Question

A circle of radius $4$ cm is inscribed in an equilateral triangle T. Find the area of T (in cm${}^2$)

• A. $42\sqrt{3}$
• B. $45\sqrt{3}$
• C. $48\sqrt{3}$
• D. $50\sqrt{3}$

Answer 1

The correct option is C $48\sqrt{3}$

Solution:

To find:

Area of the $△ABC=?$

Solution:

Let $a$ be side length of the equilateral triangle.

So, height $AD=\frac{\sqrt{3}}{2}a$

We know that that centroid of the triangle divides the median into $2:1$ ratio.

So, $OD=4cm=\frac{1}{3}\times \frac{\sqrt{3}}{2}a$

or, $a=8\sqrt{3}cm$

Area of the $△ABC=\frac{\sqrt{3}}{4}{a}^2=\frac{\sqrt{3}}{4}(8\sqrt{3}{)}^2=48c{m}^2$

Hence, C is the correct option.