Question

In the given figure, ABDE and BCDE are parallelograms. If $ar\left(CFD\right)=x,thenar\left(ACDE\right):ar\left(BED\right)$ equals

• A.
  $4:1$
• B.
  $3:2$
• C.
  $4:2$
• D.
  $3:1$

Answer 1

The correct option is D

$3:1$
Given that, ABDE and BCDE are parallelogram.

$\therefore AB=DEandBC=DE$

In parallelogram ABDE, BE is a diagonal and we know that a diagonal divides a parallelogram in two triangles of equal area.

$\therefore Ar\left(\Delta ABE\right)=Ar\left(\Delta BED\right)...\left(i\right)$

Also, we know that if a parallelogram and a triangle lie on the same base and between the same parallels, then the area of the triangle is half of the area of the parallelogram.

$\therefore Ar\left(\Delta BED\right)=\frac{1}{2}Ar\left(BCDE\right)...\left(ii\right)Now,Ar\left(ACDE\right)=Ar\left(\Delta ABE\right)+Ar\left(BCDE\right)=Ar\left(\Delta BED\right)+2Ar\left(\Delta BED\right)=3Ar\left(\Delta BED\right)\therefore \frac{Ar\left(ACDE\right)}{Ar\left(\Delta BED\right)}=\frac{3}{1}$

Hence, the correct answer is option (d).