10.158. Three circles with radii $R_{1}=6 \mathrm{~cm}, R_{2}=7 \mathrm{~cm}, R_{3}=8 \mathrm{~cm}$ touch each other pairwise. Determine the area of the triangle whose vertices coincide with the centers of these circles.
## Solution.

The segments connecting the centers of these circles are: $a=R_{1}+R_{1}=6+7=13$ (cm), $b=R_{2}+R_{3}=7+8=15$ (cm), $c=R_{1}+R_{3}=$ $=6+8=14$ (cm). Then the area of this triangle can be found using Heron's formula, where $p=\frac{a+b+c}{2}=\frac{13+15+14}{2}=21$ (cm):

$$
S=\sqrt{p(p-a)(p-b)(p-c)}=\sqrt{21(21-13)(21-14)(21-15)}=84\left(\mathrm{~cm}^{2}\right)
$$

Answer: $84 \mathrm{~cm}^{2}$.