following statements is not true?
(a) A tangent to a circle intersects the circle exactly at one point.
(b) The point common to a circle and its tangent is called the point of contact.
(c) The tangent at any point of a circle is perpendicular to the radius of the circle through the point of contact.
(d) A straight line can meet a circle at one point only.

Answer:

(d)A straight line can meet a circle at one point only.
This statement is not true because a straight line that is not a tangent but a secant cuts the circle at two points.

Page No 536:

Question 28:

Which of the following statements is not true?
(a) A line which intersects a circle at two points, is called a secant of the circle.
(b) A line intersecting a circle at one point only is called a tangent to the circle.
(c) The point at which a line touches the circle is called the point of contact.
(d) A tangent to the circle can be drawn from a point inside the circle.

Answer:

(d) A tangent to the circle can be drawn from a point inside the circle.
This statement is false because tangents are the lines drawn from an external point to the circle that touch the circle at one point.

Page No 536:

Question 29:

Assertion (A)
At point P of a circle with centre O and radius 12 cm, a tangent PQ of length 16 cm is drawn. Then, OQ = 20 cm.

Reason (R)
The tangent at any point of a circle is perpendicular to the radius through the point of contact.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R)is true.

Answer:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

$In ∆ O P Q, ∠ O P Q = 90^{0} . ∴ O Q^{2} = O P^{2} + P Q^{2} \Rightarrow O Q = \sqrt{O P^{2} + P Q^{2}} = \sqrt{12^{2} + 16^{2}} = \sqrt{144 + 256} = \sqrt{400} = 20 cm$

Page No 537:

Question 30:

Assertion (A)
If two tangents are drawn to a circle from an external point, they subtend equal angles at the centre.

Reason (R)
A parallelogram circumscribing a circle is a rhombus.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R)is true.

Answer:

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).

Assertion :-
We know that if two tangents are drawn to a circle from an external point, they subtend equal angles at the centre.

Reason:-

Given, a parallelogram ABCD circumscribes a circle with centre O.
$A B = B C = C D = A D$
We know that the tangents drawn from an external point to circle are equal .
$∴ A P = A S \dots \dots . (i) [tangents from A] B P = B Q \dots \dots \dots . (i i) [tangents from B] C R = C Q \dots \dots \dots . (i i i) [tangents from C] D R = D S \dots \dots \dots . . (i v) [tangents from D] ∴ A B + C D = A P + B P + C R + D R = A S + B Q + C Q + D S [from (i), (i i), (i i i) and (i v)] = (A S + D S) + (B Q + C Q) = A D + B C Thus, (A B + C D) = (A D + B C) \Rightarrow 2 A B = 2 A D [∵ opposite sides of a parallelogram are equal] \Rightarrow A B = A D ∴ C D = A B = A D = B C$
Hence, ABCD is a rhombus.

Page No 537:

Question 31:

Assertion (A)
In the given figyre, a quadrilateral ABCD is drawn to circumscribe a given circle, as shown.
Then, AB + BC = AD + DC.
Figure

Reason (R)
In two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R)is true.

Answer:

(d) Assertion (A) is false and Reason (R)is true.
Assertion (A) is false $[Since the correct result is A B + C D = A D + B C]$
$Since the length of tangents drawn from an external point are equal, therefore : A P = A S B P = B Q C Q = C R R D = S D ∴ A B + C D = (A P + P B) + (C R + D R) = A S + B Q + C Q + S D = (A S + S D) + (B Q + C Q) = A D + B C$

Page No 541:

Question 1:

In the given figure, O is the centre of a circle, PQ is a chord and the tangent PT at P makes an angle of 50° with PQ. Then, ∠POQ = ?

