# NCERT Solutions Class 10 Maths Chapter 10

## NCERT Solutions for Class 10 Mathematics Chapter 10 Circles

Practicing questions from NCERT Class 10 Mathematics Chapter 10- Circles is important to prepare for your board exams. To help students with solving NCERT book questions and cross-check their answers, Extramarks provides NCERT Solutions for Class 10 Mathematics Chapter 10. These solutions are prepared by subject experts to help students solve NCERT textbook questions with accuracy and confidence. They can rely on these reference materials for quick revision and last-minute preparation also.

## NCERT Solutions for Class 10 Mathematics Chapter 10 Circles

NCERT Solutions for Class 10 Mathematics Chapter 10 Circles are prepared as per the CBSE guidelines. The subject matter experts have kept the language simple and explained the answers with relevant examples, wherever possible. Students can access these solutions for online and offline studies.

## Get Complete Details about Chapter 10 Mathematics Class 10 NCERT Solutions

Chapter 10 Circles of NCERT Class 10 Mathematics is a branch of Unit 4 Geometry. Unit 4 includes 4 multiple choice questions of 1 mark each, 2 short answer questions of 3 marks each, and 2 long type questions carrying 6 marks each. To ensure that you score 22/22 in questions from this chapter refer to NCERT solutions for Class 10 Mathematics Chapter 10. Only checking out NCERT solutions will not help. It requires practicing the concepts and tips as conveyed in the guide.

Other details about Class 10 Chapter 10 Circles

Chapter 10 Circles of NCERT Class 10 Mathematics consists of the following sub-topics:

• Introduction
• Tangent to a circle
• Number of Tangents from a point on a circle
• Summary

Furthermore, Chapter 10 Circles of NCERT Class 10 Mathematics comes with a set of extensive exercises. These include:

Exercise 10.1 – A total of 4 Questions out of which 1 short format question, 1 Fill in the blanks and 2 long format questions.

Exercise 10.2 – A total of 13 Questions out of which 10 long format questions, 4 descriptive types and 2 short format questions.

NCERT Solutions Class 10 Mathematics Chapter 10 will guide the students appropriately not only to master the concepts but also to solve the problems with utmost precision. This in turn will give students the much-needed confidence to solve questions in their final examination with the same accuracy.

## NCERT Solutions for Class 10 Mathematics

The NCERT Solutions for Class 10 Mathematics are available for the following chapters:

[Include Chapter wise Pages]

Chapter 1 – Real Numbers

Chapter 2 – Polynomials

Chapter 3 – Pair of Linear Equations in Two Variables

Chapter 4 – Quadratic Equations

Chapter 5 – Arithmetic Progressions

Chapter 6 – Triangles

Chapter 7 – Coordinate Geometry

Chapter 8 – Introduction to Trigonometry

Chapter 9 – Some Applications of Trigonometry

Chapter 10 – Circles

Chapter 11 – Constructions

Chapter 12 – Areas Related to Circles

Chapter 13 – Surface Areas and Volumes

Chapter 14 – Statistics

Chapter 15 – Probability

## Why Should Students Choose Extramarks?

Here are some of the reasons why students should pick  Extramarks’ study material:

• Extramarks’ NCERT Solutions Class 10 Mathematics Chapter 10 are prepared by subject matter experts.
• The answers are written in simple and easy-to-comprehend language.
• The solutions are prepared as per CBSE guidelines.
• The solutions are self-explanatory and do not require any external assistance for studying.

## Related Questions

Q1. Fill in the blanks:

(i) A tangent to a circle intersects it in _______ point(s).

(ii) A line intersecting a circle in two points is called a _____.

(iii) A circle can have ______ parallel tangents at the most.

(iv) The common point of a tangent to a circle and the circle is called _____.

Ans.

(i) A tangent to a circle intersects it in one point.

(ii) A line intersecting a circle in two points is called a secant.

(iii) A circle can have two parallel tangents at the most.

(iv) The common point of a tangent to a circle and the circle is called the point of contact.

Q2. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC

Ans. From this figure, we can conclude a few points which are:

(i) DR = DS

(ii) BP = BQ

(iii) AP = AS

(iv) CR = CQ

Since they are tangents on the circle from points D, B, A, and C respectively.

