Question

In the diagram, $ABCD$ is a rectangle with $AE=EF=FB$. What is the ratio of the areas of $\Delta CEF$ and that of the rectangle ?

• A. $\frac{1}{6}$
• B. $\frac{1}{8}$
• C. $\frac{1}{9}$
• D. $\frac{1}{4}$

Answer 1

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

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

Let length of rectangle be $l$ units & breadth of $b$ units.

Now, $ar\Delta CBE=\frac{1}{2}\times BE\times BC=\frac{1}{2}\times \frac{2l}{3}\times b=\frac{lb}{3}$

$ar\Delta CBF=\frac{1}{2}\times BF\times BC=\frac{1}{2}\times \frac{l}{3}\times b=\frac{lb}{6}$

$\therefore$ $ar\Delta CEF=ar\Delta CBE-ar\Delta CBF=\frac{lb}{3}-\frac{lb}{6}=\frac{lb}{6}$

Now, $\frac{ar\Delta CEF}{arABCD}=\frac{\frac{lb}{6}}{lb}=\frac{1}{6}or1:6$