Question

In the given figure, AD || BC, AC and BD intersect at H. E is a point on BD such that DE < HE. F and G are points such that AEDF and ACGD are parallelograms. If $ar\left(ABEF\right)=80c{m}^2,$ then ar(ACGD) equals

• A.
  $160c{m}^2$
• B.
  $80c{m}^2$
• C.
  $240c{m}^2$
• D.
  $40c{m}^2$

Answer 1

The correct option is A

$160c{m}^2$
$Ar\left(ABEF\right)=Ar\left(\Delta ABE\right)+Ar\left(\Delta AEF\right)\dots ..\left(i\right)$

Now, triangles on the same base and between same parallels are equal in area.

$\therefore Ar\left(\Delta AEF\right)=Ar\left(\Delta AED\right)\dots ..\left(ii\right)\text{From (i) and (ii), we get}Ar\left(ABEF\right)=Ar\left(\Delta ABE\right)+Ar\left(\Delta AED\right)=Ar\left(\Delta ABD\right)\dots ..\left(iii\right)$

Now, if a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.

$\therefore Ar\left(\Delta ABD\right)=\frac{1}{2}Ar\left(ACGD\right)...\left(iv\right)\text{From (iii) and (iv), we get}Ar\left(ABEF\right)=\frac{1}{2}Ar\left(ACGD\right)\Rightarrow Ar\left(ACGD\right)=2\times Ar\left(ABEF\right)=2\times 80\left(Given\right)=160c{m}^2$


Hence, the correct answer is option (a).