In the figure, a square $OABC$ is inscribed in a quadrant $OPBQ$. If $OA = 20\ cm$, find the area of the shaded region. (Use $\pi = 3.14$)
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Given:

A square OABC inscribing a quadrant OPBQ. $OA=20\ cm$.

To do:

We have to find the area of shaded region.

Solution:

$OABC$ is a square and $OA=AB=BC=CA=20\ cm$

Join $OB$.

In $\vartriangle OAB$,

$OB^{2} =OA^{2} +AB^{2}$

$\Rightarrow \ OB^{2} =20^{2} +20^{2}$

$\Rightarrow OB^{2} =400+400$

$\Rightarrow OB=\sqrt{800}$

$\Rightarrow OB=20\sqrt{2}$

Here, $OB$ is the radius $r$ of the given quadrant.

Area of the given quadrant $A_{1} =\frac{\theta }{360^{o}} \times \pi r^{2}$

$A_{1} =\frac{90^{o}}{360^{o} } \times 3.14\times \left( 20\sqrt{2}\right)^{2}$

$\Rightarrow A_{1} =\frac{1}{4} \times 3.14\times 800$

$\Rightarrow A_{1} =628cm^{2}$

Area of the square $A_{2} =( side)^{2}$

$\Rightarrow A_{2} =20\times 20$

$\Rightarrow A_{2} =400cm^{2}$

Area of the shaded region, $A=$Area of the quadrant, $A_{1} -$Area of the square, $A_{2}$

$A=628-400$

$\Rightarrow A=228cm^{2}$

Therefore, the area of the shaded region is $228cm^{2}$.