A circular disc of radius 7 cm is inscribed inside an equilateral triangle. What is the approximate area of the remaining portion of the triangle?

Correct option is A

Given :

Radius of inscribed circle r = 7 cm

Formula Used :

Inradius of equilateral triangle: r $$= \dfrac{a\sqrt{3}}{6}$$​​

Area of equilateral triangle: A $$= \dfrac{\sqrt{3}}{4}a^2$$​​

Area of circle: A = $$\pi r^2$$​​

Solution :
a =$$\dfrac{6r}{\sqrt{3}} = \dfrac{6 \times 7}{\sqrt{3}} = 14\sqrt{3}$$​​

​$$\text{Area of triangle} = \dfrac{\sqrt{3}}{4}(14\sqrt{3})^2 = 147\sqrt{3} \approx 254.6 \text{ cm}^2$$​​

​$$\text{Area of circle} = \pi(7)^2 = 49\pi \approx 153.9 \text{ cm}^2$$​​

​$$\text{Remaining area} = 254.6 - 153.9 \approx 100.7 \text{ cm}^2$$​​