A right-angled triangle has sides of length 6,8 and 10 . A circle is drawn so that the area inside the circle but outside this triangle equals the area inside the triangle but outside the circle. The radius of the circle is closest to
(A) 2.9
(B) 2.8
(C) 3.8
(D) 3.1
(E) 3.6

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-118.jpg?height=236&width=236&top_left_y=717&top_left_x=1454)
Let the area that is inside the triangle and inside the circle be $A$, and let the area outside the triangle but inside the circle be $B$. Then $B$ is also equal to the area outside the circle but inside the triangle.

We can then see that $A+B$ is equal to the area of the circle and is also equal to the area of the

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-231.jpg?height=271&width=271&top_left_y=987&top_left_x=1488)
triangle, so the circle and triangle have the same area.

Thus, if $r$ is the radius of the circle, then $\pi r^{2}=\frac{1}{2}(6)(8)$ or $\pi r^{2}=24$ or $r=\sqrt{\frac{24}{\pi}} \approx 2.8$.

Answer: (B)