Now, adding the LHS and RHS of the above equations we get,

DR+BP+AP+CR = DS+BQ+AS+CQ

By rearranging them we get,

(DR+CR) + (BP+AP) = (CQ+BQ) + (DS+AS)

By simplifying,

Q3. In Fig. 10.13, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that ∠ AOB = 90°.

Ans. From the figure given in the textbook, join OC. Now the triangles △OPA and △OCA are similar using SSS congruence as:

(i) OP = OC They are the radii of the same circle

(ii) AO = AO It is the common side

(iii) AP = AC These are the tangents from point A

So, △OPA ≅ △OCA

Similarly,

△OQB ≅ △OCB

So,

∠POA = ∠COA … (Equation i)

And, ∠QOB = ∠COB … (Equation ii)

Since the line POQ is a straight line, it can be considered as the diameter of the circle.

So, ∠POA +∠COA +∠COB +∠QOB = 180°

Now, from equations (i) and equation (ii) we get,

2∠COA+2∠COB = 180°

∠COA+∠COB = 90°

∴∠AOB = 90°

Q.1 How many tangents can a circle have?

Ans.

A circle can have infinite tangents.

Q.2 Fill in the blanks:
(i) A tangent to a circle intersects it in ­______ point (s).
(ii) A line intersecting a circle in two points is called a _________.
(iii) A circle can have _______ parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called _______.

Ans.

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

Q.3

$\begin{array}{l}\text{A tangent PQ at a point P of a circle of radius 5 cm meets a line through}\\ \text{the centre O at a point Q so that OQ = 12 cm. Length PQ is}\\ \left(\text{A}\right)\text{12 cm \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\text{B}\right)\text{13 cm}\\ \left(\text{C}\right)\text{8.5 cm \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\text{D}\right)\text{}\sqrt{\text{119}}\text{ cm}\end{array}$

Ans.

$\begin{array}{l}\text{We know that the line drawn from the center of the circle}\\ \text{to the tangent is perpendicular to the tangent.}\\ \therefore \text{OP}\perp \text{PQ}\end{array}$

$\begin{array}{l}\text{By applying Pythagoras theorem in ΔOPQ, we get}\\ {\text{OP}}^{\text{2}}{\text{+PQ}}^{\text{2}}{\text{= OQ}}^{\text{2}}\\ {\text{or 5}}^{\text{2}}{\text{+PQ}}^{\text{2}}{\text{= 12}}^{\text{2}}\\ {\text{or PQ}}^{\text{2}}\text{= 144}-\text{25 = 119}\\ \text{or PQ =}\sqrt{\text{119}}\text{cm.}\\ \text{Hence, the correct option is (D).}\end{array}$

Q.4 Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.

Ans.

Here, we draw a circle centred at O and having AB as diameter. We draw two lines CD and XY parallel to diameter AB. CD is a secant and XY is a tangent to the circle at point P.

Q.5 From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(A) 7 cm (B) 12 cm
(C) 15 cm (D) 24.5 cm

Ans.

$\begin{array}{l}\text{We know that the line drawn from the center of the circle}\\ \text{to the tangent is perpendicular to the tangent.}\\ \therefore \text{OP}\perp \text{PQ}\\ \text{By applying Pythagoras theorem in Δ\hspace{0.17em}OPQ, we get}\\ {\text{OP}}^{\text{2}}{\text{+PQ}}^{\text{2}}{\text{= OQ}}^{\text{2}}\\ {\text{or OP}}^{\text{2}}{\text{+24}}^{\text{2}}{\text{= 25}}^{\text{2}}\\ {\text{or OP}}^{\text{2}}\text{= 625}-\text{576 = 49}\\ \text{or OP = 7 cm.}\\ \text{Therefore, radius of the circle is 7 cm.}\\ \text{Hence, the correct option is (A).}\end{array}$

Q.6

$\begin{array}{l}\text{In the following figure, if TP and TQ are the two tangents to a circle with centre O so that}\\ \angle \text{POQ}=\text{110°, then}\angle \text{PTQ is equal to}\\ \text{(A) 60° (B) 70°}\\ \text{(C) 80° (D) 90°}\end{array}$

Ans.

