Question

In the given figure, if $AE=\frac{1}{4}AC$and $BD=\frac{1}{3}BC.$ then the ratio of areas of$\Delta EDCand\Delta ABE$ is

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

Answer 1

The correct option is B

$2:1$
Given,
$AE=\frac{1}{4}AC,BD=\frac{1}{3}BC$
Let, $BF\perp ACandEG\perp BC.Now,AE=\frac{1}{4}AC\Rightarrow 4AE=AE+EC\Rightarrow 3AE=EC\Rightarrow \frac{AE}{EC}=\frac{1}{3}...\left(i\right)Also,BD=\frac{1}{3}BC\Rightarrow BC-CD=\frac{1}{3}BC\Rightarrow 3BC–3CD=BC\Rightarrow 2BC=3CD\Rightarrow \frac{CD}{BC}=\frac{2}{3}...\left(ii\right)Now,Ar\left(\Delta ABE\right)=\frac{1}{2}\times AE\times BFand,Ar\left(\Delta BEC\right)=\frac{1}{2}\times EC\times BF\therefore \frac{Ar\left(\Delta ABE\right)}{Ar\left(\Delta BEC\right)}=\frac{AE}{EC}=\frac{1}{3}\left[From\right(i\left)\right]....\left(iii\right)Also,Ar\left(\Delta EDC\right)=\frac{1}{2}\times CD\times EGand,Ar\left(\Delta BEC\right)=\frac{1}{2}\times BC\times EG\therefore \frac{Ar\left(\Delta EDC\right)}{Ar\left(\Delta BEC\right)}=\frac{CD}{BC}=\frac{2}{3}\left[From\right(ii\left)\right]....\left(iv\right)\text{From (iii) and (iv), we get}\frac{Ar\left(\Delta ABE\right)}{Ar\left(\Delta BEC\right)}+\frac{Ar\left(\Delta EDC\right)}{Ar\left(\Delta BEC\right)}=\frac{1}{3}+\frac{2}{3}\Rightarrow \frac{Ar\left(\Delta ABE\right)}{Ar\left(\Delta EDC\right)}=\frac{1}{2}\Rightarrow Ar\left(\Delta EDC\right):Ar\left(\Delta ABE\right)=2:1$

Hence, the correct answer is option (b).