Question : PQRS is a square whose side is 16 cm. What is the value of the side (in cm) of the largest regular octagon that can be cut from the given square?
Option 1: $8-4\sqrt2$
Option 2: $16+8\sqrt2$
Option 3: $16\sqrt2-16$
Option 4: $16-8\sqrt2$
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Correct Answer: $16\sqrt2-16$
Solution : Given: PQRS is a square whose side is 16 cm. Let ER = RD = $a$ cm So, the sides of the octagon ED = CD = (16 – 2a) cm From $\triangle$ERD, $(16-2a)^2=a^2+a^2$ ⇒ $256-64a+4a^2=2a^2$ ⇒ $2a^2-64a+256=0$ ⇒ $a^2-32a+128=0$ Solving this we get, ⇒ $a = \frac{-(-32)+\sqrt{(32^2 -4×128)}}{2}$ ⇒ $a=16-8\sqrt2$ cm So, the sides of the octagon = $16-2(16-8\sqrt2)=16\sqrt2-16$ cm Hence, the correct answer is $16\sqrt2-16$.
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