Question

One side of an equilateral triangle is $30$ cm. The mid-points of its sides are joined to form another triangle whose mid-points are in turn joined to form still another triangle. This process continuous indefinitely, then which of the following is/are correct?

• A. Sum of altitudes (One altitude per triangle) of all such triangles is $30\sqrt{3}$ cm.
• B. Sum of areas of all such triangles is $300\sqrt{3}$ sq. cm.
• C. Sum of radius of all the circumcircles of such triangles is $10\sqrt{3}$ cm.
• D. Sum of areas of all the incircles inscribed in such triangles is $100\pi$ sq. cm.

Answer 1

Answer: (A), (B), (D)

Sum of areas of all the incircles inscribed in such triangles is $100\pi$ sq. cm.

Let $a$ be the sides of $\Delta ABC$, then side of $\Delta A_1{B}_1{C}_1$ will be $\frac{a}{2}$ and so on.
$\text{Given,}a=30\text{cm}$
Sum of heights of all the triangles
$=\sqrt{3}\left(\frac{a}{2}\right)+\sqrt{3}\left(\frac{a}{4}\right)+\sqrt{3}\left(\frac{a}{8}\right)+...\infty$
$=\sqrt{3}a\left(\frac{\frac{1}{2}}{1-\frac{1}{2}}\right)$
$=\sqrt{3}a=30\sqrt{3}\text{cm}[\because a=30]$

Sum of areas of all the triangles
$=\frac{\sqrt{3}}{4}{a}^2+\frac{\sqrt{3}}{4}{\left(\frac{a}{2}\right)}^2+\frac{\sqrt{3}}{4}{\left(\frac{a}{4}\right)}^2+...\infty$
$=\frac{\sqrt{3}}{4}\left(\frac{a^2}{1-\frac{1}{4}}\right)$
$=\frac{\sqrt{3}}{3}{a}^2=300\sqrt{3}\text{sq. cm}$

Sum of radius of all the circumcircles
$=\frac{a}{\sqrt{3}}+\frac{1}{\sqrt{3}}\left(\frac{a}{2}\right)+\frac{1}{\sqrt{3}}\left(\frac{a}{4}\right)+...\infty$
$=\frac{a}{\sqrt{3}}\left(\frac{1}{1-\frac{1}{2}}\right)$
$=\frac{2a}{\sqrt{3}}=20\sqrt{3}\text{cm}$

Radius of incircle$=\frac{a\sqrt{3}}{6}$
$\therefore$ Area of incircle$=\frac{\pi a^2}{12}$
Sum of areas of all the incircles
$=\frac{\pi }{12}{a}^2+\frac{\pi }{12}{\left(\frac{a}{2}\right)}^2+\frac{\pi }{12}{\left(\frac{a}{4}\right)}^2+...\infty$
$=\frac{\pi }{12}\left(\frac{a^2}{1-\frac{1}{4}}\right)$
$=\frac{\pi a^2}{9}=100\pi \text{sq. cm}$