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Question
AB is a chord of a circle with centre O and radius 4 cm. AB is of length 4 cm and divides the circle into two segments. Find the area of the minor segment.
Solution
We know that the area of minor segment of angle `theta` in a circle of radius r is,
`A={(pitheta)/360^@-"sin"theta/2 "cos"theta/2}r^2`
It is given that the chord AB divides the circle in two segment.
We have OA=4 cm and AB=4 cm. so,
`AL=(AB)/2 cm`
`=4/2 cm`
= 2 cm
Let `angleAOB=2theta`. Then,
`angleAOL=angleBOL`
`=theta`
In`triangleOLA`,We have
`sintheta=(AL)/(OA)`
`=2/4`
`=1/2`
`theta="sin"^(-1)1/2`
`=30^@`
Hence, `angleAOB=60^@`
Now using the value of r and `theta`, we will find the area of minor segment
`A={(pixx60^@)/360^@-"sin"60^@/2"cos"60^@/2}xx4xx4`
`={pi/6-sin30^@cos30^@}xx16`
`={(16xxpi)/6-1/2xxsqrt(3)/2xx16}`
`={(8pi)/3-4sqrt(3)}cm^2`
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