QJEEM24Ma8ddd: JEE Main 2024 Mathematics - Class 11 JEEM24Ma8ddd Solution: Let be an equilateral triangle. A new triangle is formed by…

$\triangle A B C$ 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)

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