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Question

(A) The side length of the $1^{	ext{st}}$ triangle is $a$. The side length of the $2^{	ext{nd}}$ triangle is $a/2$,the $3^{	ext{rd}}$ is $a/4$,and

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The side length of the $1^{	ext{st}}$ triangle is $a$. The side length of the $2^{	ext{nd}}$ triangle is $a/2$,the $3^{	ext{rd}}$ is $a/4$,and so on.Sum of perimeters $P = 3a + 3(a/2) + 3(a/4) + \dots = 3a(1 + 1/2 + 1/4 + \dots) = 3a 	imes \frac{1}{1 - 1/2} = 3a 	imes 2 = 6a$.Thus,$a = P/6$.Sum of areas $Q = \frac{\sqrt{3}}{4}a^2 + \frac{\sqrt{3}}{4}(a/2)^2 + \frac{\sqrt{3}}{4}(a/4)^2 + \dots = \frac{\sqrt{3}}{4}a^2(1 + 1/4 + 1/16 + \dots) = \frac{\sqrt{3}}{4}a^2 	imes \frac{1}{1 - 1/4} = \frac{\sqrt{3}}{4}a^2 	imes \frac{4}{3} = \frac{\sqrt{3}a^2}{3} = \frac{a^2}{\sqrt{3}}$.Substituting $a = P/6$ into the expression for $Q$:$Q = \frac{(P/6)^2}{\sqrt{3}} = \frac{P^2}{36\sqrt{3}}$.Therefore,$P^2 = 36\sqrt{3}Q$.

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