circleseuclidean-geometrygeometrytriangles
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!
From the Cosine Law,
$$AB^2+AC^2-BC^2=2\cdot AB\cdot AC\cdot \cos A \implies AB = 5 \text{ or } 16$$ The area of $\triangle ABC$ can be written as the following two ways: $$S_{ABC}=\frac{1}{2}\cdot AB\cdot AC\cdot \sin A = \frac{1}{2}\cdot(AB+AC+BC)\cdot r$$ where $r$ is the radius of the incircle of $\triangle ABC$. This will give us $$r=\frac{AB\cdot AC\cdot \sin60^\circ}{AB+AC+BC}$$ By plugging in the values of $AB, AC, BC$ and $\sin A$, we can get $r$. Further, because $BP=BQ,\ CQ=CR,\ AR=AP$, and $$BP+AP=AB\\BQ+CQ=BC\\AR+RC=AC$$ we know that $$BQ = \frac{BA+BC-AC}{2}$$Therefore, the following two cases are:
Case 1: $AB=5$, then $r=\frac{7\sqrt{3}}{6}$ and $BQ=\frac{3}{2}$.
Case 2: $AB = 16$, then $r = 3\sqrt{3}$ and $BQ = 7$.
$$( a - r ) + (b - r ) = 2 R $$
Best Answer
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 $$\frac12\cdot7\cdot11=\frac12h(\sqrt{170}-x)+\frac12hx\implies\text{altitude}=h=\frac{77}{\sqrt{170}}.$$