Right triangle inscribed in a square. Find the square area? I hope it's valid to ask for &quot;a more neat solution&quot; of a problem on this network, despite the fact that I don't have a strict definition of the word &quot;neat&quot;.
Here is the square and the right triangle inscribed in it.

I did the following:
$$AC = Ah + hB$$
$$4\sin\theta = 4\cos\theta + 3\sin\theta$$
So
$$\tan\theta = 4$$
But
$$\sin\theta = \frac{\tan\theta}{\sqrt{1+\tan^2\theta}}$$
Therefore
$$AC = 4\sin\theta = \frac{16}{\sqrt {17}}$$
$$\text{Area} = \left(\frac{16}{\sqrt {17}}\right)^2$$
Without using trigonometry:



enjoy...

By coloring triangles the same I am trying to emphasize the similar triangles, which are not necessarily equal (although red and blue triangles are). I don't mean that the triangles with the same color has the same area, be careful.
Noting CAh and hBf are similar triangles in the linear ratio 4:3, redraw the original figure in a 4x4 square. The square of its hypotenuse Ch² is therefore 17 (see figure) - which is too large, it should be 4² or 16. Since the area of any similar figure is proportional to the square of any of its linear dimensions, simply scale the 4x4 square by 16/17 to find the area of the original figure: 16²/17

redrawn figure