Question

Given an equilateral triangle T1 with side 24 cm, a second triangle T2 is formed by joining the midpoints of the sides of T1. Then a third triangle T3 is formed by joining the midpoints of the sides of T2. If this process of forming triangles is continued, the sum of the areas, in sq cm, of infinitely many such triangles T1, T2, T3,... will be

• A. 164√3
• B. 188√3
• C. 248√3
• D. 192√3

Answer 1

The Correct Option is D

An equilateral triangle with side length \( s \) has an area given by the formula:

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

Given an equilateral triangle \( T_1 \) with side 24 cm, its area is:

\( A_1 = \frac{\sqrt{3}}{4} \times 24^2 = 144\sqrt{3} \, \text{sq cm} \)

When the midpoints of the sides of an equilateral triangle are joined, the resulting triangle is similar to the original and has sides half the length. Consequently, the area of the new triangle is \(\left(\frac{1}{2}\right)^2= \frac{1}{4}\) of the original.

Thus, the area of triangle \( T_2 \) is:

\( A_2 = \frac{1}{4} A_1 = \frac{1}{4} \times 144\sqrt{3} = 36\sqrt{3} \, \text{sq cm} \)

Similarly, the area of triangle \( T_3 \) is: 

\( A_3 = \frac{1}{4} A_2 = \frac{1}{4} \times 36\sqrt{3} = 9\sqrt{3} \, \text{sq cm} \)

This process continues indefinitely forming a geometric series where each subsequent area is a quarter of the previous one. The series of areas \( A_1, A_2, A_3, \ldots \) forms a geometric series with first term \( a = 144\sqrt{3} \) and common ratio \( r = \frac{1}{4} \).

The sum of an infinite geometric series is given by:

\( S = \frac{a}{1-r} \)

Substituting the values:

\( S = \frac{144\sqrt{3}}{1-\frac{1}{4}} = \frac{144\sqrt{3}}{\frac{3}{4}} = 144\sqrt{3} \times \frac{4}{3} = 192\sqrt{3} \, \text{sq cm} \)

Therefore, the sum of the areas of infinitely many triangles \( T_1, T_2, T_3, \ldots \) is \( 192\sqrt{3} \, \text{sq cm} \).