NCERT Solutions for Class 10 Maths Chapter 10 - Circles

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Chapter 10 - Circles Exercise Ex. 10.1

Solution 1

Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 2
Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles
Solution 3
Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles
Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles
Solution 4
Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles

Chapter 10 - Circles Exercise Ex. 10.2

Solution 1
Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles
Solution 2

Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 3
Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles
Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles
Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles

Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 4
Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles
Solution 5
Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles
Solution 6
Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles
Solution 7
Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles

Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles
Solution 8
Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles
Solution 9
Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles

Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles


Solution 10
Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles
Solution 11

Since, ABCD is a parallelogram,

AB = CD                                                                (i)

BC = AD                                                                (ii)

Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles

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
Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles

Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles

Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles
Solution 13
Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles
Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles
Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles
Ncert Solutions Cbse Class 10 Mathematics Chapter - Circles