Square inside a right angled triangle.

Question

If you have a square inscribed inside of a right angled triangle. Call the sides of the square $k$. The hypotenuse is $z$. How would you express either of the two other sides of the triangle in terms of $z$ and $k$? The corner of the square touches the hypotenuse-there isn't a side of the square against the hypotenuse.

Answer 1

Hint:

Call one angle of the right triangle $\theta$. The square produces two more smaller right triangles whose hypotenuses add up to $z$, i.e. $$k\sec\theta+k\csc \theta =z \\ \frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}=\frac zk \\ \frac{1+2t}{t^2}=\frac{z^2}{k^2}$$ where $t=\sin\theta \cos\theta$. You can solve for $t$ and consequently $\theta$ using this equation. Once you have $\theta$, the required sides will be $k+k\tan\theta$ and $k+k\cot\theta$.

Answer 2

Hint:

If the other sides are $x$ and $y$ then you have $$\sqrt{x^2+y^2}=z$$ $$\sqrt{(x-k)^2+k^2} +\sqrt{(y-k)^2+k^2} =z$$

which are two equations in the two unknowns. You should expect multiple solutions and so should check for spurious solutions introduced by squaring