The length of a side of an equilateral triangle is 8 cm. The area of the region lying between the circum circle and the incircle of the triangle is (Use `pi = (22)/(7)`)

Options

A. `50(1)/(7) cm^(2)`
B. `50(2)/(7) cm^(2)`
C. `75 (1)/(7) cm^(2)`
D. `75(2)/(7) cm^(2)`

Answer

To find the area of the region lying between the circumcircle and the incircle of an equilateral triangle with a side length of 8 cm, we will follow these steps:

### Step 1: Calculate the circumradius (R) and inradius (r) of the equilateral triangle.

For an equilateral triangle:
- The circumradius \( R \) is given by the formula:
\[ R = \frac{a}{\sqrt{3}} \]
- The inradius \( r \) is given by the formula:
\[ r = \frac{a}{2\sqrt{3}} \]
where \( a \) is the length of a side of the triangle.

Given \( a = 8 \) cm:
- Circumradius \( R \):
\[ R = \frac{8}{\sqrt{3}} \text{ cm} \]
- Inradius \( r \):
\[ r = \frac{8}{2\sqrt{3}} = \frac{4}{\sqrt{3}} \text{ cm} \]

### Step 2: Calculate the area of the circumcircle and the incircle.

The area \( A \) of a circle is given by the formula:
\[ A = \pi R^2 \]

- Area of the circumcircle:
\[ A_{circum} = \pi R^2 = \pi \left(\frac{8}{\sqrt{3}}\right)^2 = \pi \cdot \frac{64}{3} \]

- Area of the incircle:
\[ A_{in} = \pi r^2 = \pi \left(\frac{4}{\sqrt{3}}\right)^2 = \pi \cdot \frac{16}{3} \]

### Step 3: Find the area of the region between the circumcircle and the incircle.

To find the area between the circumcircle and the incircle, subtract the area of the incircle from the area of the circumcircle:
\[ A_{between} = A_{circum} - A_{in} \]
Substituting the areas we calculated:
\[ A_{between} = \left(\pi \cdot \frac{64}{3}\right) - \left(\pi \cdot \frac{16}{3}\right) \]
Factoring out \( \pi \):
\[ A_{between} = \pi \left(\frac{64}{3} - \frac{16}{3}\right) = \pi \cdot \frac{48}{3} = \pi \cdot 16 \]

### Step 4: Substitute the value of \( \pi \).

Using \( \pi = \frac{22}{7} \):
\[ A_{between} = \frac{22}{7} \cdot 16 = \frac{352}{7} \text{ cm}^2 \]

### Final Answer:
The area of the region lying between the circumcircle and the incircle of the triangle is:
\[ \frac{352}{7} \text{ cm}^2 \]