# Square inside a right angled triangle.

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.

• Does another corner of the square coincide with the right angle of the triangle? May 30, 2020 at 20:15

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 $$k+k\cot\theta$$.

• I'm not sure how to solve for t and theta. Could you provide this solution? May 30, 2020 at 20:37
• @tomm0334 Can you solve the quadratic equation for $t$? You have to remember to take the positive root. After you have obtained $t$, $$\sin\theta \cos\theta =t \implies \frac 12 \cdot 2\sin\theta\cos\theta=t \implies \sin 2\theta =2t \implies \theta =\frac 12 \sin^{-1}(2t)$$ May 30, 2020 at 20:45
• I get a math error. I got t=4. When I put it in, I get math error on the calculator. May 30, 2020 at 21:06
• @tomm0334 $t=4$ is impossible. Why don’t you try calculating it by hand? May 31, 2020 at 9:47
• I don't understand how to solve by hand? May 31, 2020 at 10:20

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