Let ABC be an equilateral triangle. A new triangle is formed byn joining the middle points of all sides of the triangle ABC and then same process is repeated infinitely many times. If P is the sum ofn perimeters and Q is the sum of areas of all the triangles formedn in this process, then

Correct option is (1) P^2=36 sqrt(3) Q Area of first Delta=fracsqrt(3 a^2)4 Area of second Delta=fracsqrt(3) a^24 frac a^24=fracsqrt(3) a^216 Area of third Delta=fracsqrt(3) a^264 sum of area =fracsqrt3 a^24(1+(1 / 4)+(1 / 16) ... ) Q=fracsqrt(3) a^24 frac1(3 / 4)=frac a^2sqrt(3)perimeter of 1^ st Delta=3 aperimeter of 2^ nd Delta \ =frac3 a2perimeter of 3^ rd Delta=frac3 a4\ P=3 a(1+(1 / 2)+(1 / 4)+ ... ) P=3 a * 2=6 a a=(P / 6) Q=frac1sqrt(3) * frac P^236 P^2=36 sqrt(3) Q\

Let ABC be an equilateral triangle. A new triangle is ... | Sequence & Series (Mathematics)

Correct option is (1) $\mathrm{P}^2=36 \sqrt{3} \mathrm{Q}$ $\\$ Area of first $\Delta=\frac{\sqrt{3 a^2}}{4}$ $\\$ Area of second $\Delta=\frac{\sqrt{3} \mathrm{a}^2}{4} \frac{\mathrm{a}^2}{4}=\frac{\sqrt{3} \mathrm{a}^2}{16}$ $\\$ 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}

\begin{aligned}& \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{P}{6} \\& \mathrm{Q}=\frac{1}{\sqrt{3}} \cdot \frac{\mathrm{P}^2}{36} \\& \mathrm{P}^2=36 \sqrt{3} \mathrm{Q}\end{aligned}

Explanation

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