Question

Find the area of a sector whose arc length is $30\pi $ cm and angle of the sector is $40^\circ$.

Answer 1

Hint: First, change the angle from degree to the radian. Then, find the radius of the circle by using the length of the arc and the angle of the sector. After that, find the area of the sector of the circle by using the radius of the circle and the angle of the sector.

Complete step-by-step answer:
Given:- Arc length = $30\pi $ cm
Angle of the sector= $40^\circ$
Let S be the length of the arc, $\theta $ be the angle of the sector and A be the area of the sector.
Convert the angle of the sector in radians.
As, we know that $1^\circ = \dfrac{\pi }{{180}}$. Then,
$40^\circ = 40 \times \dfrac{\pi }{{180}}$
Cancel out the common factor from the numerator and the denominator,
$40^\circ = \dfrac{{2\pi }}{9}$
Since, the arc length is the product of the radius and the angle of the sector. Then,
$S = r\theta $
Put S= $30\pi $ cm and $\theta = \dfrac{{2\pi }}{9}$,
$30\pi = r \times \dfrac{{2\pi }}{9}$
Divide $30\pi $ by $\dfrac{{2\pi }}{9}$ to get the radius of the circle.
$r = \dfrac{{30\pi }}{{2\pi }} \times 9$
Cancel out the common factors from both numerator and denominator,
$r = 135\,cm$
Now, find the area of the segment,
$A = \dfrac{1}{2}{r^2}\theta $
Put the values of r and $\theta $, we get,
$A = \dfrac{1}{2} \times 135 \times 135 \times \dfrac{{2\pi }}{9}$
Cancel out the common factors from both numerator and denominator. Multiply the remaining parts,
$A = 2025\pi \,c{m^2}$

Hence, the area of the segment is $2025\pi \,c{m^2}$.

Note: The students might make mistakes by not transforming the angle from degree to radians and when they calculate the area of the segment. As the radius is obtained by using the angle of sector. The students use $A = \pi {r^2}$ to calculate the area which gives the area of the whole circle instead of the area of the segment.