Question

In the given figure O is the centre of the circle of radius 28cm. Find the area of the minor segment ASB.

Answer 1

Hint: In this question, we have to find the area of the minor segment ASB. First step is to find the area of sector ASB by using the formula for finding the area of the area of sector. Then find the area of $\vartriangle {\text{OAB}}$ by using the formula $\dfrac{1}{2}{{\text{r}}^2}\sin \theta $ and finally subtract the area of $\vartriangle {\text{OAB}}$ from area of sector ASB.

Complete step-by-step answer:

In the question, we have a circle with centre as O and segment ASB which makes an angle of ${45^0}$ at the centre.

And we have to find the area of segment ASB.
From question fig, r = 28 cm
Angle suspended by minor segment ASB at centre = ${45^0}$
Area of segment ASB is calculated by using the formula given below:
Area of segment = Area of sector – Area of $\vartriangle {\text{OAB}}$. (1)
So first of all we will find the area of sector ASB.
We know that the area of sector is given as:
Area of sector = \[\pi {{\text{r}}^2} \times \dfrac{\theta }{{{{360}^0}}}\] .

Where r= radius of the circle .
          $\theta $ = Angle suspended by sector at the centre in degree.
$\therefore $ Area of minor sector ASB = \[\pi {{\text{r}}^2} \times \dfrac{\theta }{{{{360}^0}}}\].
Putting the values of r and $\theta $ in above equation, we get:
Area of sector ASB = \[\pi {{\text{r}}^2} \times \dfrac{\theta }{{{{360}^0}}} = \pi \times {28^2} \times \dfrac{{{{45}^0}}}{{{{360}^0}}} = 98\pi {\text{ c}}{{\text{m}}^2}.\]
Now, we will find the area of $\vartriangle {\text{OAB}}$

The area of $\vartriangle {\text{OAB}}$ is given by:
Area of $\vartriangle {\text{OAB}}$= $\dfrac{1}{2}{{\text{r}}^2}\sin \theta $.
Putting the values of r and $\theta $ in above equation, we get:
Area of $\vartriangle {\text{OAB}}$= $\dfrac{1}{2}{{\text{r}}^2}\sin \theta = \dfrac{1}{2} \times {28^2} \times \sin {45^0} = 277.18{\text{ c}}{{\text{m}}^2}$.
Putting the values in equation 1, we get:
Area of minor segment ASB = Area of sector – Area of $\vartriangle {\text{OAB}}$.

Area of minor segment ASB= $98\pi {\text{ c}}{{\text{m}}^2} - 277.18{\text{ c}}{{\text{m}}^2} = \left( {307.72 - 277.18} \right){\text{c}}{{\text{m}}^2} = 30.54{\text{ c}}{{\text{m}}^2}.$

Note: In this type of question, you should know the definition of sector of a circle and segment of a circle. You should remember the formula for finding the area of sector and area of segment of a circle. You should be careful while taking the angle suspended by sector at the centre as there are two types of sectors, major and minor sector.