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


AP = AQ


BR = BQ


CR = CS


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.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Bonus Questions on CirclesBonus Questions on CirclesBonus Questions on Circles40 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses