# 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. (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 $\Delta AOB$
$= \frac{1}{2}{r^2}\theta \pm \frac{1}{2}{r^2}\sin \theta$
$= \frac{1}{2}{r^2}(\theta – \sin \theta )$ 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 $15$cm makes an angle of ${60^ \circ }$at the center of the circle. Find the area of the major and minor segment.

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

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

Example:

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

Solution:

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