JEE Main 2024 — 6 April, Shift 2 — Question 1

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

Area of second $\Delta=\frac{\sqrt{3} \mathrm{a}^{2}}{4} \frac{1}{4}=\frac{\sqrt{3} \mathrm{a}^{2}}{16}$

Area of third $\Delta=\frac{\sqrt{3} a^{2}}{64}$

sum of area $=\frac{\sqrt{3} \mathrm{a}^{2}}{4}\left(1+\frac{1}{4}+\frac{1}{16} \ldots\right)$

$\mathrm{Q}=\frac{\sqrt{3} \mathrm{a}^{2}}{4} \frac{1}{\frac{3}{4}}=\frac{\mathrm{a}^{2}}{\sqrt{3}}$

perimeter of $1^{\text {st }} \Delta=3 \mathrm{a}$

perimeter of $2^{\text {nd }} \Delta=\frac{3 \mathrm{a}}{2}$

perimeter of $3^{\text {rd }} \Delta=\frac{3 \mathrm{a}}{4}$

$\mathrm{P}=3 \mathrm{a}\left(1+\frac{1}{2}+\frac{1}{4}+\ldots\right)$

$P=3 a .2=6 a$

$a=\frac{P}{6}$

$\mathrm{Q}=\frac{1}{\sqrt{3}} \cdot \frac{\mathrm{P}^{2}}{36}$

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