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 the sum of areas of all the triangles

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Solution

Q.
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 the sum of areas of all the triangles formed in this process, then: [2024]

1 $P^{2} = 6 \sqrt{3} Q$

2 $P^{2} = 72 \sqrt{3} Q$

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

4 $P = 36 \sqrt{3} Q^{2}$

Ans.
(3)

$∆ A B C is an equilateral triangle of side' a' unit.$

$Perimeter of ∆ A B C = 3 a$

$Area of ∆ A B C = \frac{\sqrt{3}}{4} a^{2}$

Now, $∆ D E F$ is made by joining midpoints of the sides of $∆ A B C$

$∴$ $D E = E F = D F = \frac{a}{2}$

$Perimeter of ∆ D E F = \frac{3 a}{2}$

$Area of ∆ D E F = \frac{\sqrt{3}}{4} \times \frac{a^{2}}{4} = \frac{\sqrt{3}}{16} a^{2}$

$∴$ $P = 3 a + \frac{3 a}{2} + \frac{3 a}{4} + \dots$

$= 3 a [1 + \frac{1}{2} + \frac{1}{2^{2}} + \dots] = \frac{3 a}{1 - \frac{1}{2}} = 6 a$

Now, $Q = \frac{\sqrt{3}}{4} a^{2} [1 + \frac{1}{4} + \frac{1}{4^{2}} + \dots] = \frac{\frac{\sqrt{3}}{4} a^{2}}{1 - \frac{1}{4}} = \frac{4}{3} \times \frac{\sqrt{3}}{4} a^{2}$

$= \frac{1}{\sqrt{3}} a^{2} = \frac{1}{\sqrt{3}} {(\frac{P}{6})}^{2}$ $[∵ P = 6 a]$

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

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