$\begin{array}{l}\text{We know that the line drawn from the center of the circle}\\ \text{to the tangent is perpendicular to the tangent}\text{.}\\ \therefore \text{OP}\perp \text{PT and OQ}\perp \text{TQ}\\ \text{In quadrilateral OPTQ,}\angle \text{P =}\angle \text{Q = 90°}\text{.}\\ \text{Now, we have}\\ \text{}\angle \text{P+}\angle \text{Q+}\angle \text{T+}\angle \text{O = 360°}\\ \text{or 90°+90°+110°+}\angle \text{T = 360°}\\ \text{or 290°+}\angle \text{T = 360°}\\ \text{or}\angle \text{T = 360°-290° = 70°}\\ \text{Therefore,}\angle \text{PTQ = 70°}\\ \text{Hence, the correct option is (B)}\text{.}\end{array}$

Q.7

$\begin{array}{l}\text{If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°,}\\ \text{then}\angle \text{POA is equal to}\\ \text{(A) 50° (B) 60°}\\ \text{(C) 70° (D) 80°}\end{array}$

Ans.

$\begin{array}{l}\text{We know that the line drawn from the center of the circle}\\ \text{to the tangent is perpendicular to the tangent.}\\ \therefore \text{OA}\perp \text{AP and OB}\perp \text{PB}\\ \text{In quadrilateral OAPB,}\angle \text{A =}\angle \text{B = 90°.}\\ \text{Now, we have}\\ \text{}\angle \text{P+}\angle \text{A+}\angle \text{B+}\angle \text{O = 360°}\\ \text{or 80°+90°+90°+}\angle \text{O = 360°}\\ \text{or 260°+}\angle \text{O = 360°}\\ \text{or}\angle \text{O = 360°}-\text{260° = 100°}\\ \text{Therefore,}\angle \text{AOB = 100°}\\ \text{Now, in Δ\hspace{0.17em}AOP and Δ\hspace{0.17em}BOP, we have}\\ \text{AO = BO [Radius]}\\ \text{AP = BP [Tangents drawn from P]}\\ \text{OP = OP}\\ \therefore \text{Δ\hspace{0.17em}AOP}\cong \text{Δ\hspace{0.17em}BOP [By SSS congruence criterion]}\\ \text{So,}\angle \text{AOP =}\angle \text{BOP \hspace{0.17em}[By CPCT]}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{AOB =}\angle \text{AOP+}\angle \text{BOP = 100°}\\ \text{or}\angle \text{AOP+}\angle \text{AOP = 100° [}\angle \text{AOP =}\angle \text{BOP]}\\ \text{or \hspace{0.17em}\hspace{0.17em}2}\angle \text{AOP = 100°}\\ \text{or \hspace{0.17em}\hspace{0.17em}}\angle \text{AOP = 50°}\\ \text{or \hspace{0.17em}\hspace{0.17em}}\angle \text{POA = 50°}\\ \text{Hence, the correct option is (A).}\end{array}$

Q.8 Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Ans.

$\begin{array}{l}\text{Let AB is a diameter and PQ and RS be two tangents at}\\ \text{the ends of diameter AB.}\\ \text{We know that the line drawn from the center of the circle}\\ \text{to the tangent is perpendicular to the tangent.}\\ \therefore \text{OA}\perp \text{RS and OB}\perp \text{PQ}\\ \text{Thus,we have}\\ \text{}\angle \text{OAR =}\angle \text{OAS = 90° =}\angle \text{OBP =}\angle \text{OBQ}\\ \text{or}\angle \text{OAR =}\angle \text{OBQ and}\angle \text{OAS =}\angle \text{OBP}\\ \text{i.e., alternate interior angles are equal.}\\ \text{Therefore, by converse of parallel lines axiom, RS||PQ.}\end{array}$

Q.9 Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Ans.

Let us consider a circle with centre O. Let AB be a tangent which touches the circle at P.

Let the perpendicular to AB at P does not pass through the centre O. Let it pass through another point R. We join OP and PR.

$\begin{array}{l}\text{Perpendicular to AB at P passes through R}\text{.}\\ \therefore \angle \text{RPB = 90°}\dots \text{(1)}\\ \text{We know that the line joining the centre and point of}\\ \text{contact to the tangent to a circle are perpendicular to}\\ \text{each other}\text{.}\\ \therefore \angle \text{OPB = 90°}\dots \text{(2)}\\ \text{From (1) and (2), we get}\\ \angle \text{OPB =}\angle \text{RPB}\\ \text{From figure, we find that}\\ \angle \text{RPB <}\angle \text{OPB}\\ \therefore \text{}\angle \text{OPB =}\angle \text{RPB is not possible}\text{. It is only possible when the}\\ \text{line RP coincides with OP}\text{.}\\ \text{Therefore, the perpendicular at the point of contact to}\\ \text{the tangent to a circle passes through the centre}\text{.}\end{array}$

Q.10 The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Ans.

