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…

Area of first $\Delta=\frac{\sqrt{3} \mathrm{a}^2}{4}$Δ=43​a2​ Area of second $\Delta=\frac{\sqrt{3} \mathrm{a}^2}{4} \frac{\mathrm{a}^2}{4}=\frac{\sqrt{3} \mathrm{a}^2}{16}$Δ=43​a2​4a2​=163​a2​ Area of third $\Delta=\frac{\sqrt{3} \mathrm{a}^2}{64}$Δ=643​a2​ sum of area $=\frac{\sqrt{3} \mathrm{a}^2}{4}\left(1+\frac{1}{4}+\frac{1}{16} \ldots\right)$=43​a2​(1+41​+161​…) $\begin{aligned} & \mathrm{Q}=\frac{\sqrt{3} \mathrm{a}^2}{4} \frac{1}{\frac{3}{4}}=\frac{\mathrm{a}^2}{\sqrt{3}} \\ & \text { perimeter of } 1^{\text {st }} \Delta=3 \mathrm{a} \\ & \text { perimeter of } 2^{\text {nd }} \Delta=\frac{3 \mathrm{a}}{2} \\ & \text { perimeter of } 3^{\text {rd }} \Delta=\frac{3 \mathrm{a}}{4} \\ & \mathrm{P}=3 \mathrm{a}\left(1+\frac{1}{2}+\frac{1}{4}+\ldots\right) \\ & \mathrm{P}=3 \mathrm{a} \cdot 2=6 \mathrm{a} \\ & \mathrm{a}=\frac{\mathrm{P}}{6} \\ & \mathrm{Q}=\frac{1}{\sqrt{3}} \cdot \frac{\mathrm{P}^2}{36} \\ & \mathrm{P}^2=36 \sqrt{3} \mathrm{Q}\end{aligned}$​Q=43​a2​43​1​=3​a2​ perimeter of 1st Δ=3a perimeter of 2nd Δ=23a​ perimeter of 3rd Δ=43a​P=3a(1+21​+41​+…)P=3a⋅2=6aa=6P​Q=3​1​⋅36P2​P2=363​Q​