(a) 130°
(b) 100°
(c) 90°
(d) 75°

Answer:

$(b) 100 ° Given, ∠QPT = 50^{0} Now, ∠OPT = 90^{0} (S ince tangents drawn from an external point a re perpendicular to t he radius at point of contact) ∴ ∠OPQ = (∠OPT - ∠QPT) = (90^{0} - 50^{0}) = 40^{0} OP = OQ (Radii of the same c ircle) \Rightarrow ∠OPQ = ∠OQP = 40^{0} In △ POQ, ∠ POQ + ∠ OPQ + ∠OQP = 180^{0} \Rightarrow ∠ POQ + 40^{0} + 40^{0} = 180^{0} \Rightarrow ∠ POQ = 180^{0} - (40^{0} + 40^{0}) \Rightarrow ∠POQ = 180^{0} - 80^{0} \Rightarrow ∠POQ = 100^{0}$

Page No 541:

Question 2:

If the angle between two radii of a circle is 130°, then the angle between the angles at the ends of the radii is
Figure

(a) 65°
(b) 40°
(c) 50°
(d) 90°

Answer:

$(c) 50 ° OA and OB a re the two r adii of a circle with centre O. Also, AP and BP are the tangents to the c ircle . Given, ∠ AOB = 130^{0} Now, ∠ OAB = ∠ OBA = 90^{0} (S ince tangents drawn from an external point a re perpendicular to the radius at p oint of c ontact) In quadrilateral OAPB, ∠ AOB + ∠ OAB + ∠ OBA + ∠ APB = 360^{0} \Rightarrow 130^{0} + 90^{0} + 90^{0} + ∠ APB = 360^{0} \Rightarrow ∠ APB = 360^{0} - (130^{0} + 90^{0} + 90^{0}) \Rightarrow ∠ APB = 360^{0} - 310^{0} \Rightarrow ∠ APB = 50^{0}$

Page No 541:

Question 3:

If tangents PA and PB from a point P to a circle with centre O are drawn, so that ∠APB = 80°, then ∠POA = ?

(a) 40°
(b) 50°
(c) 80°
(d) 60°

Answer:

(b) 50°

$From ∆ O P A and ∆ O P B, O A = O B (R adi i of the same circle) O P (C ommon side) P A = P B (S ince tangents drawn from an external point to a circle are equal) ∴ ∆ O P A ≅ ∆ O P B (SSS rule) ∴ ∠ A P O = ∠ B P O ∴ ∠ A P O = \frac{1}{2} ∠ A P B = 40^{0} A nd ∠ O A P = 90^{0} (S ince tangents drawn from an external point are perpendicular to the radius at point of contact) Now, in ∆ O A P, ∠ A O P + ∠ O A P + ∠ A P O = 180^{0} \Rightarrow ∠ A O P + 90^{0} + 40^{0} = 180^{0} \Rightarrow ∠ A O P = 180^{0} - 130^{0} = 50^{0}$

Page No 541:

Question 4:

In the given figure, AD and AE are the tangents to a circle with centre O and BC touches the circle at F. If AE = 5 cm, then perimeter of ∆ABC is

(a) 15 cm
(b) 10 cm
(c) 22.5 cm
(d) 20 cm

Answer:

(b) 10 cm
Since the tangents from an external point are equal, we have:
$A D = A E, C D = C F, B E = B F Perimeter of ∆ A B C = A C + A B + C B = (A D - C D) + (C F + B F) + (A E - B E) = (A D - C F) + (C F + B F) + (A E - B F) = A D + A E = 2 A E = 2 \times 5 = 10 cm$

Page No 541:

Question 5:

Find the length of the tangent drawn to a circle with radius 7 cm from a point 25 cm away from the centre of the circle.

Answer:

$Since, triangle O P T is a right - angled triangle, by P ythagoras' T heorem : O P^{2} = O T^{2} + T P^{2} \Rightarrow T P^{2} = O P^{2} - O T^{2} \Rightarrow T P = \sqrt{O P^{2} - O T^{2}} \Rightarrow T P = \sqrt{25^{2} - 7^{2}} \Rightarrow T P = \sqrt{625 - 49} \Rightarrow T P = \sqrt{576} \Rightarrow T P = 24 cm$
The length of the tangent to the circle is 24 cm.

