Question

ABCD is a llgm X and Y are mid points of BC and CD respectively. Prove that ar(∆AXY)=⅜ar(llgmABCD)​

Answer 1

Hello mate

Given a parallelogram ABCD where X and Y are the mid points of the sides BC and CD respectively. Prove that ar (∆AXY) =3/8 (ar ABCD)?

ABCD is a parallelogram where X and Y are the midpoints of BC and CD. respectively. Prove that ar(triangle AXY) =(3/8) ar (ABCD).

Draw XM parallel to AB, so M is midpoint of AD. Also draw YN parallel to AD, so N is midpoint of AB.

Ar (triangle AXY) = Ar(ABCD) - ar(ABX) - ar(ADY) - ar(XYC)_

ar(ABX) = (1/2) of parallelogram ABXM = (1/4) Ar(ABCD)

ar(ADY) = (1/2) of parallelogram ANYD = (1/4) Ar(ABCD)

ar(XYC)_= (1/2) of parallelogram OXCY = (1/8) Ar(ABCD)

Therefore, Ar (triangle AXY) = Ar(ABCD) - (1/4) Ar(ABCD) - (1/4) Ar(ABCD) - (1/8) Ar(ABCD) = Ar(ABCD)[1 - (1/4) - (1/4) - (1/8)] = Ar(ABCD)[8-2-2-1]/8

= (3/8) ar (ABCD). Proved.