# Inscribed circle in a right-angled triangle

In right-angled $$\triangle ABC$$ with catheti $$a = 11\,\text{cm}, b=7\,\text{cm}$$ a circle has been inscribed. Find the radius and altitude from $$C$$ to the hypotenuse.

I found that the hypotenuse is $$c = \sqrt{170}$$ and the radius is $$r = \frac{a+b-c}{2}=\frac {18 - \sqrt{170}}{2}$$. I think that the altitude $$CH=$$ the sum of radii of circles inscribed in $$\triangle ABC, \triangle AHC, \triangle HBC$$ but I don't understand how I should calculate them. Thank you in advance!

• By the height do you mean the altitude of the triangle through the right angle? – TheSimpliFire Apr 21 '19 at 13:23
• @TheSimpleFire, yes. – Katherine Apr 21 '19 at 13:29

Let $$h$$ be the altitude of the triangle, whose line splits $$\sqrt{170}$$ into $$x$$ and $$\sqrt{170}-x$$. We know that the total area of the triangle is $$\frac12\cdot7\cdot11=\frac12h(\sqrt{170}-x)+\frac12hx\implies\text{altitude}=h=\frac{77}{\sqrt{170}}.$$
I assume you mean that $$CH$$ is the altitude from $$C$$ to the hypotenuse.
Since $$\triangle ABC$$ is a right triangle, and since $$\triangle AHC$$ and $$\triangle HBC$$ each share one angle with $$\triangle ABC$$, the three triangles are similar. We have $$\frac{CH}{AC} = \frac{BC}{AB}$$ and (equivalently) $$\frac{CH}{BC} = \frac{AC}{AB}.$$
The radii of the three incircles are proportional to $$AB,$$ $$AC,$$ and $$BC,$$ so their sum is \begin{align} \frac{a+b-c}{2} + \frac{a(a+b-c)}{2c} + \frac{b(a+b-c)}{2c} &= \frac{(a+b+c)(a+b-c)}{2c} \\ &= \frac{(a+b)^2 - c^2}{2c} \\ &= \frac{ab}{c}, \end{align} which is indeed equal to the altitude.