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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
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