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A chord of a circle of radius 12cm subtends an angle of ${120^0}$at the center. Find the area of the corresponding segment of the circle.
$\left[ {{\text{take = 3}}{\text{.14 and }}\sqrt 3 = 1.73} \right]$.

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Answer
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Hint: The area of the corresponding segment of the circle is the subtended angle at the center by the chord to the complete revolution which is ${360^0}$ at the center multiplied by the area of the circle. Use this concept to find the area of the corresponding segment.

Complete step-by-step answer:
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As we know that the area (A) of the circle is $\pi {r^2}$ square unit, where r is the radius of the circle.
Now it is given that the chord AB as shown in figure subtends an angle 120 degree at the center of circle as shown in the figure.
And we know that the complete revolution at the center is 360 degree.
So the area of the corresponding segment of the circle is the ratio of subtended angle at the center by the chord to the complete revolution at the center multiplied by the area of the circle.
Let the area of the corresponding segment be S square unit.
$ \Rightarrow S = \dfrac{{{{120}^0}}}{{{{360}^0}}} \times \pi {r^2}$ Square unit.
Now it is given that the radius (r) of the circle is 12 cm, so substitute this value in the above equation we have,
$ \Rightarrow S = \dfrac{{{{120}^0}}}{{{{360}^0}}} \times \left( {3.14 \times {{\left( {12} \right)}^2}} \right)$ Square unit. $\left[ {\because \pi = 3.14} \right]$
Now simplify the above equation we have,
$ \Rightarrow S = \dfrac{{{{120}^0}}}{{{{360}^0}}} \times \left( {3.14 \times {{\left( {12} \right)}^2}} \right)$
$ \Rightarrow S = \dfrac{1}{3} \times \left( {3.14 \times {{\left( {12} \right)}^2}} \right) = 48 \times 3.14 = 150.72{\text{ c}}{{\text{m}}^2}$
So this is the required area of the corresponding sector.

Note: Whenever we face such types of problems the key concept is to build the diagrammatic representation of the provided information as it helps figuring out the area which we are dealing with. Chord refers to the line segment joining two points lying on the circle.