ધારોકે mathrm{ABC} એક સમબાજુ ત્રિકોણ છે. આપેલ | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

Area of first $\Delta=\frac{\sqrt{3} \mathrm{a}^2}{4}$

Area of second $\Delta=\frac{\sqrt{3} a^2}{4} \frac{a^2}{4}=\frac{\sqrt{3} a^2}{16}$

Area of third $\Delta=\frac{\sqrt{3} \mathrm{a}^2}{64}$

sum of area $=\frac{\sqrt{3} a^2}{4}\left(1+\frac{1}{4}+\frac{1}{16} \ldotsight)$

$\mathrm{Q}=\frac{\sqrt{3} \mathrm{a}^2}{4} \frac{1}{\frac{3}{4}}=\frac{\mathrm{a}^2}{\sqrt{3}}$

perimeter of $1^{	ext {st }} \Delta=3 \mathrm{a}$

perimeter of $2^{	ext {nd }} \Delta=\frac{3 a}{2}$

perimeter of $3^{	ext {rd }} \Delta=\frac{3 \mathrm{a}}{4}$

$ \mathrm{P}=3 \mathrm{a}\left(1+\frac{1}{2}+\frac{1}{4}+\ldotsight) $

$ \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}$

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