Let centre of the circle be at O and point of contact of the tangent from A be P.

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
null.

Q.11 Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Ans.

Let centre of the circles be at O and point of contact of the tangent PQ be A.

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

$\begin{array}{l}\text{We know that the line drawn from the center of the circle}\\ \text{to the tangent is perpendicular to the tangent}\text{.}\\ \therefore \text{OA}\perp \text{PQ}\\ \text{It is given that OP = OQ = 5}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{cm and OA = 3 cm}\\ \text{By applying Pythagoras theorem in Δ}\text{\hspace{0.17em}}\text{OAP, we get}\\ {\text{OA}}^{\text{2}}{\text{+PA}}^{\text{2}}{\text{= OP}}^{\text{2}}\\ {\text{or 3}}^{\text{2}}{\text{+PA}}^{\text{2}}{\text{= 5}}^{\text{2}}\\ {\text{or PA}}^{\text{2}}\text{= 25}-\text{16 = 9}\\ \text{or PA = 3 cm}\\ \text{Similarly, we have AQ = 3 cm}\\ \text{PQ = PA+AQ = 3+3 = 6 cm}\\ \text{Therefore, length of the chord PQ is 6 cm}\text{.}\end{array}$

Q.12

$\begin{array}{l}\text{A quadrilateral ABCD is drawn to circumscribe a circle}\\ \text{(see the following figure). Prove that}\\ \mathrm{AB}+\mathrm{CD}=\mathrm{AD}+\mathrm{BC}.\end{array}$

Ans.

$\begin{array}{l}\text{We have to prove that}\\ \text{AB+CD = AD+BC}\\ \text{We know}\text{ }\text{that lengths of tangents drawn from a point to a}\\ \text{circle are equal}\text{.}\\ \text{Therefore, from figure, we have}\\ \text{DR = DS, CR = CQ,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{AS = AP,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{BP = BQ}\\ \text{Now,}\\ \text{LHS = AB+CD = (AP+BP)+(CR+DR)}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{= (AS+BQ)+(CQ+DS)}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{= AS+DS+BQ+CQ}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{= AD+BC}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{\hspace{0.17em}}\text{= RHS}\end{array}$

Q.13

$\begin{array}{l}\text{In the following figure. XY and X’Y’ are two parallel}\\ \text{Tangents to a circle with center O and another }\\ \text{tangent AB with point of contact C intersecting XY at}\\ \text{A and X’Y’ at B. Prove that }\angle {\text{AOB = 90}}^{\circ }\text{.}\end{array}$

Ans.

$\begin{array}{l}\text{Let us join points O and C.}\\ \text{In Δ\hspace{0.17em}OPA and Δ\hspace{0.17em}OCA,}\\ \text{OP = OC (Radii of the given circle)}\\ \text{AP = AC (Tangents from point A)}\\ \text{AO = AO (Common side)}\\ \therefore \text{Δ\hspace{0.17em}OPA}\cong \text{Δ\hspace{0.17em}OCA (SSS congruence criterion)}\\ \text{So,}\angle \text{POA =}\angle \text{COA}...\text{(1)}\\ \text{Similarly, we have Δ\hspace{0.17em}OQB}\cong \text{Δ\hspace{0.17em}OCB and}\\ \text{}\angle \text{QOB =}\angle \text{COB}...\text{(2)}\\ \text{Now, POQ is a straight line. So, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{POA+}\angle \text{COA+}\angle \text{QOB+}\angle \text{COB = 180°}\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{COA+}\angle \text{COA+}\angle \text{COB+}\angle \text{COB = 180° [From (1) and (2)]}\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2}\left(\angle \text{COA+}\angle \text{COB}\right)\text{= 180°}\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{COA+}\angle \text{COB =}\frac{\text{180°}}{\text{2}}\text{= 90°}\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{AOB = 90°}\end{array}$

Q.14 Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Ans.

