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Solution

Diameter of circle = d

Side of biggest square = d

Area of the biggest square is = side × side

=d $\times$ d=d2

Diagonal of smallest square = d

Let side = a

d2=a2+a2                 {using Pythagoras theorem}

d2=2a2

$\frac{d}{\sqrt{2}}=a$

Area of smallest square  $\frac{d}{\sqrt{2}} \times \frac{d}{\sqrt{2}}=\frac{d^{2}}{2}$

Here we found that area of outer square is not 4 times the area of inner square.

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