Question

In the given figure, ABCD is a parallelogram. If DC is produced to E and AF = FB, then the area of $\Delta AFE$ cannot be equal to

• A.
  $\frac{1}{4}ar\left(ABCD\right)$
  
• B.
  $ar\left(\Delta ADC\right)$
  
• C.
  $\frac{1}{2}ar\left(\Delta ADC\right)$
  
• D.
  $\frac{1}{2}ar\left(\Delta ACB\right)$
  

Answer 1

The correct option is B

$ar\left(\Delta ADC\right)$

Given that ABCD is a parallelogram.
$\therefore AB||CDandAD||BC$



Join B to E.

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.

Since,$\Delta ABE$ and parallelogram ABCD lie on the same base AB and between the same parallels AB and DE.

$\therefore Ar\left(\Delta ABE\right)=\frac{1}{2}\left[Ar\right(ABCD\left)\right]...\left(i\right)$


Also, ΔADC and parallelogram ABCD lie on the same base DC and between the same parallels AB and DC.

$\therefore Ar\left(\Delta ADC\right)=\frac{1}{2}\times Ar\left(\Delta ABCD\right)\dots ..\left(ii\right)\text{From (i) and (ii), we get}Ar\left(\Delta ABE\right)=Ar\left(\Delta ADC\right)\dots ..\left(iii\right)\text{Now, given that}AF=FB.\text{Therefore, F is the median of}\Delta ABE.\text{Since median divides the triangle into two equal halves.}\therefore Ar\left(\Delta ABE\right)=\frac{1}{2}\left(\Delta AFE\right)...\left(vi\right)\text{From (iii) and (iv), we get}\frac{1}{2}Ar\left(\Delta AFE\right)=Ar\left(\Delta ADC\right)\Rightarrow Ar\left(DeltaAFE\right)=2Ar\left(\Delta ADC\right)\Rightarrow Ar\left(\Delta AFE\right)\ne Ar\left(\Delta ADC\right)$

Hence, the correct answer is option (b).