Question

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

Answer 1

Given is that

$ABCD$ is a rectangle

$AE=EF=FB$

Now to find the ratio of $\Delta CEF$ in the $\square ABCD$

Now,

area of $\Delta CEF=$ area of $\Delta CEB-$ area of $\Delta CFB$

$\left\{\begin{array}{c}AB=EF=FB\\ EB=2EF\end{array}\right\}$

area of $\Delta CEF=\frac{1}{2}\times EB\times BC-\frac{1}{2}EB\times BC$

area of $\Delta CEF=\frac{1}{2}\left(\frac{2AB}{3}\right)\times BC-\frac{1}{2}\left(\frac{AB}{3}\right)\times BC$

$\{AB=EF=FB\}$

area of $\Delta CEF=\frac{1}{2}[\frac{AB}{3}\times BC]$

area of $\Delta CEF=\frac{1}{6}AB\times BC$

area of $\Delta CEF=\frac{1}{6}$

area of $\square ABCD$

{$AB\times BC$ is area of rectangle}