Question

What is the area of the given figure EFGH and what kind of shape is it?

Answer 1

It is given that there are 4 right triangles: $\Delta HAE,\Delta EBF,\Delta FCG$, and $\Delta GDH$.
For $\Delta HAE,$
$H{E}^2=H{A}^2+A{E}^2$ (Using the Pythagorean theorem)
$\Rightarrow H{E}^2={12}^2+5^2=144+25=169$
$\Rightarrow H{E}^2=169$
Or
$HE=\sqrt{169}=\sqrt{13\times 13}=13\text{inches}$
$\Delta EBF,\Delta FCG$, and $\Delta GDH$ have the same measures of legs as $\Delta HAE$.
$\Rightarrow$ Hypotenuse of $\Delta EBF,\Delta FCG$, and $\Delta GDH$ is the same as $\Delta HAE$.
Hence,
$HE=EF=FG=GH=13\text{in}$
$\therefore EFGH$ is a square of side length 13 in. (As all the sides of EFGH is equal and one angle is ${90}^{\circ })$
Then, the area of $EFGH=13\times 13=169\text{sq inches}$. (Area of square = Side Side)
$\to$ Option C is correct.