# NCERT Solutions for Class 10 Maths Chapter 10 - Circles

Does the sound of Class 10 Maths intrigue you? Or, does it scare you away? It’s not easy to handle the Board pressure for Class 10. But, we can definitely help you tackle the syllabus for CBSE Class 10 Maths. Scoring more marks in Mathematics CBSE Class 10 has never been this easy before. You just have to start by referring to the CBSE class 10 maths notes.

Our study materials for CBSE Class 10 Mathematics are created by subject experts who teach students Maths. Our materials will help you to score high marks in your Mathematics examination as we focus on providing solutions which simplify the complex fundamentals of a subject. At TopperLearning, we believe in providing quality solutions at a low cost, and we strictly follow the latest CBSE syllabus; our study materials are revised from time to time. Our TopperLearning packages involve all the study resources for CBSE Class 10 such as solved question papers, video lessons and revision notes which will help you to score high marks. We also provide free NCERT and RD Sharma textbook solutions which provide students with step-by-step solutions.

Page / Exercise

## Chapter 10 - Circles Exercise Ex. 10.1

Solution 1 Solution 2 Solution 3  Solution 4 ## Chapter 10 - Circles Exercise Ex. 10.2

Solution 1 Solution 2 Solution 3    Solution 4 Solution 5 Solution 6 Solution 7  Solution 8 Solution 9  Solution 10 Solution 11

Since, ABCD is a parallelogram,

AB = CD                                                                (i) Now, it can be observed that:

DR = DS                 (tangents on circle from point D)

CR = CQ                 (tangents on circle from point C)

BP = BQ                 (tangents on circle from point B)

AP = AS                 (tangents on circle from point A)

Adding all the above four equations,

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

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

CD + AB = AD + BC                                            (iii)

From equation (i) (ii)  and (ii):

2AB = 2BC

AB = BC

AB = BC = CD = DA

Hence, ABCD is a rhombus.

Solution 12   Solution 13    