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.

Answer 1

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`