Area of a Segment

A segment is a portion of a circle which cut off by a straight line not passing through the centre. The segment, smaller than the semi-circle is called the minor segment and the segment larger than the semi-circle is called the major segment.


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(a) Area of the minor segment when angle \theta and radius r are given:
            Area of segment  = Area of sector AOBC  \pm Area of \Delta  AOB
                                        = \frac{1}{2}{r^2}\theta \pm \frac{1}{2}{r^2}\sin \theta
                                        = \frac{1}{2}{r^2}(\theta - \sin \theta )


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Now area of major segment  = Area of circle  - Area of minor segment
                                              =  \frac{1}{2}{r^2}(2\pi - \theta + \sin \theta )

Example:

A chord AB of a circle of radius 15cm makes an angle of {60^  \circ }at the centre of the circle. Find the area of the major and minor segment.

Solution:


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            Given that,\angle AOB = {60^  \circ }, radius, r = 15cm  

            \therefore Area of the sector OAB  =  \frac{\theta }{{360}} \times \pi {r^2}
                                                           = \frac{{60}}{{360}} \times 3.1415 \times 15 \times  15 = 117.75Square cm
            \therefore Area of \Delta  OAB  = \frac{1}{2}{r^2}\sin \theta
                                              = \frac{1}{2} \times 15 \times 15 \times \sin {60^  \circ } = \frac{{225 \times 1.73}}{4} = 97.31Square cm
            Area of minor segment   = Area of sector OAB - Area of  \Delta OAB
                                                   = 117.75 - 97.31 = 20.44 Square cm
            Area of the circle  = \pi  {r^2}
                                          = 3.1415 \times {(15)^2} = 3.1415 \times 225 =  706.5 Square cm
            Area of major segment   = Area of the circle  - Area of minor segment
                                                   = 706.5 - 20.4 = 686.1 Square cm.

(b) Area of segment when height and length of the chord of the segment are given:


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            Let r = radius of the circle
                  h = height of the segment
                  c = length of the chord

            We note that ODB is a right triangle; the hypotenuse is OB = r and the other two sides are OD = r - h and BD = c/2
            \therefore      By Pythagorean Theorem
                        {\left(  {\frac{c}{2}} \right)^2} + {\left( {r - h} \right)^2} = {r^2}
                        {\left(  {\frac{c}{2}} \right)^2} + {r^2} - 2rh + {h^2} = {r^2}
                        {\left(  {\frac{c}{2}} \right)^2} - 2rh + {h^2} = 0
            Solving for r, c and h, we obtain the following formulas
                        r = \frac{{{{\left( {\frac{c}{2}} \right)}^2} +  {h^2}}}{{2h}} --- (1)
                        h = r \pm \sqrt  {{r^2} - {{\left( {\frac{c}{2}} \right)}^2}} --- (2)
                        c = 2\sqrt {h\left(  {2r - h} \right)} --- (3)
Note:
            r + \sqrt {{r^2} - {{\left( {\frac{c}{2}}  \right)}^2}} Gives the height of the major segment
            r - \sqrt {{r^2} -  {{\left( {\frac{c}{2}} \right)}^2}} Gives the height of the minor segment
            Many formulas are given for finding the approximate area of a segment. Two of the common methods are:
Method-I: A = \frac{{2hc}}{3}  + \frac{{{h^2}}}{{2c}} = \frac{h}{{6c}}\left( {3{h^2} + 4{c^2}} \right)
Note: If the height of the segment is less \frac{1}{{10}}of radius of the circle, then A  = \frac{{2hc}}{3}.
Method-II: A = \frac{{4{h^2}}}{3}\sqrt  {\frac{{2r}}{h} - 0.608}

Example:

If the chord of the segment of a circle is 66cm and the height of the segment is 10cm, find the radius of the circle.

Solution:

Given that, c = 66cm, h = 10cm
            \therefore      r = \frac{{{{\left( {\frac{c}{2}} \right)}^2} +  {h^2}}}{{2h}}
                            = \frac{{{{(33)}^2} + {{(10)}^2}}}{{2 \times 10}} =  \frac{{1089 + 100}}{{20}} = 59.45cm

Example:

Find the area of the segment whose chord is 10cm, and whose height is 1.5cm.

Solution:

Given that h = 1.5cm, c = 10cm
            \therefore      A = \frac{{2hc}}{3} + \frac{{{h^2}}}{{2c}}
                               = \frac{2}{3}(1.5)(10) + \frac{{{{(1.5)}^2}}}{{2  \times 1}} = 10.17Square cm.

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