$\begin{array}{l}\text{We have the above figure as per given information.}\\ \text{We know that a line from the centre of a circle is perpendicular}\\ \text{to the point of contact of a tangent.}\\ \text{Therefore, in quadrilateral PAOB, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{OAP+}\angle \text{OBP+}\angle \text{AOB+}\angle \text{APB = 360°}\\ \text{or 90°+90°+}\angle \text{O+}\angle \text{P = 360°}\\ \text{or}\angle \text{AOB+}\angle \text{APB = 360°-180° = 180°}\\ \text{or}\angle \text{AOB+}\angle \text{APB = 180°}\end{array}$

Q.15 Prove that the parallelogram circumscribing a circle is a rhombus.

Ans.

$\begin{array}{l}\text{It is given that ABCD is a rhombus. Therefore, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB = CD}...\text{(1)}\\ \text{and \hspace{0.17em}AD = BC}...\text{(2)}\\ \text{We know that lengths of tangents drawn from a point to a}\\ \text{circle are equal.}\\ \text{Therefore, from figure, we have}\\ \text{DR = DS, CR = CQ,\hspace{0.17em}\hspace{0.17em}AS = AP,\hspace{0.17em}\hspace{0.17em}BP = BQ}\\ \text{Now,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB+CD = (AP+BP)+(CR+DR)}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB+CD\hspace{0.17em}\hspace{0.17em}= (AS+BQ)+(CQ+DS)}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB+CD\hspace{0.17em}\hspace{0.17em}= AS+DS+BQ+CQ}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB+CD\hspace{0.17em}\hspace{0.17em}= AD+BC}...\text{(3)}\\ \text{From equations (1), (2) and (3), we get}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB+AB = AD+AD}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2AB = 2AD}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB = AD}...\text{(4)}\\ \text{On comparing equations (1), (2) and (4), we get}\\ \text{AB = BC = CD = AD}\\ \text{Hence,ABCD is a rhombus.}\end{array}$

Q.16 A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see the following figure). Find the sides AB and AC.

Ans.

$\begin{array}{l}\text{Let the given circle touch the sides AB and AC of the triangle}\\ \text{at points E and F respectively and length of the line segment}\\ \text{AF be x.}\\ \text{In Δ\hspace{0.17em}ABC, we have}\\ \text{CF = CD = 6\hspace{0.17em}cm [Tangents on the circle from point C]}\\ \text{BE = BD = 8\hspace{0.17em}cm \hspace{0.17em}[Tangents on the circle from point B]}\\ \text{AE = AF = x \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[Tangents on the circle from point A]}\\ \text{AB = AE+EB = x+8}\\ \text{BC = BD+DC = 8+6 = 14 cm}\\ \text{CA = CF+FA = 6+x}\\ \text{Now, perimeter of Δ\hspace{0.17em}ABC = 2s = AB+BC+CA}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}= (x+8)+14+(x+6)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}= 2x+28 = 2(x+14)}\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}s = x+14}\\ \text{Area of Δ\hspace{0.17em}ABC =}\sqrt{\text{s(s}-\text{a)(s}-\text{b)(s}-\text{c)}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=}\sqrt{\text{(x+14)(x+14}-\text{x}-\text{8)(x+14}-\text{14)(x+14}-\text{6}-\text{x)}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=}\sqrt{\text{(x+14)48x}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}= 4}\sqrt{{\text{3(x}}^{\text{2}}\text{+14x)}}\\ \text{Area of Δ\hspace{0.17em}OBC =}\frac{\text{1}}{\text{2}}\text{OD×BC =}\frac{\text{1}}{\text{2}}{\text{×4×14 = 28 cm}}^{\text{2}}\\ \text{Area of Δ\hspace{0.17em}OCA =}\frac{\text{1}}{\text{2}}\text{OF×AC =}\frac{\text{1}}{\text{2}}\text{×4×(6+x) = 12+2x}\\ \text{Area of Δ\hspace{0.17em}OAB =}\frac{\text{1}}{\text{2}}\text{OE×AB =}\frac{\text{1}}{\text{2}}\text{×4×(8+x) = 16+2x}\\ \text{Now,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Area of Δ\hspace{0.17em}ABC = Area of Δ\hspace{0.17em}OBC+Area of Δ\hspace{0.17em}OCA+Area of Δ\hspace{0.17em}OAB}\\ \text{or 4}\sqrt{{\text{3(x}}^{\text{2}}\text{+14x)}}\text{= 28+12+2x+16+2x = 56+4x = 4(x+14)}\\ {\text{or 3x}}^{\text{2}}\text{+42x =}{\left(\text{x+14}\right)}^{\text{2}}{\text{= x}}^{\text{2}}\text{+28x+196}\\ {\text{or 3x}}^{\text{2}}\text{+42x}-{\text{x}}^{\text{2}}-\text{28x = 196}\\ {\text{or 2x}}^{\text{2}}\text{+14x = 196}\\ {\text{or 2(x}}^{\text{2}}\text{+7x}-\text{98) = 0}\\ {\text{or x}}^{\text{2}}\text{+7x}-\text{98=0}\\ {\text{or x}}^{\text{2}}\text{+14x}-\text{7x-98=0}\\ \text{or x(x+14)}-\text{7(x+14)=0}\\ \text{or (x+14)(x}-\text{7)=0}\\ \text{or x=7 [Length can not be negative, so x¹}-\text{14.]}\\ \text{Hence, \hspace{0.17em}AB=x+8=7+8=15 cm}\\ \text{\hspace{0.17em}​​​​​​​​​​\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}CA=6+x=6+7=13 cm}\end{array}$

