# NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2

The subject of Mathematics is used by scientists to find quantitative solutions to experimental laws. A wide variety of fields are affected by it, including Computer Science, Social Sciences, Finance, Medicine, Natural Sciences, and Engineering. Therefore, many students opt for it for their higher education. Mathematical concepts introduced in Class 10 form the basis for higher-level courses such as Engineering, Architecture, etc. It is crucial for students to possess strong concepts in Class 10 so that they will not only score well but also have those fundamentals applied in the future. However, Mathematics is a subject that many students find intimidating. Therefore, Extramarks provides them with NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2. Mathematics is an area of knowledge that includes such topics as numbers formulas and related structures, shapes and the spaces in which they are contained, quantities and their changes.

Chapter 10 of the Class 10 Mathematics NCERT textbook is Circles. The chapter contains two exercises and has a significant weightage in the Class 10 Mathematics board examinations. The concepts of Exercise 10.2 Class 10 Maths can be challenging for students as they are a bit complicated. Therefore, they require proper step-by-step solutions to be able to comprehend the concepts of the chapter. Extramarks provides students with the NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2 so that they can have access to properly detailed and straightforward solutions without having to look anywhere else. Mathematics defines a circle as an ellipse with zero eccentricity and coincident foci. In addition, it may also be referred to as the locus of the points drawn at an equidistant distance from the centre. The radius is the distance between the centre and the outer line of a circle. Circles are divided into two equal parts by their diameter, which is also twice their radius. Essentially, a circle is a 2D shape that is measured in terms of its radius. A circle divides the plane into two regions, which are the interior and exterior regions. It is the same as a line segment. It is a two-dimensional figure that has a perimeter and an area. The circumference, or distance around a circle, is also called its perimeter. In a 2D plane, the area of a circle is the region bounded by it. Students can refer to the NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2 for a better understanding of the concepts of the chapter.

The NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2 guides students to understand the concepts involved in the chapter Circles. They can refer to these solutions for a better understanding of the properties and  concepts used in the chapter. Consequently, NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2 assist students to score better in their board examinations. Class 10 Maths Chapter 10 Exercise 10.2 deals with the existence of the tangents to a circle and some of the properties of a circle. The chapter introduces students to some complex terms such as Tangents, Tangents to a Circle, The Number of Tangents from a Point on the Circle, and much more. The diagrams and geometrical calculations in this chapter, make it very interesting for students. The NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2 contain some tricky conceptual questions that can help students  clear all their doubts and queries. Extramarks recommends that students practise these solutions to know the alternative calculation methods that can be used in the chapter.

## NCERT Solutions for Class 10 Maths Chapter 10 Circles (Ex 10.2) Exercise 10.2

The NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2 are one of the best resources for preparing for the Class 10 Mathematics board examination. These solutions provide answers to all the questions in Circle Class 10 Exercise 10.2. Practising the NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2 can help students solve any complicated questions that can appear in the board examinations. These solutions are updated according to the latest examination pattern and provide students with an idea of the types of questions that they can encounter in the examinations. The NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2 aid students in memorising the essential formulas and properties that are included in the chapter. Students should thoroughly review the NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2 in order to have a thorough understanding of the chapter’s concepts and calculations.These solutions enhance the conceptual clarity of students and help them perform well in their Class 10 Mathematics board examinations.

Practising the NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2 can help students gain a better understanding of the chapter’s important points and achieve higher grades in their examinations. These solutions provide precise and appropriate answers to the questions mentioned in the NCERT textbooks. To get a better grip on the concepts of this chapter, Extramarks provides students with the NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2. Furthermore, these solutions are very helpful for students who struggle with the subject of Mathematics. The NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2 mainly help students in understanding the fundamentals of the Number of Tangents from a point on a Circle and various other concepts related to this topic. These solutions are available in accordance with the latest examination pattern. The NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2 help students understand the practical applications of the chapter Circles. It is therefore very important for them to learn these solutions effectively. Extramarks recommends students thoroughly review these solutions prior to their examinations.

## Access NCERT Solutions for Class 10 Mathematics Chapter 10 – Circles

Students of Class 10 are encouraged to familiarise themselves with the NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2 in order to prepare well for the in-school exams and competitive exams. Consequently, students should also practise various extra questions from the chapter to be able to solve any complicated problem that they can encounter in the board examinations. Therefore, Extramarks recommends students practise a number of important questions along with the NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2. Moreover, these solutions assist students in clarifying their doubts since they provide detailed explanations of the logic used behind the solution. Therefore, In order to help students resolve all their queries, Extramarks provides them with the NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2. Circles are not just a concept of the Class 10 Mathematics curriculum; they have variouspractical applications and are an importantimportant part of the real-world scenario. This chapter discusses the units of physical quantities and methods of evaluating them, while the other section discusses measurement errors and significant figures. By practising the NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2 students can get a better understanding of the concept of measurement. Circles have significantly contributed to civilisation. For instance, the invention of the wheel transformed society and the methods of transportation. They are also used by seismologists to determine and locate the centre of earthquakes. Furthermore, architects use circles to design buildings. There are various other real-life applications of Circles. Students must thoroughly go through the NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2 to have a better understanding of the concepts and applications of Circles.

The CBSE curriculum is followed by the schools that are affiliated with the CBSE.NCERT textbooks are compiled by professionals who are highly knowledgeable and proficient in their subjects. In order to build students’ basic concepts, NCERT textbooks are the best educational tools. These textbooks, however, do not provide answers to the questions they pose. It can be challenging for students to find authentic answers to those questions. In order to provide young learners with the best education possible, Extramarks provides them with comprehensive study materials for all academic years and subjects. To have a successful academic future, students should be introduced to comprehensive learning in smaller classes. Therefore, Extramarks provides students with the NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2. The subject of Mathematics encourages logical reasoning, creative thinking, critical thinking, problem-solving ability, abstract thinking, and much more in students.

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## NCERT Solutions for Class 10 Maths Chapter 10 Circles Exercise 10.2

Subject experts at Extramarks have compiled the NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2. Students can easily understand these solutions, as they are compiled in simple and straightforward language. Learners who find Mathematics Chapter 10 complicated can greatly benefit from the NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2. These solutions are well-structured and properly detailed in an easy language so that students can comprehend them easily. Students can easily download and access the NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2 on a variety of devices. These solutions can improve the conceptual clarity of students. They can prepare for any examinations using the NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2 provided by Extramarks. Students who wish to pursue Mathematics for their further studies must thoroughly review the NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2.

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### NCERT Solutions for Class 10 Mathematics Chapter 10 Exercises

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Q.1 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.2

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

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.3

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

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.4 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.5 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.6 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.7  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.8

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

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.9

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

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.10 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.11 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.12 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.13 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}$