Q. 114.7( 18 Votes )

In Figure, a circ

Answer :

Given that the sides of the quadrilateral ABCD are as follows: S

AB = 6 cm, BC = 9 cm, CD = 8 cm.

Let P, Q, R, S be the points where the tangent touches the sides AD, AB, BC, and CD respectively. P R

From a point outside of a circle, two tangents to the circle are always equal. Q

Therefore, DP = DS




Thus, adding the left sides and right sides separately and equating them, we get:

(DP + AP) + (BR + CR) = (DS + CS) + (AQ + BQ)

AD + BC = CD + AB

AD = CD + AB - BC

= 8 + 6 – 9

= 5

AD = 5 cm.

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