The question is as such:
Triangle ABC is a equilateral triangle of side length 8 cm. Each arc shown in the diagram is an arc of a circle with the opposite vertex of the triangle as its center. The total area enclosed within the entire figure (in $\text{cm}^2$) is ____
I drew the large circles and managed to find the area using the formula for area of a sector and so forth my answer is $32(\pi - \sqrt{3})$.
Could someone please verify this answer as it is not available online?
Area of a sector of a circle: $$\frac{\theta}{360} \cdot \pi r^2$$
Now, considering each sector, as $\triangle ABC$ is equilateral, each angle is $60^{\circ}$, and each side is $8$ cm.
Area of one sector = $$\frac{60}{360} \cdot \pi \cdot 8^2 = \frac{1}{6} \cdot \pi \cdot 64$$
Since there are three sectors of the same size, total area of $3$ sectors is equivalent to 3 $\cdot$ area of one sector $$3 \cdot \frac{1}{6} \cdot 64 \cdot \pi = 32 \pi$$
But we notice, that on including the sector three times, we have also included the triangle three times, however in the figure, there is only one triangle, so we will need to subtract the area of $2$ triangles.
Area of an equilateral triangle = $$\frac{\sqrt{3}}{4} \cdot a^2 = \frac{\sqrt{3}}{4} \cdot 8^2 = 16\sqrt{3}$$
Area of $2$ triangles is $16\sqrt{3} \cdot 2 = 32\sqrt{3}$
On subtracting the area of two triangles from the area of 3 sectors (which was our goal):$$\\ 32\pi - 32\sqrt{3} \implies 32(\pi - \sqrt{3}) \tag*{$\blacksquare$}$$