An equilateral triangle is drawn by joining the mid-points of a given equilateral triangle, A third equilateral triangle is drawn inside the second in the same manner and the process is continued indefinitely. If the side of the first equilateral triangle is `3^(1//4)` inch, then the sum of the areas in sq. inch of all these triangles is

Options

A. `sqrt(3)`
B. 3
C. `3 sqrt(3)`
D. 1

Answer

To find the sum of the areas of all the equilateral triangles formed by repeatedly joining the midpoints of the previous triangle, we can follow these steps:

### Step 1: Calculate the area of the first triangle
The formula for the area \( A \) of an equilateral triangle with side length \( s \) is given by:
\[ A = \frac{\sqrt{3}}{4} s^2 \]
Given that the side length of the first triangle is \( s = 3^{1/4} \), we can substitute this value into the formula:
\[ A_1 = \frac{\sqrt{3}}{4} \left(3^{1/4}\right)^2 \]
Calculating \( \left(3^{1/4}\right)^2 \):
\[ \left(3^{1/4}\right)^2 = 3^{1/2} = \sqrt{3} \]
Thus, the area of the first triangle becomes:
\[ A_1 = \frac{\sqrt{3}}{4} \cdot \sqrt{3} = \frac{3}{4} \text{ square inches} \]

### Step 2: Calculate the area of the second triangle
The area of the second triangle \( A_2 \) is one-fourth of the area of the first triangle:
\[ A_2 = \frac{1}{4} A_1 = \frac{1}{4} \cdot \frac{3}{4} = \frac{3}{16} \text{ square inches} \]

### Step 3: Calculate the area of the third triangle
Similarly, the area of the third triangle \( A_3 \) is one-fourth of the area of the second triangle:
\[ A_3 = \frac{1}{4} A_2 = \frac{1}{4} \cdot \frac{3}{16} = \frac{3}{64} \text{ square inches} \]

### Step 4: Generalize the area of the \( n \)-th triangle
We can see a pattern here:
- \( A_1 = \frac{3}{4} \)
- \( A_2 = \frac{3}{16} = \frac{3}{4} \cdot \left(\frac{1}{4}\right) \)
- \( A_3 = \frac{3}{64} = \frac{3}{4} \cdot \left(\frac{1}{4}\right)^2 \)

In general, the area of the \( n \)-th triangle can be expressed as:
\[ A_n = \frac{3}{4} \left(\frac{1}{4}\right)^{n-1} \]

### Step 5: Find the sum of the areas of all triangles
The total area \( S \) of all triangles is the sum of an infinite geometric series:
\[ S = A_1 + A_2 + A_3 + \ldots = \frac{3}{4} + \frac{3}{16} + \frac{3}{64} + \ldots \]
This can be expressed as:
\[ S = \frac{3}{4} \left(1 + \frac{1}{4} + \left(\frac{1}{4}\right)^2 + \ldots\right) \]
The series inside the parentheses is an infinite geometric series with the first term \( a = 1 \) and common ratio \( r = \frac{1}{4} \). The sum of an infinite geometric series is given by:
\[ \text{Sum} = \frac{a}{1 - r} = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \]
Thus, substituting back into our sum:
\[ S = \frac{3}{4} \cdot \frac{4}{3} = 1 \text{ square inch} \]

### Final Answer
The sum of the areas of all these triangles is \( \boxed{1} \) square inch.

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