1. Let P$$_{1}$$ be the circle of radius R. A square Q$$_{1}$$ is inscribed in P$$_{1}$$ such that all the vertices of the square Q$$_{1}$$ lie on the circumference of P$$_{1}$$. Another circle P$$_{2}$$ is inscribed in Q$$_{1}$$. Another Square Q$$_{2}$$ is inscribed in the circle P$$_{2}$$. Circle P$$_{3}$$ is inscribed in the square Q$$_{2}$$ and so on. If S$$_{N}$$ is the area between Q$$_{N}$$ and P$$_{N+1}$$, where N represents the set of natural numbers, then the ratio of sum of all such S$$_{N}$$ to that of the area of the square Q$$_{1}$$ is :
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By: anil on 05 May 2019 02.39 am
Let us draw the diagram according to the information given, There will be infinite shaded areas as shown in the figure. Area of circle P$$_{1}$$ = $$pi*R^2$$ Area of square Q$$_{1}$$ = $$(sqrt{2}*R)^2$$ = $$2R^2$$ Area of circle P$$_{2}$$ = $$pi*(dfrac{R}{sqrt{2}})^2=pi*R^2/2$$
Area of square Q$$_{2}$$ = $$R^2$$
Area of circle P$$_{3}$$ = $$pi*(dfrac{R}{2})^2=pi*R^2/4$$
Therefore, S$$_{N}$$ = [$$2R^2-pi*dfrac{R^2}{2}$$]+[$$R^2-pi*dfrac{R^2}{4}$$]+... S$$_{N}$$ = $$(2R^2+R^2+dfrac{R^2}{2}+... )$$- $$(pi*dfrac{R^2}{2}+pi*dfrac{R^2}{4}+pi*dfrac{R^2}{8}...)$$
S$$_{N}$$ = $$4R^2-pi*R^2$$
Therefore, $$dfrac{S_{N}}{Q_{1}}$$ = $$dfrac{4R^2-pi*R^2}{2R^2}$$ = $$dfrac{4 - pi}{2}$$. Hence, option A is the correct answer.
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Area of square Q$$_{2}$$ = $$R^2$$
Area of circle P$$_{3}$$ = $$pi*(dfrac{R}{2})^2=pi*R^2/4$$
Therefore, S$$_{N}$$ = [$$2R^2-pi*dfrac{R^2}{2}$$]+[$$R^2-pi*dfrac{R^2}{4}$$]+... S$$_{N}$$ = $$(2R^2+R^2+dfrac{R^2}{2}+... )$$- $$(pi*dfrac{R^2}{2}+pi*dfrac{R^2}{4}+pi*dfrac{R^2}{8}...)$$
S$$_{N}$$ = $$4R^2-pi*R^2$$
Therefore, $$dfrac{S_{N}}{Q_{1}}$$ = $$dfrac{4R^2-pi*R^2}{2R^2}$$ = $$dfrac{4 - pi}{2}$$. Hence, option A is the correct answer.