Progressions

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

  1. 164√3
  2. 188√3
  3. 248√3
  4. 192√3

Answer

Any equilateral triangle formed by joining the midpoints of the sides of another equilateral triangle will have its side equal to half the side of the second equilateral triangle.

So, side of T1 = 24 cm, side of T2 = 12 cm, side of T3 = 6 cm and so on.

Area of Equilateral Triangle = √3/4 × a2

Sum of the areas of all triangles

= √3/4 × (242 + 122 + 62 + ...)

This is geometric progression with common ratio of 1/4.

= 3√4 × [576/(1 - 1/4)]

= 192√3

The correct option is B.