Question

In the figure alongside, OACB is a quadrant of a circle with center O. The radius is 3.5 cm and OD = 2 cm. Calculate the area of the shaded portion. Take \[\pi ={}^{22}/{}_{7}\].

Answer 1

Hint: The area of the shaded region can be obtained by subtracting the area of triangle BOD from the area of the quarter of the circle. By doing so, you will get the area of the shaded region.

Complete step-by-step answer:
From the figure, we can say that OBCD is a quarter of a circle. What we need to do is find the area of the shaded region given in the figure.
We have been given the length of OB as 3.5 cm and OD = 2cm.
We can say that OA is also the radius of a quarter of the circle. We can take the radius of the quarter as ‘r’.
\[OA=r=3.5cm\].
We know the area of the circle \[=\pi {{r}^{2}}\].
A circle can be divided into 4 equal quarters. Thus, the area of each quarter will be \[{{{}^{1}/{}_{4}}^{th}}\] of the area of the circle.
\[\therefore \] Area of one quarter \[=\dfrac{1}{4}\times \] area of the circle
Area of quarter \[=\dfrac{\pi {{r}^{2}}}{4}\].
Thus, the area of quarter OBCA \[=\dfrac{1}{4}\pi {{r}^{2}}\].
We know that r =3.5 cm.
\[\therefore \]Area of quarter OBCA
\[\begin{align}
  & =\dfrac{1}{4}\pi \times {{\left( 3.5 \right)}^{2}} \\
 & =\dfrac{1}{4}\times \dfrac{22}{7}\times 3.5\times 3.5=9.625c{{m}^{2}} \\
\end{align}\]
From the figure, we can say that \[\angle BOD={{90}^{\circ }}\].
Now let us find the area of \[\Delta BOD\].
We know that the area of triangle \[=\dfrac{1}{2}\times base\times height\]
Here, base = OB = 3.5 cm and height = OD = 2cm.
\[\therefore \] Area of \[\Delta BOD=\dfrac{1}{2}\times OD\times OB=\dfrac{1}{2}\times 2\times 3.5=3.5c{{m}^{2}}\]
\[\therefore \] Area of \[\Delta BOD=3.5c{{m}^{2}}\].
Thus, to get the area of the shaded portion, we need to subtract the area of \[\Delta BOD\] from the area of quarter OBCD.
Area of shaded portion = Area of quarter OBCA - Area of \[\Delta BOD\]
\[=9.625-3.5=6.125c{{m}^{2}}\]
Thus, we got the required area as \[6.125c{{m}^{2}}\].

Note: We have found the area of quarter as \[{}^{1}/{}_{4}\pi {{r}^{2}}\]. But the quarter can also be considered as the sector of a circle. The central angle is \[{{90}^{\circ }}\]. Thus, by the sector formula, we can find the area of OBCA.
Sector formula \[=\dfrac{\theta }{360}\times \pi {{r}^{2}}\]
We know \[\theta ={{90}^{\circ }}\], i.e. \[\angle AOD={{90}^{\circ }}\].
\[\therefore \] Area of sector OBCD \[=\dfrac{90}{360}\pi {{r}^{2}}=\dfrac{1}{4}\pi {{r}^{2}}\], which is equal to the area of one quarter of a circle.