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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!

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  • $\begingroup$ By the height do you mean the altitude of the triangle through the right angle? $\endgroup$
    – TheSimpliFire
    Apr 21, 2019 at 13:23
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    $\begingroup$ @TheSimpleFire, yes. $\endgroup$ Apr 21, 2019 at 13:29

2 Answers 2

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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 $$\frac12\cdot7\cdot11=\frac12h(\sqrt{170}-x)+\frac12hx\implies\text{altitude}=h=\frac{77}{\sqrt{170}}.$$

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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.

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