Q.17 Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Ans.

$\begin{array}{l}\text{Let ABCD be a quadrilateral circumscribing a circle centred at O}\\ \text{such that it touches the circle at points P, Q, R and S. Let us join}\\ \text{the vertices of the quadrilateral ABCD to the centre of the circle.}\\ \text{In ΔOAP and Δ\hspace{0.17em}OAS, we have}\\ \text{AP = AS [Tangents drawn from point A]}\\ \text{OP = OS \hspace{0.17em}[Radii of the same circle]}\\ \text{OA = OA [Common side]}\\ \text{Δ\hspace{0.17em}OAP}\cong \text{Δ\hspace{0.17em}OAS [SSS congruence criterion]}\\ \text{Thus,}\angle \text{POA =}\angle \text{AOS or \hspace{0.17em}\hspace{0.17em}}\angle \text{1 =}\angle \text{2}\\ \text{Similarly,}\angle \text{3 =}\angle \text{4,}\angle \text{5 =}\angle \text{6,\hspace{0.17em} \hspace{0.17em}}\angle \text{7 =}\angle \text{8}\\ \text{Now, from the above figure, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{1+}\angle \text{2+}\angle \text{3+}\angle \text{4+}\angle \text{5+}\angle \text{6+\hspace{0.17em}}\angle \text{7+}\angle \text{8 = 360°}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(}\angle \text{1+}\angle \text{2)+(}\angle \text{3+}\angle \text{4)+(}\angle \text{5+}\angle \text{6)+(}\angle \text{7+}\angle \text{8) = 360°}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(}\angle \text{2+}\angle \text{2)+(}\angle \text{3+}\angle \text{3)+(}\angle \text{6+}\angle \text{6)+(\hspace{0.17em}}\angle \text{7+}\angle \text{7) = 360°}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2}\left(\angle \text{2+}\angle \text{3+}\angle \text{6+}\angle \text{7}\right)\text{= 360°}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\angle \text{2+}\angle \text{3}\right)\text{+}\left(\angle \text{6+}\angle \text{7}\right)\text{=}\frac{\text{360°}}{\text{2}}\text{= 180°}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\angle \text{2+}\angle \text{3}\right)\text{+}\left(\angle \text{6+}\angle \text{7}\right)\text{= 180°}\\ \text{or}\angle \text{AOB+}\angle \text{COD = 180°}\\ \text{Similarly,\hspace{0.17em}we\hspace{0.17em}\hspace{0.17em}have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\angle \text{1+}\angle \text{8}\right)\text{+}\left(\angle \text{4+}\angle \text{5}\right)\text{= 180°}\\ \text{or}\angle \text{AOD+}\angle \text{BOC = 180°}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{AOB+}\angle \text{COD = 180° =}\angle \text{AOD+}\angle \text{BOC}\\ \text{i.e., opposite sides of a quadrilateral circumscribing a circle}\\ \text{subtend supplementary angles at the centre of the circle.}\end{array}$

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