Question

What is the area of an equilateral triangle whose inscribed circle has radius R?

• A. \(4R^2\)
• B. \(4\sqrt{3}R^2\)
• C. \((3 + 2\sqrt{2})R^2\)
• D. \(3\sqrt{3}R^2\)

Answer 1

The Correct Option is D

To find the area of an equilateral triangle whose inscribed circle has radius \( R \), we begin by understanding the relationship between the radius of the inscribed circle (inradius) and the side length of the triangle.

For an equilateral triangle with side length \( a \), the inradius \( r \) is given by:

\( r = \frac{a \sqrt{3}}{6} \)

We are given that the inradius \( r = R \). Therefore,

\( R = \frac{a \sqrt{3}}{6} \)

Solving for \( a \), we get:

\( a = \frac{6R}{\sqrt{3}} = 2\sqrt{3}R \)

Now, the area \( A \) of an equilateral triangle with side length \( a \) is given by:

\( A = \frac{\sqrt{3}}{4}a^2 \)

Substituting \( a = 2\sqrt{3}R \) into the area formula:

\( A = \frac{\sqrt{3}}{4}(2\sqrt{3}R)^2 \)

Calculating further:

\( A = \frac{\sqrt{3}}{4} \times 4 \times 3R^2 \)

\( A = 3\sqrt{3}R^2 \)

Thus, the area of the equilateral triangle in terms of \( R \) is \( 3\sqrt{3}R^2 \).

Therefore, the correct answer is \( 3\sqrt{3}R^2 \) .