6. The base and leg of an isosceles triangle are in the ratio $4: 3$. The length of the height to the base is $20 \mathrm{~cm}$. What is the length of the radius of the circle inscribed in this triangle?
First method:

Sketch:

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![](https://cdn.mathpix.com/cropped/2024_05_30_64eca78245e6b7f11e3dg-4.jpg?height=434&width=668&top_left_y=1893&top_left_x=291)

Let $a$ be the length of the base, $b$ be the length of the leg, and $r$ be the radius of the inscribed circle of the triangle.

Triangles $\triangle C C_{1} B$ and $\triangle S D C$ are similar by the $\mathrm{AA}$ (Angle-Angle) similarity criterion:

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the triangles are right-angled, and angle $\angle D C S$ is common to both.

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Therefore, the ratio of the corresponding sides is: $\left|C_{1} B\right|:|B C|=|S D|:|S C|$.

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$\left|C_{1} B\right|=\frac{a}{2},|S C|=20-r$,

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so $\frac{a}{2}: b=r:(20-r)$.

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Since $a: b=4: 3$, then $2: 3=r:(20-r)$,

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from which we get $r=8 \mathrm{~cm}$.

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