In the given figure, a square OABC has been inscribed in the quadrant OPBQ.

Solution:

(b) 228 cm2
Join OB. 
Now, OB is the radius of the circle.

We have :

$O B^{2}=O A^{2}+A B^{2} \quad$ [By Pythagoras' theorem]

$\Rightarrow O B^{2}=\left\{(20)^{2}+(20)^{2}\right\} \mathrm{cm}^{2}$

$\Rightarrow O B^{2}=(400+400) \mathrm{cm}^{2}$

$\Rightarrow O B^{2}=800 \mathrm{~cm}^{2}$

 

$\Rightarrow O B=20 \sqrt{2} \mathrm{~cm}$

Hence, the radius of the circle is $20 \sqrt{2} \mathrm{~cm}$.

Now,

Area of the shaded region = Area of the quadrant -">−Area of the square OABC

$=\left|\left(\frac{1}{4} \times 3.14 \times 20 \sqrt{2} \times 20 \sqrt{2}\right)-(20 \times 20)\right| \mathrm{cm}^{2}$

$=\left|\left(\frac{1}{4} \times \frac{314}{100} \times 800\right)-400\right| \mathrm{cm}^{2}$

$=(628-400) \mathrm{cm}^{2}$

 

$=228 \mathrm{~cm}^{2}$