Question

During Durga Puja, for the purpose of lighting, one puja pandal in Kolkata used many identical structures made of wooden sticks. The design of the structures was as follows: each structure was constructed with the help of six wooden sticks by combining an isosceles triangular structure, and a square structure, with the bases of both structures being the same. Let us take one such structure. Call the triangle $PAB$, with $PA = PB$, and the square $ABCD$, with $AB$ being the same wooden stick as a

Accepted Answer

C. Square side $AB = 0.5	ext{ m}$. $CD = 0.5	ext{ m}$, so $CE = EF = FD = \frac{0.5}{3} = \frac{1}{6}	ext{ m}$. $	riangle PFE \sim 	riangle PAB$. Let $PE = x$, $EB = y$. Then $\frac{PE}{PB} = \frac{EF}{AB} \implies \frac{x}{x+y} = \frac{1/6}{1/2} = \frac{1}{3} \implies y = 2x$. In right triangle $	riangle ECB$, $EB^2 = EC^2 + CB^2 = \left(\frac{1}{6}ight)^2 + \left(\frac{1}{2}ight)^2 = \frac{1}{36} + \frac{1}{4} = \frac{10}{36} \implies EB = \frac{\sqrt{10}}{6}$. Since $y = 2x$, $PE = \frac{EB}{2} = \frac{\sqrt{10}}{12}$. Side $PB = PE + EB = \frac{3\sqrt{10}}{12} = \frac{\sqrt{10}}{4}	ext{ m}$. Total length of sticks per structure: $= PA + PB + AB + BC + CD + DA = 2\left(\frac{\sqrt{10}}{4}ight) + 4(0.5) = \frac{\sqrt{10}}{2} + 2	ext{ m}$. For 20 structures $= 20\left(2 + \frac{\sqrt{10}}{2}ight) = 10(4 + \sqrt{10})	ext{ m}$.

Length of wooden sticks for puja pandal structure

Explanation

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