Question

In the given figure,$AB||DCandAD||BC.$ If $ar\left(ABCD\right)=Landar\left(ACED\right)=M,$ then $ar\left(ADC\right):ar\left(DEC\right)$ equals

• A.
  $L:(2M-L)$
• B.
  $L:(M-L)$
• C.
  
  $L:2(M-L)$
  
• D.
  $M:L$

Answer 1

The correct option is A

$L:(2M-L)$
Given: Ar(ABCD) = L and Ar(ACED) = M

Since $AB||DCandAD||BC.$
Therefore, ABCD is a parallelogram.

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

$\therefore Ar\left(ADC\right)=\frac{1}{2}Ar\left(ABCD\right)\Rightarrow Ar\left(ADC\right)=\frac{L}{2}.......\left(i\right)Also,Ar\left(ACED\right)=Ar\left(\Delta ADC\right)+Ar\left(\Delta DEC\right)\Rightarrow M=\frac{L}{2}+Ar\left(\Delta DEC\right)\Rightarrow Ar\left(\Delta DEC\right)=M-\frac{L}{2}=\frac{2M-L}{2}\text{From (i) and (ii), we get}\frac{Ar\left(\Delta ADC\right)}{Ar\left(\Delta DEC\right)}=\frac{\frac{L}{2}}{\frac{2M-L}{2}}=\frac{L}{2M-L}\therefore Ar\left(\Delta ADC\right):Ar\left(\Delta DEC\right)=L:(2M-L)$
Hence, the correct answer is option (a).