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 \((16/\sqrt{17})^2 = \frac{256}{17}\).