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:

• A. $P^2 = 36\sqrt{3}Q$
• B. $P^2 = 36\sqrt{3}Q$
• C. $P^2 = 36\sqrt{3}Q$
• D. $P^2 = 36\sqrt{3}Q$

Answer 1

The Correct Option is A

\[\text{Area of first } \triangle = \frac{\sqrt{3}}{4} a^2\]
\[\text{Area of second } \triangle = \frac{\sqrt{3}}{4} \cdot \frac{a^2}{4} = \frac{\sqrt{3}a^2}{16}\]
\[\text{Area of third } \triangle = \frac{\sqrt{3}}{4} \cdot \frac{a^2}{16} = \frac{\sqrt{3}a^2}{64}\]
\[\text{Sum of areas} = \frac{\sqrt{3}a^2}{4} \left( 1 + \frac{1}{4} + \frac{1}{16} + \cdots \right)\]
The sum of this infinite geometric series is:
\[Q = \frac{\sqrt{3}}{4} \cdot a^2 \cdot \frac{1}{1 - \frac{1}{4}} = \frac{\sqrt{3}}{4} \cdot a^2 \cdot \frac{4}{3} = \frac{\sqrt{3}}{3} a^2\]
Perimeter Calculations:
\[\text{Perimeter of first } \triangle = 3a\]
\[\text{Perimeter of second } \triangle = 3 \cdot \frac{a}{2} = \frac{3a}{2}\]
\[\text{Perimeter of third } \triangle = 3 \cdot \frac{a}{4} = \frac{3a}{4}\]
\[P = 3a \left( 1 + \frac{1}{2} + \frac{1}{4} + \cdots \right)\]
The sum of this infinite geometric series is:
\[P = 3a \cdot \frac{1}{1 - \frac{1}{2}} = 3a \cdot 2 = 6a\]
Final Calculations:
\[a = \frac{P}{6}\]
\[Q = \frac{1}{\sqrt{3}} \cdot \frac{P^2}{36}\]
\[P^2 = 36 \sqrt{3} Q\]
Answer: \((1)\; P = 36\sqrt{3}Q\)