Question

How do you find the radius with the area of a sector of a circle with a sector with its area $ 12\pi {\text{c}}{{\text{m}}^2} $ and the angle $ {120^\circ } $ ?

Answer 1

Hint: Since we have to find the radius of the circle when the area and angle is given, we have to basically find an value of radius which will satisfy the given values. So you have to use the formula for the area of the sector and substitute all the given values and then solve the formed equation ro get the value of unknown.

Complete step-by-step answer:
Generally we are asked to find the area of the sector, but this question has given the area of the sector when the radii make an angle of $ {120^\circ } $ and told us to find the length of the radii. The length of both the radii will be the same.
So we will first start the sum by mentioning down the formula for radius of a sector.
The area of the sector = $ \dfrac{\theta }{{{{360}^\circ }}} \times \pi {r^2} $ , where $ \theta $ is the angle between the two radii at the centre and r is the radius. We have to find the radius.
The area of the circle given to us is $ 12\pi {\text{c}}{{\text{m}}^2} $ and the angle given to us is $ {120^\circ } $ . Substituting this values in the formula we get
 $ 12\pi = \dfrac{{{{120}^\circ }}}{{{{360}^\circ }}} \times \pi {r^2} $
Bringing all the known terms on one sides and keeping the unknowns on one side we get
 $ {r^2} = \dfrac{{{{360}^\circ }}}{{{{120}^\circ }}} \times \dfrac{1}{\pi } \times 12\pi $
Solving the above equation we get
 $ {r^2} = 36 $
Taking positive square root on both sides as radius cannot be negative we get
 $ r = 6 $
hence the value of radius should be 6 to satisfy all the given conditions above. $ r = 6 $
So, the correct answer is “ $ r = 6 $ ”.

Note: As in the above question we are asked to find the radius of sector, similarly, we can be asked to find the angle when the rest of the information is given. The process remains the same as done above. we just have to solve the equation and simplify it till we are satisfied with the last answer.