Question

e and f are mid points of sides ab and cd of parallelogram abcd . if the area of parallelogram abcd is 36cm² . find the area of triangle apd . plzzz help me with this

Answer 1

In ||gm ABCD and triangle ABP-

Since they are on same base and between same parallels.Therefore
ar triangle=1/2ar of parallelogram

Thus ar APB=18 sqcm

Also DFAE=18 sqcm because. E and F are mid point.
We can prove ADE congruent to BEP.by AAS

so ar triangleABP=27sqcm

because ar APB=ar APE+ADE

HOPE IT WORKS....OR ASK ME AGAIN..

Answer 2

area is 18 cm²

area of ∆ AED is ¼ of total area of parallelogram.
means area of ∆AED=9cm²

area of APD=2× area of AED

= 2×9= 18cm²