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ABCD and EFGA are the squares of side 4 cm each. In square ABCD, DMB and PMQ are the arcs of circles with centres at A and C respectively. In square AEFG, the shaded region is enclosed by two arcs of circles with centres at A and F respectively. What is the ratio of the shaded regions of the squares ABCD and AEFG respectively.

(a) \(\frac{2+π(\sqrt{2}-2)}{(π-2)}\)

(b) \(\frac{(π-2)}{2(\sqrt{2}+1-π)}\) 

(c) \(\frac{4}{3}\)

(d) None of these

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Answer : (a) = \(\frac{2+π(\sqrt{2}-2)}{(π-2)}\)

Area of square ABCD = Area of square AEFG = 16 cm2 

Area of quadrant ADMB = \(\frac{1}{4}\) × π × (4)2 = 4π cm2  

Radius of the smaller quadrant CPMQ = CM = AC – AM

=4√2 - 4 = 4 ( √2 - 1) cm.

∴ Area of the smaller quadrant 

Area of the shaded region inside the square ABCD 

= Area of sq. ABCD – (Area of quadrant ADMB + Area of quadrant CPMQ)

= 16 – [4π + 4π(3 – 2 √2 )] 

= 16 – [4π (1 + 3 – 2 √2 )]

= 16 – [4π (4 – 2 √2 )] 

= 8 [2 – 2π + √2 π] 

Now area of quadrants AEG and EFG 

= 2 × area of quadrant AEG = 2 × \(\frac{1}{4}\) × π × (4)2 = 8π

∴ Area of shaded region inside the square AEFG = Sum of the areas of quadrants AEG and EFG – Area of square AEFG

= 8π – 16 = 8(π – 2)

∴ Required ratio = \(\frac{8(2- 2π + \sqrt{2}\pi)}{8(π-2)}\) 

\(\frac{2+π(\sqrt{2}-2)}{(π-2)}\)

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