Question

A chord of a circle of radius $12cm$ subtends an angle of $120°$ at the centre. Find the area of the corresponding segment of the circle. (Use $\pi =$$3.14$and $\surd 3=1.73$)

Answer 1

Step 1: Find the area of minor sector of the circle

Radius, $r=12cm$

Draw a perpendicular $OD$on chord $AB$and it will bisect chord $AB$.

So, $AD=DB$

Radius $=r=12cm$

Area of triangle $=\frac{1}{2}\times base\times height$

Area of segment=Area of the sector -the area of a triangle

The area of the minor sector $=\frac{\theta }{360}\times \pi r^2$

$=\frac{120°}{360°}\times 3.14\times {12}^2$

$=150.72c{m}^2$

Step 2: Find the area of $\Delta AOB$

From the $\Delta AOB$,

$\angle OAB=\frac{180°-(120°)}{2}=30°\left[\because OA=OB=r,\text{and in triangle the angle opposite to equal sides are equal}\right]$

Now, $C\mathrm{os}30°=\frac{AD}{OA}$

$=\frac{\sqrt{3}}{2}=\frac{AD}{12}$

Or, $AD=6\surd 3cm$

We know that $OD$bisects $AB$.

So,

$AB=2\times AD=12\surd 3cm$

Now, $S\mathrm{in}30°=\frac{OD}{OA}$

$=\frac{1}{2}=\frac{OD}{12}$

$\therefore OD=6cm$

So, the area of $\Delta AOB=\frac{1}{2}\times base\times height$

Here, base $AB=12\sqrt{3}$and

Height$=OD=6$

So, area of $\Delta AOB=\frac{1}{2}\times 12\sqrt{3}\times 6=36\sqrt{3}cm=62.28c{m}^2$

Step 3: Find the area of Minor segment

$\text{∴ Area of the corresponding Minor segment = Area of the Minor sector – Area of ΔAOB}$

$=150.72c{m}^2–62.28c{m}^2$

$=88.44c{m}^2$

Step 4: Find the area of major segment

Area of major segment $=$Area of circle $-$Area of minor segment

$$\begin{gathered} ={\mathrm{\pi r}}^2-88.44 \\ =3.14\times {12}^2-88.44 \\ =452.16-88.44 \\ =363.72{\mathrm{cm}}^2 \end{gathered}$$

Hence, the area of the corresponding segment of the circle are $88.44c{m}^2$ and $363.72c{m}^2$