NCERT Class 10 Maths Solutions Chapter - 10 Circles, Ex - 10.2

Ex - 10.2

Question 1.  

Solution



Question 2.  

Solution



Question 3.


Solution






Question 4.  

Solution



Question 5.  

Solution



Question 6.  

Solution



Question 7.  

Solution




Question 8.  

Solution



Question 9.  

Solution




Question 10.  

Solution



Question 11.  Prove that the parallelogram circumscribing a circle is a rhombus.

Solution

Since, ABCD is a parallelogram,

AB = CD                                                                (i)

BC = AD                                                                (ii)

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.

Question 12.  

Solution

Question 13.  Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Solution