Question

A circle has a radius of 6 in. The circumscribed equilateral triangle will have an area of:

Answer 1

area of triangle = √3/4*6²
                         =√3/4*36
                         =√3*9 
                         =9√3
ans is =9√3

Answer 2

Answer:

Area of the circumscribed equilateral triangle is $27\sqrt{3}\text{ inches}^2$

Step-by-step explanation:

Given : A circle has a radius of 6 in.

To find : The circumscribed equilateral triangle will have an area of ?

Solution :

The radius of the circle is 6 inches.

First we draw a rough image for clarification.

Refer the attached figure below.

Let APB be the equilateral triangle in which a circle with center C is form

According to question and figure,

Side a = 6 in.

∠A = ∠B = 30°

∠C = 120°

In ΔOBC,

Applying sine rule,

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\\\\\text{Taking first and third terms}\\\\\implies \frac{a}{\sin A}=\frac{c}{\sin C}\\\\\implies c = a\times (\frac{\sin C}{\sin A})\\\\\implies c = 6\times (\frac{\sin 120}{\sin 30})\\\\\implies c =6\sqrt{3}$

Now, we get the side of the equilateral triangle we find the area

$Area=\frac{\sqrt{3}}{4}\times Side^2\\\\Area = \frac{\sqrt{3}}{4}\times c^2\\\\Area=\frac{\sqrt{3}}{4}\times 6^2\\\\\implies Area = 27\sqrt{3}\text{ inches}^2$

Therefore, Area of the circumscribed equilateral triangle is $27\sqrt{3}\text{ inches}^2$