Question

p is the midpoint of cd in parralelogram abcd then prove ar(abcd):ar(pbc) =4:1​

Answer 1

Construction : make F the midpoint of AB, and join F and P.

FPBC is a parallelogram since, BF = PC (midpoints of equal sides) and FB is parallel to PC

Triangle PBC and Parallelogram FPBC are on the same base (BC) and between the same parallels (FB and PC)

   Therefore, Ar(PBC) = 1/2 Ar(FPBC)  (i)

Parallelogram FPBC is equal to half of the area of Parallelogram ABCD, as Area of Parallelogram is Base x Height and FPBC has same height and half the base.

   Hence ar(FPBC) = 1/2 ar(ABCD)  (ii)

Using (i) and (ii)

 ar(PBC) = 1/2 ar(FPBC)

= ar(PBC) = 1/4 ar(ABCD)

Therefore, ar(PBC) : ar(ABCD)  = 1 : 4