Question

The area of an equilateral \( \Delta ABC \) is \( 169\sqrt{3} \text{ cm}^2 \).  

Find the in-radius of \( \Delta ABC \).

• A. $10/\sqrt{3}$
• B. $13/\sqrt{3}$
• C. $12/\sqrt{3}$
• D. $11/\sqrt{3}$

Hint:

In an equilateral triangle, the circum-radius ($R$) is exactly double the in-radius ($r$). So, $R = a/\sqrt{3}$ and $r = a/2\sqrt{3}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
An equilateral triangle has all sides equal. The in-radius ($r$) is the radius of the circle that touches all three sides internally. It is related to the side length ($a$) of the triangle.

Step 2: Key Formula or Approach:
1. Area of an equilateral triangle = $\frac{\sqrt{3}}{4} a^2$
2. In-radius ($r$) of an equilateral triangle = $\frac{a}{2\sqrt{3}}$

Step 3: Detailed Explanation:
First, we find the side ($a$) using the given area: \[ \frac{\sqrt{3}}{4} a^2 = 169\sqrt{3} \] \[ a^2 = 169 \times 4 \] \[ a = \sqrt{169 \times 4} = 13 \times 2 = 26 \text{ cm} \] Now, we calculate the in-radius ($r$): \[ r = \frac{a}{2\sqrt{3}} \] \[ r = \frac{26}{2\sqrt{3}} \] \[ r = \frac{13}{\sqrt{3}} \text{ cm} \]

Step 4: Final Answer:
The in-radius of the triangle is $13/\sqrt{3}$.