Question

Calculate the area of the designed region in the figure, common between the two quadrants of the circle of radius 8 cm each.

Answer 1

Hint: Given is a square of side 8cm. The area of the designed region can be found by dividing it into \[I\] and \[II\]. Subtract the area of the triangle from the area of sector. Similarly for region \[II\] also. Total area will sum the area of region \[I\] and \[II\].

Complete step-by-step answer:
Consider the figure that has been given to us, it’s a square with sides 8 cm. We know in a square all sides are the same and the angles are \[{{90}^{\circ }}\]. Now let us mark the square as ABCD.
Now, we need to find the area of the designed region in the given figure.

Let us mark the region we need to find as\[I\]and\[II\].
Now, let us find the area of the first quadrant.
Thus area of designed region =
Area of the Region\[I\]+ Area of the Region\[II\] \[\to (1)\]

\[\text{Area of Region I = Area of sector ABC - area of }\vartriangle \text{ABC}\text{.}\] \[\to (2)\]

We know the formula for calculating for \[area\text{ }of\text{ }sector=\dfrac{\theta }{360}\pi {{r}^{2}}\]

Now for the quadrant radius, \[r=8cm\](from the figure)

We know the \[area\text{ }of\text{ traingle}=\dfrac{1}{2}bh\]

Now let us substitute all these values in (2). Here\[\theta ={{90}^{\circ }}\], as the angle of a square is\[{{90}^{\circ }}\], as given in figure. Here \[b=h=8cm\]
\[\begin{align}
  & area\text{ }of\text{ Region I}=\dfrac{\theta }{360}\pi {{r}^{2}}-\dfrac{1}{2}bh \\
 & \\
 & \text{ = }\dfrac{90}{360}\times \dfrac{22}{7}\times {{8}^{2}}-\dfrac{1}{2}\times 8\times 8 \\
 & \\
 & \text{ = }\dfrac{1}{4}\times \dfrac{22}{7}\times 8\times 8-\dfrac{1}{2}\times 8\times 8 \\
 & \\
 & \text{ = }\dfrac{352}{7}-32 \\
 & \\
 & area\text{ }of\text{ Region I}=\left( \dfrac{352}{7}-32 \right)c{{m}^{2}}. \\
\end{align}\]

Now the area of the Region\[\text{II}\] is the same as the area of the 1st quadrant. But we could still check it.

\[\begin{align}
  & Area\text{ of Region }II=Area\text{ of sector ADC - Area of}\vartriangle ADC \\
 & \\
 & Area\text{ of Region }II=\dfrac{\theta }{360}\times \pi {{r}^{2}}-\dfrac{1}{2}bh \\
\end{align}\]

Here, \[\theta ={{90}^{\circ }},r=8cm,b=8cm,h=8cm\]

Thus let us substitute all these values in the above expression.

\[\begin{align}
  & Area\text{ of Region }II=\dfrac{90}{360}\times \dfrac{22}{7}\times {{8}^{2}}-\left( \dfrac{1}{2}\times 8\times 8 \right) \\
 & \\
 & \text{ =}\left( \dfrac{1}{4}\times \dfrac{22}{7}\times 8\times 8 \right)-\left( 4\times 8 \right) \\
 & \\
 & \text{ =}\left( \dfrac{352}{7}-32 \right)c{{m}^{2}} \\
\end{align}\]

Thus\[area\text{ }of\text{ }the\text{ }designed\text{ }region\text{ }=Area\text{ }of\text{ }region\text{ }I\text{ }+\text{ }area\text{ }of\text{ }region\text{ }II\]
\[\begin{align}
  & =\left( \dfrac{352}{7}-32 \right)+\left( \dfrac{352}{7}-32 \right) \\
 & \\
 & =2\left( \dfrac{352}{7}-32 \right) \\
 & \\
 & =2\left( \dfrac{352-224}{7} \right) \\
 & \\
 & =\dfrac{256}{7} \\
\end{align}\]
Hence we got the area of the shaded region as\[\dfrac{256}{7}c{{m}^{2}}\].


Note: You can also find the area of the designed region by.

     \[\begin{align}
  & =Area\text{ }of\text{ }the\text{ }{{1}^{st}}quadrant\text{ }+\text{ }Area\text{ }of\text{ }the\text{ }{{2}^{nd}}quadrant\text-Area\text{ }of\text{ }the\text{ }square \\
 & \\
 & =area\text{ }of\text{ }\sec tor\text{ }ABC\text{ }+\text{ }area\text{ }of\text{ }\sec tor\text{ }ADC\text{ }-\text{ }\left( side \right){{~}^{2}}~~~ \\
 & \\
 & =\dfrac{\theta }{360}~\times \text{ }\pi {{r}^{2}}+~\dfrac{\theta }{360}~\times \text{ }\pi {{r}^{2}}~-{{8}^{2}} \\
 & \\
 & =2\times \dfrac{90}{360}\times \dfrac{22}{7}\times {{8}^{2}}-{{8}^{2}} \\
 & \\
 & =\dfrac{11}{7}\times 64-64 \\
 & \\
 & =64\left( \dfrac{11}{7}-1 \right) \\
 & \\
 & =\dfrac{64\times 4}{7} \\
 & \\
 & =\dfrac{256}{7}c{{m}^{2}}. \\
\end{align}\]