Question

In the given figure, ABCD is a parallelogram F, E, G and H are the mid-points of AD, AB, BC and DC respectively. The ar(EBGHDF) is equal to

• A.
  $\frac{1}{2}ar\left(ABCD\right)$
  
• B.
  $\frac{3}{4}ar\left(ABCD\right)$
  
• C.
  $\frac{4}{5}ar\left(ABCD\right)$
  
• D.
  $\frac{1}{8}ar\left(ABCD\right)$
  

Answer 1

The correct option is B

$\frac{3}{4}ar\left(ABCD\right)$

Given that ABCD is a parallelogram.

We know that area of the triangle formed by joining the mid-points of the sides of a triangle is equal to one-fourth of area of the given triangle.




$\therefore Ar\left(\Delta AFE\right)=\frac{1}{4}\times Ar\left(\Delta ADB\right)\dots ..\left(i\right)Similarly,Ar\left(\Delta CGH\right)=\frac{1}{4}\times Ar\left(\Delta CBD\right)\dots ..\left(ii\right)$

Now, 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 the area of the parallelogram.

$\therefore Ar\left(\Delta ABD\right)=\frac{1}{2}Ar\left(ABCD\right)andAr\left(\Delta CBD\right)=\frac{1}{2}Ar\left(ABCD\right).....\left(iii\right)\therefore Ar\left(EBGHDF\right)=Ar\left(ABCD\right)-Ar\left(\Delta AFE\right)-Ar\left(\Delta CGH\right)=Ar\left(ABCD\right)-\frac{Ar\left(\Delta ADB\right)}{4}-\frac{Ar\left(\Delta CBD\right)}{4}\left[from\right(i\left)and\right(ii\left)\right]=Ar\left(ABCD\right)-\frac{Ar\left(ABCD\right)}{8}-\frac{Ar\left(ABCD\right)}{8}\left[From\right(iii\left)\right]=\frac{3}{4}Ar\left(ABCD\right)$


Hence, the correct answer is option (b).