Question

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

(1) \(\mathrm{P}^{2}=36 \sqrt{3} \mathrm{Q}\)

(2) \(\mathrm{P}^{2}=6 \sqrt{3} \mathrm{Q}\)

(3) \(P=36 \sqrt{3} Q^{2}\)

(4) \(\mathrm{P}^{2}=72 \sqrt{3} \mathrm{Q}\)

Answer 1

Correct option is (1) \(\mathrm{P}^{2}=36 \sqrt{3} \mathrm{Q}\)

Area of first \(\Delta=\frac{\sqrt{3} \mathrm{a}^{2}}{4}\)

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 \(3^{\text {rd }} \Delta=\frac{3 \mathrm{a}}{4}\)

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

\( \mathrm{P}=3 \mathrm{a} .2=6 \mathrm{a}\)

\(a=\frac{P}{6}\)

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

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