Page No 542:

Question 6:

In the given figure, PA and PB are the tangents to a circle with centre O. Show that the points A, O, B, P are concyclic.

Answer:

$Here, O A = O B A nd O A ⊥ A P, O A ⊥ B P (S ince tangents drawn from an external point are perpendicular to the radius at the point of contact) ∴ ∠ O A P = 90^{0}, ∠ O B P = 90^{0} ∴ ∠ O A P + ∠ O B P = 90^{0} + 90^{0} = 180^{0} ∴ ∠ A O B + ∠ A P B = 180^{0} (S ince, ∠ O A P + ∠ O B P + ∠ A O B + ∠ A P B = 360^{0}) S um of opposite angle of a quadrilateral is 180 ° . Hence, A, O, B and P are concyclic .$

Page No 542:

Question 7:

In the given figure, PA and PB are tangents, such that PA = 9 cm and ∠APB = 60°. Find the length of chord AB.

Answer:

The lengths of tangents drawn from a point to a circle are equal.
So, PA= PB.
$S u p p o s e, ∠ P A B = ∠ P B A = x Then, in ∆ P A B : ∠ P A B + ∠ P B A + ∠ A P B = 180^{0} \Rightarrow x + x + 60^{0} = 180^{0} \Rightarrow 2 x = 120^{0} \Rightarrow x = \frac{120^{0}}{2} = 60^{0} ∴ The angles of ∆ P A B are 60^{0} . ∴ It is an equilateral triangle . ∴ A B = P A = P B = 9 cm$
The length of chord AB is 9 cm.

Page No 542:

Question 8:

Two tangents BC and BD are drawn to a circle with centre O, such that ∠CBD = 120°. Prove that OB = 2BC.

Answer:

$Here, O B is the bisector of ∠ C B D . (T wo tangents are equally inclined to the line segment joining the centre to that point) ∴ ∠ C B O = ∠ D B O = \frac{1}{2} ∠ C B D = 60^{0} ∴ From ∆ B O D, ∠ B O D = 30^{0} Now, from right - angled ∆ B O D, \frac{B D}{O B} = \sin 30^{0} \Rightarrow \frac{B D}{O B} = \frac{1}{2} \Rightarrow O B = 2 B D \Rightarrow O B = 2 B C (S ince tangents from an external point are equal, i . e . B C = B D) ∴ O B = 2 B C$

Page No 542:

Question 9:

Fill in the blanks.
(i) A line intersecting a circle at two distinct points is called a ....... .
(ii) A circle can have ....... parallel tangents at the most.
(iii) The common point of a tangent to a circle and the circle is called the ....... .
(iv) A circle can have ...... tangents.

Answer:

(i) A line intersecting a circle at two distinct points is called a secant.
(ii) A circle can have two parallel tangents at the most.
(iii) The common point of a tangent to a circle and the circle is called the point of contact.
(iv) A circle can have infinite tangents.

Page No 542:

Question 10:

Prove that the length of two tangents drawn from an external point to a circle are equal.

Answer:

Given two tangents AP and AQ are drawn from a point A to a circle with centre O.
$To prove : A P = A Q Join O P, O Q and O A . A P is tangent at P and O P is the radius . ∴ O P ⊥ A P (since tangents drawn from an external point are perpendicular to the radius at the point of contact) Similarly, O Q ⊥ A Q In the right ∆ O P A and ∆ O Q A, we have : O P = O Q [radii of the same circle] ∠ O P A = ∠ O Q A (= 90^{0}) O A = O A [common side] ∴ ∆ O P A ≅ ∆ O Q A [By R . H . S - Congruence] Hence, A P = A Q$

Page No 542:

Question 11:

Prove that the tangents drawn at the ends of the diameter of a circle are parallel.

