Question

In the below diagram ABCD is a rectangle with AE = EF = FB What is the ratio of the areas of $△CEF$ and that of the rectangle?

• A. $\frac{1}{6}$
• B. $\frac{1}{8}$
• C. $\frac{1}{9}$
• D. None of these

Answer 1

The correct option is A $\frac{1}{6}$

Let length of rectangle be ${}^{\prime }{l}^{\prime }$ and breadth be ${}^{\prime }{b}^{\prime }$. Then

$AB=l$ & $BC=b$. Also

$AE=EF=FB$ $\therefore$ $AE=EF=FB=$$\frac{l}{3}$

ar $ar△CEF=ar△BEC-ar△BFC=\frac{1}{2}\times \frac{2l}{3}\times b-\frac{1}{2}\times \frac{l}{3}\times b$

$=\frac{lb}{3}-\frac{lb}{6}=\frac{lb}{6}$

Now, $\frac{ar△CEF}{ar△ABCD}=\frac{lb/6}{lb}=\frac{1}{6}=1:6$