Subject: General Aptitude · Chapter: Area · Exam: · Marks: · Difficulty:
$A B C D$ is a rectangle and $E$ and $F$ are the mid-points of $A D$ and $D C$ respectively. Then the ratio of the areas of EDF and AEFC would be
(a)1 : 2
(b)1 : 3
(c)1 : 4
(d)2 : 3
Answer
Answer (as printed): B
Explanation
Let $A D=x$ and $D C=y$. Then, $A E=E D=\frac{x}{2}$ and $D E=F C=\frac{y}{2}$. area $(\triangle E D F)=\left(\frac{1}{2} \times \frac{x}{2} \times \frac{y}{2}\right)=\frac{x y}{8}$. $\operatorname{area}(\operatorname{trap} . A E F C)=\operatorname{area}(\triangle A D C)-\operatorname{area}(\triangle E D F)$ $$=\frac{x y}{2}-\frac{x y}{8}=\frac{3 x y}{8} .$$ ∴ Required ratio $=\frac{x y}{8}: \frac{3 x y}{8}=1: 3$.
Explanation as extracted from the printed page; notation may be imperfect.