Question

Let ABCD be a convex quadrilateral with AB CD = 10,
BC = 14, and AD = 265. Assume that the diagonals
of ABCD intersect at point P, and that the sum of the
areas of triangles APB and CPD equals the sum of the
areas of triangles BPC and APD. If A be the area of
quadrilateral ABCD, find A/2.​

Answer 1

Answer:

What is natural number

Answer 2

Answer:

We know that,

Area of a Δ is (1/2)*base*height.

So with reference to the figure,

Ar(ΔAPD)=(1/2)*x*DP...................(1)

Ar(ΔBPC)=(1/2)*y*BP..................(2)

Since height for ΔAPB is x and for ΔCPD is y

So Ar(ΔAPB)=(1/2)*x*BP...................(3)

Ar(ΔAPD)=(1/2)*y*DP..................(4)

Thus from (1),(2),(3) and (4),

Ar(ΔAPD)*Ar(ΔBPC)=Ar(ΔAPB)*Ar(ΔAPD)