Let A B C be an equilateral triangle.

Question

Let $A B C$ be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle $A B C$ and the same process is repeated infinitely many times. If $$\mathrm{P}$$ is the sum of perimeters and $Q$ is be the sum of areas of all the triangles formed in this process, then :

Accepted Answer

$$\triangle A B C$$ is an equilateral triangle having side $=a$ unit

Now, perimeter $=$ sum of all sides $=3 a$

Area $$=\frac{\sqrt{3}}{4} a^2$$

Now, $$\ln \triangle D E F, D E=\frac{a}{2}=E F=D F$$

$$\begin{aligned}
& \text { Perimeter }=3 \times \frac{a}{2}=\frac{3 a}{2} \
& \text { Area }=\frac{\sqrt{3}}{4} \times\left(\frac{a}{2}\right)^2=\frac{\sqrt{3} a^2}{16}
\end{aligned}$$

Now, $$P=3 a+\frac{3 a}{2}+\frac{3 a}{4}+\cdots$$

$$Q=\frac{\sqrt{3}}{4} a^2+\frac{\sqrt{3}}{16} a^2+\frac{\sqrt{3}}{64} a^2+\cdots$$

$$P=\frac{3 a}{1-\frac{1}{2}}=3 a \times 2 \Rightarrow P=6 a \quad \text{.... (i)}$$

$$Q=\frac{\frac{\sqrt{3}}{4} a^2}{1-\frac{1}{4}}=\frac{4}{3} \times \frac{\sqrt{3}}{4} a^2 \quad Q=\frac{\sqrt{3}}{3} a^2 \quad \text{.... (ii)}$$

From equation (i) \& (ii)

$$\begin{aligned}
& P=6 a \
& Q=\frac{\sqrt{3}}{3} a^2 \
& Q=\frac{\sqrt{3}}{3} \times \frac{P^2}{36} \
& P^2=36 \sqrt{3} Q
\end{aligned}$$

Solutions · 1