# Need the derivation of area formula of a segment

• Feb 20th 2009, 08:17 AM
shaurya
Need the derivation of area formula of a segment
This is not in my syllabus. But on my geometry box i saw the formula.

We have area of segment in our syllabus but that consists of getting area of sector then subtracting the triangular area.
I tried to do some stuffs using that., using trignometry.

I first though Sin Theta gets us the Height of the triangle from the angle Theta. Like in 60 Degree it would be Root 3/2 but then i concluded its wrong.

Can anyone derive and explain the formula to me?
I am curious to know about it.
• Feb 20th 2009, 08:24 AM
Plato
Have a look at this: Circular Segment -- from Wolfram MathWorld
Is that anything close to what you are asking?
• Feb 20th 2009, 08:31 AM
shaurya
i dont think so. Because the last formula is different. Another thing, i dont know integration yet.
• Feb 20th 2009, 08:59 AM
Plato
Quote:

Originally Posted by shaurya
i dont think so. Because the last formula is different. Another thing, i dont know integration yet.

Well then, you must do a better job in describing the problem.
The description that you gave leads to the area of a sector or Circular Segments
So reread the question and post a clarification.

BTW: it is pi for $\pi$. We like apple pie.
• Feb 20th 2009, 09:12 AM
shaurya
The thing there written is correct and the diagram too but its outta my reach bcuz half of thing i havent yet learned.
• Feb 21st 2009, 06:29 AM
HallsofIvy
The area of a "sector of a circle", the region in a circle of radius R between two radii making angle $\theta$, is given by $\frac{\pi r^2\theta}{360}$ (when the angle is measured in degrees. The formula is simpler in radians). To see that, recall that a full circle is 360 degrees and has area $\pi r^2$ while angle $\theta$ is $\frac{\theta}{360}$ of the whole circle. Multiply $\pi r^2$ by that fraction.

To get the segment, you need to subtract off the area of that triangular part between the center of circle and the chord. That is an isosceles triangle with side length R and and angle $\theta$. If you drop a perpendicular you get two right triangles with hypotenuse R and angle $\theta/2$.

Now call the "near side" of that right triangle (part of the radius) "h" and the "opposite side" (half of the chord) "b".

$sin(\theta/2)$= "opposite side/hyptotenuse= b/R so $b= R sin(\theta/2)$.

$cos(\theta/2)$= "near side/hypotenuse= h/R so $h= R cos(\theta/2)$

The base of the entire isoscelese triangle is 2b so the area is (1/2)(2b)(h)= bh= $R^2 sin(\theta/2)cos(\theta/2)$

Now we need to use a trigonometric identity: for any angle $\phi$, $sin(2\phi)= 2 sin(\phi)cos(\phi)$. Replacing $\phi$ by $\theta/2$, $sin(\theta)= 2 sin(\theta/2)cos(\theta/2)$ so $sin(\theta/2)cos(\theta/2)= \frac{1}{2}sin(\theta)$.

That identity makes the area of the triangle $\frac{R^2}{2}sin(\theta)$

Finally, then, the area of the sector is $\frac{\pi R^2\theta}{360}- \frac{R^2}{2} sin(\theta)= R^2\left(\frac{\pi\theta}{360}- \frac{1}{2}sin(\theta)\right)$.

Be very careful about fractions when writing "in line"! What you wrote would be interpreted as $sin(\frac{\theta}{2})$ but what you want is $\frac{sin(\theta)}{2}$.