Question

​ABCD is a parallelogram and E and F are mid-points of AD and BC respectively. P is any point on AB. If area of ΔEDF=16 $c{m}^2$, then find the area of ( ΔAEP+ ΔBFP) [3 MARKS]

Answer 1

Concept : 1 Mark
Application : 1 Mark
Calculation : 1 Mark

Given that area of ΔDEF = 16 $c{m}^2$

ΔDEF and parallelogram DEFC are on same bases and same parallels.

So, ar (DEFC)= 2 X ar of ΔDEF = 32 $c{m}^2$

As E and F are mid points of AD and BC,

ar (DEFC) = ar(EABF)= 32 $c{m}^2$

Again triangle PEF and parallelogram EABF are on same bases and same parallels.

So, ar (PEF) = 1/2 X ar(EABF)=16 $c{m}^2$

Now, ar (ΔAEP) + ar (ΔPFB) + ar (PEF) = ar(EABF)

ar (ΔAEP) + ar (ΔPFB) + 16 $c{m}^2$ = 32 $c{m}^2$

ar (ΔAEP) + ar (ΔPFB) = 16 $c{m}^2$