Discussion

Explanation:

Area and perimeter of square Sis a2, 4a respectively.

Now, the side of S2 will be a22=a2

Thus, the area and perimeter of S2a22,4a2

The side of S3 will be a22×2=a2

Thus, area and perimeter of S3a24,4a22

Similarly, the area and perimeter of S4a28,4a23

∴ The required ratio

=4a+4a2+4a(2)2+4a(2)3+a2+a22+a24+a28+=4a1+12+1(2)2+1(2)3a21+12+14+18=4a1112a21112=4a221a2×2=4a×2(2+1)2a2=2(2+2)a

Hence, option (c).

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