10.296. Find the area of a right-angled triangle if the radii $R$ and $r$ of the circumscribed and inscribed circles are given.
Solution.

Let $a, b, c$ be the legs and hypotenuse of the given triangle, respectively. $c=2R$, $r=\frac{a+b-c}{2}=\frac{a+b-2R}{2}$, $a+b=2(R+r)$. But $a^{2}+b^{2}=c^{2}$, then $(a+b)^{2}-2ab=4R^{2}$, $2ab=(a+b)^{2}-4R^{2}=4(R+r)^{2}-4R^{2}=8Rr+4r^{2}$, $ab=2(2Rr+r^{2})=2r(2R+r)$. The area of the triangle $S=\frac{1}{2}ab=r(2R+r)$.

Answer: $r(2R+r)$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0777.jpg?height=519&width=830&top_left_y=673&top_left_x=267)

Fig. 10.89