# Sum no 27 plz

since the diagonal of the smaller square is the diameter of the circle

and the diameter of the circle is equal to the side of the bigger square

hence,

we now solve for diagonal

since the are of the smaller circle is 64cm sq.

therefore the side of the smaller square is 8cm

now using pythagoras theorem in any triangle formed by 2 sides of square and the diagonal

diagonal=$\sqrt{{8}^{2}+{8}^{2}}$

=8$\sqrt{2}$

hence the side of the bigger square is =8$\sqrt{2}$

hence the area is=(8$\sqrt{2}$ )

^{2}

=128 cm sq.

regards

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