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Question
If AB is a chord of length \[5\sqrt{3}\] cm of a circle with centre O and radius 5 cm, then area of sector OAB is
Options
\[\frac{3\pi}{8}c m^2 \]
\[\frac{8\pi}{3}c m^2\]
\[25 \pi cm^2\]
\[\frac{25\pi}{3}c m^2\]
Solution
We have to find the area of the sector OAB.
We have,
`AM=(5sqrt3)/2`
So,
sin` ∠ AOM=(5sqrt3)/(2(5))`
Hence,
`∠ AOM=60°`
Therefore area of the sector,
`=1/2r^2θ`
`=1/2 (25)(2pi/3)`
`=(25pi)/3 cm^2`
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