Question

A chord of a circle of radius 12 cm subtends an angle of ${{120}^{\circ }}$ at the center. Find the area of the corresponding minor segment of the circle. (Use \[\pi \] = 3.14 and \[\sqrt{3}=1.73\] )

Answer 1

Hint: To solve this problem, we should know the formulae related to the area of triangle and area of a sector of a circle. In the below diagram, we are given a circle of radius 12 cm and a chord AB which subtends an angle of ${{120}^{\circ }}$ at the centre O we are asked to find the area of the shaded region. To get that area, we should find the area of the sector formed by the radius which covers an angle of ${{120}^{\circ }}$. The area of a sector of a circle of radius r units and an included angle $\theta $is given by the formula \[Area=\dfrac{\theta }{{{360}^{\circ }}}\times \pi \times {{r}^{2}}\]. We know that the area of the triangle with two sides a, b and an included angle C is given by the formula $Area=\dfrac{1}{2}ab\sin C$. The lengths of two sides of the triangle OAB are OA = 12 cm and OB = 12 cm and the angle $\angle AOB={{120}^{\circ }}$.
\[\text{Area}\ \text{of}\ \text{segment=Area}\ \text{of}\ \text{sectorOAB}-\text{Area}\ \text{of}\ \text{the}\ \text{triangleOAB}\]

Complete step-by-step answer:

We are given a circle of radius 12 cm with the center of the circle being O. We can write from the question that the chord AB subtends an angle of ${{120}^{\circ }}$. The required area of the segment is the difference of the area of the sector OAB and the area of the triangle OAB.
We know that the area of a sector of a circle of radius r units and an included angle $\theta $is given by the formula \[Area=\dfrac{\theta }{{{360}^{\circ }}}\times \pi \times {{r}^{2}}\].
Here $\theta ={{120}^{\circ }}$ and r = 12 cm. Substituting them, we get
\[Area=\dfrac{{{120}^{\circ }}}{{{360}^{\circ }}}\times \pi \times {{\left( 12 \right)}^{2}}=\dfrac{1}{3}\times 3.14\times 144=150.72c{{m}^{2}}\]
We know that the area of the triangle with two sides a, b and an included angle C is given by the formula $Area=\dfrac{1}{2}ab\sin C$.
Let us consider the triangle $\Delta OAB$
The lengths $OA=12,\text{ }OB=12$ and the included angle $\angle AOB={{120}^{\circ }}$. Using the above formula,
Area of the triangle = $=\dfrac{1}{2}\times 12\times 12\times \sin 120=72\times \dfrac{\sqrt{3}}{2}=36\times 1.73=62.28c{{m}^{2}}$
From the relation \[\text{Area}\ \text{of}\ \text{segment=Area}\ \text{of}\ \text{sectorOAB}-\text{Area}\ \text{of}\ \text{the}\ \text{triangleOAB}\],
we can write that the required area
$\text{Area}=150.72-62.28=88.44c{{m}^{2}}$
$\therefore $ The required area of the segment is $88.44c{{m}^{2}}$

Note: Some students might confuse between the sector and the segment. Sector is a shape which comes when a radius is rotated about the center for a certain angle. The circle is also a sector with an included angle of ${{360}^{\circ }}$ and the area of the sector is directly proportional to the included angle and that is how the formula \[Area=\dfrac{\theta }{{{360}^{\circ }}}\times \pi \times {{r}^{2}}\] is arrived. A segment is the division done by a chord of the circle. The smaller part is called the minor segment and the larger part is called the minor segment.