Question : In an equilateral triangle of side 24 cm, a circle is inscribed touching its sides. The area of the remaining portion of the triangle is:
Option 1: $98.55\;\mathrm{cm^2}$
Option 2: $100 \;\mathrm{cm^2}$
Option 3: $101 \;\mathrm{cm^2}$
Option 4: $95\;\mathrm{cm^2}$
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Correct Answer: $98.55\;\mathrm{cm^2}$
Solution :
In an equilateral triangle, the radius of the inscribed circle, where $a$ is the side of the equilateral triangle.
$r = \frac{a\sqrt{3}}{6}$
Put $a=24\text{ cm}$,
$r = \frac{24\sqrt{3}}{6} = 4\sqrt{3} \text{ cm}$
The area of the inscribed circle, where $r$ is the radius of the circle.
$\text{Area}_{\text{circle}} =\pi r^2$
$\text{Area}_{\text{circle}} = \pi (4\sqrt{3})^2 = 16\pi \times 3 = 48\pi=150.85 \text{ cm}^2$
The area of the equilateral triangle,
$\text{Area}_{\text{triangle}} =\frac{\sqrt{3}}{4}a^2$
$\text{Area}_{\text{triangle}} = \frac{\sqrt{3}}{4} \times 24^2 = 144\sqrt{3}=249.4 \text{ cm}^2$
The area of the remaining portion of the triangle,
$ \text{Area}_{\text{triangle}} - \text{Area}_{\text{circle}} = 249.4 - 150.85=98.55 \text{ cm}^2$
Hence, the correct answer is $98.55 \text{ cm}^2$.
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