# Inscribed Right Triangle Here’s a fun puzzle (via Brilliant.org): What is the area of the square $$ABCD$$?

There may be a simpler approach; my solution wound up being more complicated than I expected.

Since $$\Delta AEF$$ is a right triangle, $$AE = 5$$ and $$\sin{\gamma} = 3/5$$.

Let $$(FD, BE, AD) = (a, b, c)$$. We can disregard the other segments along the perimeter. We know that the area of the $$ABCD = c^2$$, so our goal is to calculate $$c$$.

For my next step, I rearranged the triangles. Since $$AB = AD$$, we can move $$\Delta AEB$$ to connect $$\overline{AD}$$ to $$\overline{AB}$$ to create $$\Delta AFE$$. What do we know about this triangle?

$$AF = 4$$ and $$AE = 5$$. $$m\angle{FAE} = \beta + \delta = \eta – \gamma$$ (where $$\eta = 90^\circ = \pi/2$$). Using the Law of Cosines, we can calculate $$a + b = BE + DF = EF$$. Specifically, $$EF^2 = 4^2 + 5^2 – 2(4)(5)\cos{(\eta – \gamma)}$$. Using an identity, $$\cos{(\eta – \gamma)} = \sin{\gamma}$$, so $$EF^2 = 16 + 25 – 40\sin{\gamma} = 41 – 40(3/5) = 17$$ and $$EF = \sqrt{17}$$.

Using Heron’s formula, the area of $$\Delta AFE = \sqrt{(5 + 4 + \sqrt{17})(5 + 4 – \sqrt{17})(5 – 4 + \sqrt{17})(-5 + 4 + \sqrt{17})}/4$$ $$= \sqrt{(9 + \sqrt{17})(9 – \sqrt{17})(\sqrt{17} + 1)(\sqrt{17} – 1)}/4$$ $$= \sqrt{(81 – 17)(17 – 1)}/4 = \sqrt{1024}/4 = 8$$.

That means the altitude of $$\Delta AFE = 2 \times 8 / \sqrt{17} = 16 / \sqrt{17} = AD$$, and the area of the square is $$(16/\sqrt{17})^2 = \frac{256}{17}$$.