Area of a Segment

A segment is a portion of a circle which is cut off by a straight line not passing through the center. 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) The 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 the area of the major segment  = area of circle  - area of the 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 center 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 = 117.75 square 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} = 97.31 square cm
Area of the 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 the major segment  = area of the circle  - area of the minor segment
 = 706.5 - 20.4 = 686.1 square cm.

(b) The area of a segment when the height and length of the chord of the segment are given:


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Let r = be the radius of the circle
h = be the height of the segment
c = be the 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}} than the 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 66 cm and the height of the segment is 10 cm, 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 10 cm and whose height is 1.5 cm.

 

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}} = 10.17 square cm.