Question

An equilateral triangle is drawn by joining the midpoints of a given equilateral triangle of side 18 cm. A third equilateral triangle is drawn inside the second in the same way. The process is constructed indefinitely. The sum of the areas of all such triangles in $c{m}^2$ is

• A. $324\sqrt{3}$
• B. $108\sqrt{3}$
• C. $54\sqrt{3}$
• D. $200\sqrt{3}$

Answer 1

The correct option is B $108\sqrt{3}$

Area of an equilateral triangle of side $a$ is given by $\sqrt{3}\frac{a^2}{4}$
Now, the side of the equilateral triangle formed by joining the midpoints of the original triangle is $\frac{a}{2}$
Hence, the corresponding area is $\sqrt{3}\frac{{\left(\frac{a}{2}\right)}^2}{4}$
Similarly, the side of next triangle inside this triangle will be $\frac{a}{4}$

Hence, its area will be $\sqrt{3}\frac{{\left(\frac{a}{4}\right)}^2}{4}$
So, the sum of all these areas is an infinite geometric series with first term as $\sqrt{3}\frac{a^2}{4}$ and with common ratio $1/4$
Sum of infinite geometric series is $\frac{firstterm}{1-commonratio}\left(\frac{a}{1-r}\right)$
Given that the side of the original triangle is $18cm$

Hence, $a=\frac{\sqrt{3}(18{)}^2}{4}=81\sqrt{3}$ and $r=\frac{1}{4}$

Putting the values of $a$ and $r$ in the formula and rationalizing, we get the answer as $108\sqrt{3}$