Radius of a circle inscribed in a right triangle is given, the hypotenuse is given, find the legs?

Question

The radius of the circle inscribed in the right triangle is 4, the hypotenuse is 20cm. How do I find the legs? Any help would be appreciated!

Answer 1

HINT:$$r = \cfrac{b+p - h}2$$,

$$b^2 + p^2 = h^2$$
Now you can solve to get answer

Answer 2

For the right triangle $ABC$ with the hypotenuse $|AB|=c=20$ and the radius of inscribed circle $r=4$, the radius of circumscribed circle is known to be found as \begin{align} R&=\tfrac{c}2=10 , \end{align}

and the semiperimeter \begin{align} \rho&=\tfrac12(a+b+c) =r+c , \end{align}

Considering the side lengths of the triangle as the roots of a general cubic polynomial in terms of semiperimeter $\rho$, inradius $r$ and circumradius $R$,

\begin{align} x^3-2\rho\,x^2+(\rho^2+r^2+4\,r\,R)\,x-4\,r\rho\,R , \end{align}

we can first simplify this cubic using the information given to

\begin{align} x^3-2(r+c) x^2+((r+c)^2+r^2+2r c) x-2(r+c)r c &= (x-c)(x^2-(2r+c)x+2(r+c)r) \\ \text{and find the legs as }\quad a,b&=r+\tfrac{c}2\pm\tfrac12\sqrt{c^2-4r(r+c)} =16,12 . \end{align}