Question

plz solve...................​

Answer 1

Answer:

We have,

ABCD as the cyclic quadrilateral in which the diagonal AC and BD.

intersect each other at point P.

also, given that,

AB=8cm,

CD=5cm

Now,

In ΔDCA and ΔAPB,

We have

∠DCP=∠ABP

∠CDP=∠PAB

Hence, 

ΔDPC∼ΔAPB (by A.A property)

According to the given question,

arΔAPBarΔDPC=(ABDC)2

⇒24arΔDPC=(85)2

⇒24arΔDPC=6425

⇒arΔDPV=6425×24

arΔDPC=9.375cm2