8. Within a circle of radius 1, three arcs of circumference, also of radius 1, are drawn, centered at the vertices of an inscribed equilateral triangle. What is the area of the shaded region?
(A) $\frac{\sqrt{3}}{4} \pi$
(B) $\pi-\frac{3 \sqrt{3}}{4}$
(C) $\pi-\frac{3 \sqrt{3}}{2}$
(D) $\frac{3 \sqrt{3}}{2}$
(E) $6-\pi$.

![](https://cdn.mathpix.com/cropped/2024_04_17_07a3c581d2eb0df332d0g-03.jpg?height=416&width=419&top_left_y=246&top_left_x=1498)
8. The answer is (D). The arcs drawn inside the circle have the same radius as the circle itself. Consequently, by equidecomposition of the circle, the area of the shaded part in Figure 1 is equal to that of the shaded part in Figure 2, which is equal to the area of the inscribed hexagon in a circumference of unit radius. The side of the hexagon is equal to the radius (i.e., it is 1); the area of the hexagon is 6 times that of the triangle \(O P Q\)

![](https://cdn.mathpix.com/cropped/2024_04_17_07a3c581d2eb0df332d0g-10.jpg?height=405&width=411&top_left_y=1234&top_left_x=1028)

Figure 1

![](https://cdn.mathpix.com/cropped/2024_04_17_07a3c581d2eb0df332d0g-10.jpg?height=411&width=417&top_left_y=1231&top_left_x=1436)

Figure 2

Consequently,

$$
\mathrm{A}_{\text {hexagon }}=6 \cdot \frac{1}{2} \cdot 1 \cdot \frac{\sqrt{3}}{2}=\frac{3 \sqrt{3}}{2} .
$$