Question

In the figure, the scale of the square ABCD is 7cm. with center 0 and radius OA sector $\vartriangle AXC$ is drawn. Find the area of the shaded region.

Answer 1

Hint: From the diagram observe that area of shaded region is equal to area of square minus area of sector. Find area of square by using the formula= \['area\text{ }of\text{ }square'={{\left( length\text{ }of\text{ }side\text{ }of\text{ }square \right)}^{2}}\] and find the area of sector by using the formula - $''area\,of\,\operatorname{s}ector=\dfrac{1}{2}{{r}^{2}}\theta ''$ where r= radius of the sector and $\theta $ is the angle of the sector.

Complete step-by-step answer:

According to the question, we have to find the area of the shaded portion in the diagram.
In the diagram, we can observe that-

$\begin{align}

  & \left( area\,of\,sector \right)+\left( area\,of\,shaded\,region \right)=\left( area\,of\,square \right) \\

 & \Rightarrow area\,of\,shaded\,region-\left( area\,of\,square \right)-\left( area\,of\,sector \right)............(i) \\

\end{align}$

For finding the area of the shaded region, we need to find the area of the square and area of the sector.

Let us find out area of the square-

We know that-area of square= ${{\left( length\,of\,side\,of\,square \right)}^{2}}$

Here the length of the side of the square 1m= 100 cm.

So, length of side of square=0.07m

$\begin{align}

  & \Rightarrow area\,of\,square={{\left( 0.07 \right)}^{2}}{{m}^{2}} \\

 &

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=0.0049\,\,\,\,\,and\,\,\,\,\,\,\,\,1

{{m}^{2}}=104c{{m}^{2}} \\

\end{align}$

So, $area\,of\,the\,square=0.0049\times 104=4900c{{m}^{2}}$

Now, let us find the area of the sector. We know that, area of sector= $\dfrac{1}{2}{{r}^{2}}\theta $ where

r = radius of the sector and

$\theta $= angle formed by the sector.

Here, r= 7cm=0.07m and

            $\theta $=$90{}^\circ =\dfrac{\pi }{4}$

So, area of the sector= $\dfrac{1}{2}\times {{\left( 0.07 \right)}^{2}}=\dfrac{\pi }{4}$

Taking $\pi =\dfrac{22}{7}$, we will get-

Area of sector= $\dfrac{1}{2}\times \left( 0.07 \right)\times \left( 0.07 \right)\times

\dfrac{1}{4}\times \dfrac{22}{7}$

$\Rightarrow Area\,of\,the\,\,sector=\dfrac{1}{2}\times \dfrac{7}{100}\times

\dfrac{7}{100}\times \dfrac{1}{4}\times \dfrac{22}{7}$

\[\] $\Rightarrow Area\,of\,\sec tor=\dfrac{77}{4\times {{10}^{4}}}{{m}^{{}}}$

We know $1{{m}^{2}}={{10}^{4}}c{{m}^{2}}$

So, area of the sector= $\left( \dfrac{77}{4\times {{10}^{4}}}\times {{10}^{4}} \right)c{{m}^{2}}$

$\Rightarrow Area\,of\,sector\,=19.25c{{m}^{2}}$

Now, we have found area of the square= \[49c{{m}^{2}}\]and area of sector=

\[19.25c{{m}^{2}}\]

Now, on putting area of the square= \[49c{{m}^{2}}\] and area of the sector=

\[19.25c{{m}^{2}}\] in equation(i) we will get-

\[\begin{align}

  & \Rightarrow Area\,\,of\,shaded\,region=49c{{m}^{2}}-19.25c{{m}^{2}} \\

 & \Rightarrow Area\,\,of\,shaded\,region=\left( 49-19.25 \right)c{{m}^{2}} \\

 &

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=29

.75c{{m}^{2}} \\

\end{align}\]

Hence, the required area of shaded region in the diagram= \[29.75c{{m}^{2}}\]


Note: Don’t forget to convert angle \[90{}^\circ \] into radian before putting it in the formula \[area\,of\,sector=\dfrac{1}{2}{{r}^{2}}\theta \]. SI unit of angle is radian.