In an equilateral triangle of side 3 cm , a circle is inscribed in which again an equliateral triagle is
inscribed and so on . This continues for an infinite
number of times . Then

Options

A. Sum of areas of all circles (in `cm^(2)) "is" sqrt(3)pi`
B. Sum of areas of all circles (in `cm^(2)) "is" pi`
C. Sum of areas of all equilateral triangles (in `cm^(2))" is " 3sqrt(3)`
D. Sum of areas of all equilateral triangles (in ` cm^(2)) "is" ssqrt(3)`

Answer

To solve the problem step by step, we will find the area of the infinite series of inscribed equilateral triangles and circles within the original triangle of side 3 cm.

### Step 1: Find the radius of the inscribed circle in the original triangle.

For an equilateral triangle, the radius \( r \) of the inscribed circle can be calculated using the formula:

\[ r = \frac{a \sqrt{3}}{6} \]

where \( a \) is the side length of the triangle.

Given \( a = 3 \) cm:

\[ r = \frac{3 \sqrt{3}}{6} = \frac{\sqrt{3}}{2} \text{ cm} \]

**Hint:** Remember that the radius of the inscribed circle in an equilateral triangle is derived from its side length.

### Step 2: Find the side length of the first inscribed triangle.

The side length \( s_1 \) of the triangle inscribed in the circle can be calculated using the formula:

\[ s_1 = 2r \sin(60^\circ) \]

Substituting \( r = \frac{\sqrt{3}}{2} \):

\[ s_1 = 2 \left(\frac{\sqrt{3}}{2}\right) \cdot \frac{\sqrt{3}}{2} = \frac{3}{2} \text{ cm} \]

**Hint:** Use the sine of 60 degrees, which is \( \frac{\sqrt{3}}{2} \), to find the side length of the triangle inscribed in the circle.

### Step 3: Find the side lengths of subsequent triangles.

The side lengths of the triangles form a geometric sequence:

- First triangle: \( s_1 = 3 \)
- Second triangle: \( s_2 = \frac{3}{2} \)
- Third triangle: \( s_3 = \frac{3}{4} \)
- Fourth triangle: \( s_4 = \frac{3}{8} \)
- And so on...

The general term for the side length of the \( n \)-th triangle is:

\[ s_n = \frac{3}{2^{n-1}} \]

**Hint:** Recognize the pattern in the side lengths; they decrease by half for each subsequent triangle.

### Step 4: Find the area of the triangles.

The area \( A_n \) of an equilateral triangle can be calculated using the formula:

\[ A_n = \frac{\sqrt{3}}{4} s_n^2 \]

Substituting \( s_n = \frac{3}{2^{n-1}} \):

\[ A_n = \frac{\sqrt{3}}{4} \left(\frac{3}{2^{n-1}}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{9}{4^{n-1}} = \frac{9\sqrt{3}}{4^n} \]

**Hint:** Use the formula for the area of an equilateral triangle and substitute the expression for \( s_n \).

### Step 5: Sum the areas of all triangles.

The total area of all triangles is given by the sum of an infinite geometric series:

\[ A = \sum_{n=1}^{\infty} A_n = \sum_{n=1}^{\infty} \frac{9\sqrt{3}}{4^n} \]

This is a geometric series with first term \( a = \frac{9\sqrt{3}}{4} \) and common ratio \( r = \frac{1}{4} \):

\[ A = \frac{a}{1 - r} = \frac{\frac{9\sqrt{3}}{4}}{1 - \frac{1}{4}} = \frac{\frac{9\sqrt{3}}{4}}{\frac{3}{4}} = 3\sqrt{3} \]

**Hint:** Use the formula for the sum of an infinite geometric series to find the total area.

### Final Result

The total area of all the inscribed equilateral triangles is:

\[ \text{Total Area} = 3\sqrt{3} \text{ cm}^2 \]