The area of an equilateral triangle ABC is 17320.5 cm
Question:

The area of an equilateral triangle $\mathrm{ABC}$ is $17320.5 \mathrm{~cm}^{2}$. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle. Find the area of the shaded region.

(Use $\pi=3.14$ and $\sqrt{\mathbf{3}}=1.73205$ )

Solution:

Area of the $\triangle \mathrm{ABC}$ (equilateral) $=17320.5 \mathrm{~cm}^{2}$

Let the side of the equilateral $\triangle \mathrm{ABC}$ be $\mathrm{x} \mathrm{cm}$.

Then, $\frac{\sqrt{3}}{4} \times x^{2}=17320.5$

$\Rightarrow \frac{1.73205}{4} \times x^{2}=17320.5(\because \sqrt{3}=1.73205)$

$\Rightarrow x^{2}=40000 \Rightarrow x=200 \mathrm{~cm}$

Then, radius of each circle = 100 cm.

Area of the sector APR

$=\frac{\mathbf{C D}}{\mathbf{3 6 0}} \times \pi \times(100)^{2} \mathrm{~cm}^{2}=\frac{\pi}{\mathbf{6}} \times \mathbf{1 0 0 0 0} \mathbf{~ c m}^{2}$

Similarly, area of the sector BPQ = area of the sector CQR

$=\frac{\pi}{6} \times 10000 \mathrm{~cm}^{2}$

Total area of regions I, II and III (i.e., non-shaded region of $\triangle \mathrm{ABC}$ )

$=3 \times \frac{\pi}{\mathbf{6}} \times \mathbf{1 0 0 0 0} \mathrm{cm}^{2}=\frac{\mathbf{1}}{\mathbf{2}} \times 3.14 \times 10000 \mathrm{~cm}^{2}$

$=15700 \mathrm{~cm}^{2}$

Then, the required area of the shaded region of $\triangle \mathrm{ABC}$

$=17320.5 \mathrm{~cm}^{2}-15700 \mathrm{~cm}^{2}=1620.5 \mathrm{~cm}^{2}$

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