Exercise 235

In the figure, the length of side $AB$ of square $ABCD$ is $\sqrt{85}$ and $BE=2$. What is the area of the inner square $EFGH$?

Solution

Since $BE^2+CE^2=AB^2$ and $BE=2$, we get $CE=9$.
The four congruent right triangles around the inner square have legs $BE=2$ and $CE=9$, so the inner square $EFGH$ has side $CE-BE$ and $[EFGH]=(CE-BE)^2=(9-2)^2=49$.