10.345. The radius of the circle inscribed in a triangle is 2 cm. A point of tangency of this circle divides one of the sides into segments of length 4 and $6 \mathrm{~cm}$. Determine the type of the triangle and calculate its area.
## Solution.

Let $O$ be the center of the circle with radius $r=2$ cm, inscribed in $\triangle ABC$ (Fig. 10.128), $M, N, K$ be the points of tangency of this circle with the sides $AB, BC, AC$ of the triangle, respectively, $AK=4 \text{ cm}, KC=6 \text{ cm}$. Then $AM=AK=4 \text{ cm}, CN=CK=6 \text{ cm}, BM=BN=x \text{ cm}$. The semiperimeter of the given triangle is $p=(x+10)$ cm and its area $S=pr=2(x+10)$ cm$^2$, or by Heron's formula:

$$
S=\sqrt{(10+x)(10+x-(x+4))(10+x-(x+6))(10+x-10)}=2 \sqrt{6 x(x+10)}
$$

We get: $2 \sqrt{6 x(x+10)}=2(x+10) ; \sqrt{6 x}=\sqrt{x+10} ; x=2$. Therefore, $S=2(x+10)=24 \text{ cm}^2$, and the sides of the triangle are 10 cm, 8 cm, and 6 cm, and since $8^2+6^2=10^2$, $\triangle ABC$ is a right triangle.

Answer: $24 \text{ cm}^2$; the triangle is a right triangle.