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Question
One side of an equilateral triangle is 18 cm. The mid-points of its sides are joined to form another triangle whose mid-points, in turn, are joined to form still another triangle. The process is continued indefinitely. Find the sum of the (i) perimeters of all the triangles. (ii) areas of all triangles.
Solution
According to the midpoint theorem, the sides of each triangle formed by joining the midpoints of an equilateral triangle are half of the sides of the equilateral triangle. In other words, the triangles formed are equilateral triangles with sides 18 cm, 9 cm, 4.5 cm, 2.25 cm, ...
\[(i) \text { Sum of the perimeters of all the triangles }, P = 3 \times 18 + 3 \times 9 + 3 \times 4 . 5 + 3 \times 2 . 25 + . . . \infty \]
\[ \Rightarrow P = 3 \times \left( 18 + 9 + 4 . 5 + 2 . 25 + . . . \infty \right)\]
\[\text { It is a G . P . with a = 18 and r } = \frac{1}{2} . \]
\[ \therefore P = 3 \times \left( \frac{18}{1 - \frac{1}{2}} \right)\]
\[ \Rightarrow P = 3 \times 36 = 108 cm\]
\[(ii) \text { Sum of the areas of all the triangles, A } = \frac{\sqrt{3}}{4} \left( 18 \right)^2 + \frac{\sqrt{3}}{4} \left( 9 \right)^2 + \frac{\sqrt{3}}{4} \left( 4 . 5 \right)^2 + . . . \infty \]
\[ \Rightarrow A = \frac{\sqrt{3}}{4}\left( \left( 18 \right)^2 + \left( 9 \right)^2 + \left( 4 . 5 \right)^2 + . . . \infty \right)\]
\[\text { It is a G . P . with a } = \left( 18 \right)^2 \text { and } r = \frac{1}{4} . \]
\[ \therefore A = \frac{\sqrt{3}}{4}\left( \frac{\left( 18 \right)^2}{1 - \frac{1}{4}} \right)\]
\[ \Rightarrow A = \frac{\sqrt{3}}{3} \times 324\]
\[ \Rightarrow A = 108\sqrt{3} {cm}^2\]
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