Geometric Sequence – Triangles

One side of an equilateral triangle is 2 cms. The mid-points of its sides are joined to form another triangle whose mid-points are joined to form still another triangle. This process continues indefinitely.

(i) Find the sum of the perimeters of all the triangles.

(a) Cannot be determined
(b) 4
(c) 12
(d) 8
(e) 10

Bonus Question

(ii) Find the sum of the areas of all the triangles.

root(x) = square root of x.

(i) Choice (C)

(ii) 4/ root(3)

The side of the outermost equilateral triangle is given to be 2 cm. The mid-pts of the sides of this triangle are joined to form another triangle. Hence the side of this triangle would be 1 cm.

Proceeding the same way, the side of the next triangle would be ½ cm.

Thus, perimeter of the outermost triangle would be (2 + 2 + 2) = 6

The perimeter of the triangle that has been formed by joining the mid-points of the outermost triangle would be (1 + 1 + 1) = 3

Therefore, the sum of the perimeters of all such triangles formed infinitely would be

6 + 3 + 3/2 + ¾ + ……………….. which is nothing but sum up to infinity of a GP with first term as 6 and common ratio as ½.

Thus, a/(1-r) = 6/(1-0.5) = 12 cm.

To find out the area of an equilateral triangle we use the formula root(3)a²/4.

Hence, the infinite GP formed would be root(3)/4 [4 + 1 + (1/4) + ……………]

 = root(3)/4 [ 4/(1-0.25)]

= 4/root(3)