Answer:

$Here, P T and Q S are the tangents to the circle with centre O and A B is the diameter .$
Now, radius of a circle is perpendicular to the tangent at the point of contact.
$∴ O A ⊥ A T a n d O B ⊥ B S (s i n c e t a n g e n t s d r a w n f r o m a n e x t e r n a l p o i n t a r e p e r p e n d i c u l a r t o t h e r a d i u s a t p o i n t o f c o n t a c t) ∴ ∠ O A T = ∠ O B Q = 90^{0} But ∠ O A T and ∠ O B Q are alternate angles . ∴ A T is parallel to B S .$

Page No 542:

Question 12:

In the given figure, if AB = AC, prove that BE = CE.
Figure

Answer:

$Given, AB = AC$
We know that the tangents from an external point are equal.
$∴ AD = AF, BD = BE and CF = C E \dots \dots .. (i) Now, AB = AC \Rightarrow AD + DB = AF + FC \Rightarrow AF + DB = AF + FC [from (i)] \Rightarrow DB = FC \Rightarrow BE = C E [from (i)] Hence proved .$

Page No 542:

Question 13:

If two tangents are drawn to a circle from an external point,show that they subtend equal angles at the centre.

Answer:

Given : A circle with centre O and a point A outside it. Also, AP and AQ are the two tangents to the circle.
: $To prove : ∠ A O P = ∠ A O Q . Proof : In ∆ A O P and ∆ A O Q, w e have : A P = A Q [tangents from a n external point are equal] O P = O Q [radii of the same circle] O A = O A [common side] ∴ ∆ A O P ≅ ∆ A O Q [by SSS - congruence] Hence, ∠ A O P = ∠ A O Q (c . p . c . t) .$

Page No 542:

Question 14:

Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.

Answer:

$Let R A and R B b e two tangents to the circle with centre O and let A B be a chord of the circle . We have to prove that ∠ R A B = ∠ R B A . ∴ Now, R A = R B (S ince tangents drawn from an external point to a circle are equal) ∴ In ∆ R A B, ∠ R A B = ∠ R B A (S ince opposite sides are equal, their base angles are also equal)$

Page No 543:

Question 15:

Prove that the parallelogram circumscribing a circle is a rhombus.

Answer:

Given, a parallelogram ABCD circumscribes a circle with centre O.
$A B = B C = C D = A D$
We know that the lengths of tangents drawn from an exterior point to a circle are equal.
$∴ A P = A S \dots \dots . (i) [tangents from A] B P = B Q \dots \dots \dots . (i i) [tangents from B] C R = C Q \dots \dots \dots . (i i i) [tangents from C] D R = D S \dots \dots \dots . . (i v) [tangents from D] ∴ A B + C D = A P + B P + C R + D R = A S + B Q + C Q + D S [from (i), (i i), (i i i) and (i v)] = (A S + D S) + (B Q + C Q) = A D + B C Thus, (A B + C D) = (A D + B C) \Rightarrow 2 A B = 2 A D [∵ opposite sides of a parallelogram are equal] \Rightarrow A B = A D ∴ C D = A B = A D = B C$
Hence, ABCD is a rhombus.

Page No 543:

Question 16:

Two concentric circles are of radii 5 cm and 3 cm, respectively. Find the length of the chord of the larger circle that touches the smaller circle.

Answer:

Given: Two circles have the same centre O and AB is a chord of the larger circle touching the
smaller circle at C; also, OA=5 cm and OC=3 cm.
$In ∆ O A C, O A^{2} = O C^{2} + A C^{2} ∴ A C^{2} = O A^{2} - O C^{2} \Rightarrow A C^{2} = 5^{2} - 3^{2} \Rightarrow A C^{2} = 25 - 9 \Rightarrow A C^{2} = 16 \Rightarrow A C = 4 cm ∴ A B = 2 A C (since perpendicular drawn from the centre of the circle bisects the chord) ∴ A B = 2 \times 4 = 8 cm$
The length of the chord of the larger circle is 8 cm.

