Let \(A B C\) be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the tria…

Let the side of the original equilateral triangle be \(a\).

When the middle points of the sides are joined, the side of the new equilateral triangle becomes half of the previous side.

So, the sides of the triangles formed are \(\frac{a}{2},\frac{a}{4},\frac{a}{8},\ldots\)

Now, perimeter of an equilateral triangle of side \(s\) is \(3s\).

Therefore, \(P=3\cdot\frac{a}{2}+3\cdot\frac{a}{4}+3\cdot\frac{a}{8}+\cdots\)

\(P=3a\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots\right)\)

\(P=3a\cdot 1\)

\(P=3a\)

Area of an equilateral triangle of side \(s\) is \(\frac{\sqrt{3}}{4}s^2\).

Therefore, \(Q=\frac{\sqrt{3}}{4}\left(\frac{a}{2}\right)^2+\frac{\sqrt{3}}{4}\left(\frac{a}{4}\right)^2+\frac{\sqrt{3}}{4}\left(\frac{a}{8}\right)^2+\cdots\)

\(Q=\frac{\sqrt{3}}{4}a^2\left(\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\cdots\right)\)

\(Q=\frac{\sqrt{3}}{4}a^2\cdot \frac{\frac{1}{4}}{1-\frac{1}{4}}\)

\(Q=\frac{\sqrt{3}}{4}a^2\cdot \frac{1}{3}\)

\(Q=\frac{\sqrt{3}a^2}{12}\)

Now, \(P^2=(3a)^2=9a^2\)

Also, \(36\sqrt{3}Q=36\sqrt{3}\cdot \frac{\sqrt{3}a^2}{12}\)

\(=36\cdot \frac{3a^2}{12}\)

\(=9a^2\)

Thus, \(P^2=36\sqrt{3}Q\)