JEE MAIN - Mathematics (2024 - 6th April Evening Shift - No. 12)

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 :
$$\mathrm{P}^2=72 \sqrt{3} \mathrm{Q}$$
$$\mathrm{P}^2=36 \sqrt{3} \mathrm{Q}$$
$$\mathrm{P}=36 \sqrt{3} \mathrm{Q}^2$$
$$\mathrm{P}^2=6 \sqrt{3} \mathrm{Q}$$

Explanation

JEE Main 2024 (Online) 6th April Evening Shift Mathematics - Sequences and Series Question 17 English Explanation

$$\triangle A B C$$ 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)

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

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