Page No 543:

Question 17:

A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.

Answer:

We know that the tangents drawn from an external point to a circle are equal.

$∴ A P = A S \dots \dots (i) [tangents from A] B P = B Q \dots \dots \dots . (i i) [tangents from B] C R = C Q \dots \dots \dots \dots . (i i i) [tangents from C] D R = D S \dots \dots \dots . . (i v) [tangents from D] ∴ A B + C D = (A P + B P) + (C R + D R) = (A S + B Q) + (C Q + D S) [using (i), (i i), (i i i) and (i v)] = (A S + D S) + (B Q + C Q) = (A D + B C) Hence, (A B + C D) = (A D + B C)$

Page No 543:

Question 18:

Prove that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Answer:

Given, a quadrilateral ABCD circumscribes a circle with centre O.
$To prove : ∠ A O B + ∠ C O D = 180^{0} and ∠ A O D + ∠ B O C = 180^{0} Join O P, O Q, O R and O S . We know that the tangents drawn from an external point of a circle subtend equal angles at the centre . ∴ ∠ 1 = ∠ 7, ∠ 2 = ∠ 3, ∠ 4 = ∠ 5 and ∠ 6 = ∠ 8 and ∠ 1 + ∠ 2 + ∠ 3 + ∠ 4 + ∠ 5 + ∠ 6 + ∠ 7 + ∠ 8 = 360^{0} [angles at a point] \Rightarrow (∠ 1 + ∠ 7) + (∠ 3 + ∠ 2) + (∠ 4 + ∠ 5) + (∠ 6 + ∠ 8) = 360^{0} 2 ∠ 1 + 2 ∠ 2 + 2 ∠ 6 + 2 ∠ 5 = 360^{0} \Rightarrow ∠ 1 + ∠ 2 + ∠ 5 + ∠ 6 = 180^{0} \Rightarrow ∠ A O B + ∠ C O D = 180^{0} and ∠ A O D + ∠ B O C = 180^{0}$

Page No 543:

Question 19:

Prove that the angles between the two tangents drawn form an external point to a circle is supplementary to the angle subtended by the line segments joining the points of contact at the centre.

Answer:

Given, PA and PB are the tangents drawn from a point P to a circle with centre O. Also, the line segments OA and OB are drawn.
$T o p r o v e : ∠ A P B + ∠ A O B = 180^{0}$
We know that the tangent to a circle is perpendicular to the radius through the point of contact.
$∴ P A ⊥ O A \Rightarrow ∠ O A P = 90^{0} P B ⊥ O B \Rightarrow ∠ O B P = 90^{0} ∴ ∠ O A P + ∠ O B P = (90^{0} + 90^{0}) = 180^{0} \dots \dots . (i)$

$But we know that the sum of all the angles of a quadrilateral is 360^{0} . ∴ ∠ O A P + ∠ O B P + ∠ A P B + ∠ A O B = 360^{0} \dots \dots \dots . (i i)$
From (i) and (ii), we get:
$∠ A P B + ∠ A O B = 180^{0}$

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Question 20:

A circle touches the side BC of ∆ABC at P and touches AB and AC produced at Q and R, respectively, as shown in the given figure. Show that $A Q = \frac{1}{2} \times (perimeter of ∆ A B C) .$
Figure

Answer:

We know that the lengths of tangents drawn from an external point to a circle are equal.
$∴ A Q = A R \dots \dots (i) [Tangents from A] B P = B Q \dots \dots . . (ii) [Tangents from B] C P = C R \dots \dots \dots . . (iii) [T angents from C] Perimeter of ∆ A B C = A B + B C + A C = A B + B P + C P + A C = A B + B Q + C R + A C [Using (i i) and (i i i)] = A Q + A R = 2 A Q [Using (i)] ∴ A Q = \frac{1}{2} (Perimeter of ∆ A B C)$

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