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 $\surd 3=1.73$)

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Solution

## Step 1: Find the area of minor sector of the circleRadius, $r=12cm$Draw a perpendicular $OD$on chord $AB$and it will bisect chord $AB$.So, $AD=DB$Radius $=r=12cm$Area of triangle $=\frac{1}{2}×base×height$Area of segment=Area of the sector -the area of a triangleThe area of the minor sector $=\frac{\theta }{360}×\pi {r}^{2}$$=\frac{120°}{360°}×3.14×{12}^{2}$$=150.72c{m}^{2}$Step 2: Find the area of $\Delta AOB$From the $\Delta AOB$,$\angle OAB=\frac{180°-\left(120°\right)}{2}=30°\left[\because OA=OB=r,\text{andintriangletheangleoppositetoequalsidesareequal}\right]$Now, $C\mathrm{os}30°=\frac{AD}{OA}$$=\frac{\sqrt{3}}{2}=\frac{AD}{12}$Or, $AD=6\surd 3cm$We know that $OD$bisects $AB$. So,$AB=2×AD=12\surd 3cm$Now, $S\mathrm{in}30°=\frac{OD}{OA}$$=\frac{1}{2}=\frac{OD}{12}$$\therefore OD=6cm$So, the area of $\Delta AOB=\frac{1}{2}×base×height$Here, base $AB=12\sqrt{3}$andHeight$=OD=6$So, area of $\Delta AOB=\frac{1}{2}×12\sqrt{3}×6=36\sqrt{3}cm=62.28c{m}^{2}$Step 3: Find the area of Minor segment$\text{∴AreaofthecorrespondingMinorsegment=AreaoftheMinorsector–AreaofΔAOB}$ $=150.72c{m}^{2}–62.28c{m}^{2}$ $=88.44c{m}^{2}$Step 4: Find the area of major segmentArea of major segment $=$Area of circle $-$Area of minor segment $={\mathrm{\pi r}}^{2}-88.44\phantom{\rule{0ex}{0ex}}=3.14×{12}^{2}-88.44\phantom{\rule{0ex}{0ex}}=452.16-88.44\phantom{\rule{0ex}{0ex}}=363.72{\mathrm{cm}}^{2}$Hence, the area of the corresponding segment of the circle are $88.44c{m}^{2}$ and $363.72c{m}^{2}$

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