The area of the square that can be

Solution:

(b) Given, radius of circle, r = OC = 8cm.

∴ Diameter of the circle = AC = 2 x OC = 2 x 8= 16 cm

which is equal to the diagonal of a square.

Let side of square be x.

In right angled $\triangle A B C, \quad A C^{2}=A B^{2}+B C^{2}$ [by Pythagoras theorem]

$\Rightarrow \quad(16)^{2}=x^{2}+x^{2}$

$\Rightarrow \quad 256=2 x^{2}$

$\Rightarrow \quad x^{2}=128$

$\therefore \quad$ Area of square $=x^{2}=128 \mathrm{~cm}^{2}$

Alternate Method

Radius of circle $(r)=8 \mathrm{~cm}$

Diameter of circle $(d)=2 r=2 \times 8=16 \mathrm{~cm}$

Since, square inscribed in circle.

 

$\therefore$ Diagonal of the squre = Diameter of circle

Now, Area of square $=\frac{(\text { Diagonal })^{2}}{2}=\frac{(16)^{2}}{2}=\frac{256}{2}=128 \mathrm{~cm}^{2}$