IPMAT Rohtak 2019 (QA) - A circle is inscribed in an equilateral triangle of side 24 cm, touching its sides. What is the area of the remaining portion of the triangle? | PYQs + Solutions

IPMAT Rohtak 2019

Geometry

Triangles

Medium

A circle is inscribed in an equilateral triangle of side 24 cm, touching its sides. What is the area of the remaining portion of the triangle?

144\sqrt{3} - 48\pi cm^2

121\sqrt{3} - 36\pi cm^2

144\sqrt{3} - 36\pi cm^2

121\sqrt{3} - 48\pi cm^2

Correct Option: 1

For an equilateral triangle with inscribed circle:

\text{Side } = a = 24 \text{ cm}

\text{Area of equilateral triangle} = \frac{\sqrt{3}}{4} \times a^2

= \frac{\sqrt{3}}{4} \times 24^2

= \frac{\sqrt{3}}{4} \times 576

= 144\sqrt{3} \text{ cm}^2

\text{Radius of inscribed circle} = \frac{a}{\sqrt{12}} = \frac{24}{\sqrt{12}} = 4\sqrt{3} \text{ cm}

\text{Area of inscribed circle} = \pi r^2

\newline

= \pi(4\sqrt{3})^2

\newline

= 48\pi \text{ cm}^2

\text{Remaining area} = \text{Area of triangle} - \text{Area of circle}

\newline

= 144\sqrt{3} - 48\pi \text{ cm}^2

Therefore, the answer is

144\sqrt{3} - 48\pi \text{ cm}^2