Find the area of the small segment whose chord is 6 cm long in a circle with a 6 cm radius.

How do i go about solving this?

Printable View

- Mar 18th 2008, 02:21 PMVitaminarea of segment
Find the area of the small segment whose chord is 6 cm long in a circle with a 6 cm radius.

How do i go about solving this? - Mar 18th 2008, 02:28 PMtopher0805
Notice that what you are describing is an equilateral triangle with sides 6cm.

Does this help? - Mar 18th 2008, 02:32 PMtopher0805
http://img2.freeimagehosting.net/uploads/087daf13a3.jpg

Sometimes it is helpful to see it.

:) - Mar 18th 2008, 02:33 PMgalactus
The area of a segment of a circle is $\displaystyle \frac{1}{2}r^{2}({\theta}-sin{\theta})$

You certainly know the radius. Use Topher's hint and you'll see theta quickly.

Then, plug them in and you're set. - Mar 18th 2008, 02:39 PMVitamin
$\displaystyle \frac{1}{2}\;6^{2}({60}-sin\;{60})\Longrightarrow 1064.4\;cm^2$

Is this correct? - Mar 18th 2008, 02:47 PMPlato
- Mar 18th 2008, 02:49 PMVitamin
$\displaystyle 6\;\pi - 9\sqrt{3}$

I'm pretty sure this is the right answer. - Mar 18th 2008, 02:51 PMtopher0805
Close! Do you know how to convert degrees to radians?

$\displaystyle

\pi radians = 180 degrees$

60 is $\displaystyle \frac {1}{3}$ of 180, so divide both sides by 3 to get:

$\displaystyle \frac {\pi}{3}$ radians $\displaystyle =$ $\displaystyle 60$ degrees.

Now just plug that in to your equation and you're done! :) - Mar 18th 2008, 02:53 PMiknowone
What type of mathematics are you allowed to use? If you can use integration I would set up the problem as the following integral:

$\displaystyle

\int_{-3}^3 \sqrt{36-x^2}-\sqrt{27} dx

$

$\displaystyle \sqrt{36-x^2}$ is the function of the circle in the upper half plane. We integrate from $\displaystyle -3 $ to [tex]3MATH] because these are the x-coordinates of the point of intersection between the line $\displaystyle x=\sqrt{27}$ and the circle. This part of this line inside the circle is a cord of length 6. This is the reason we subtract the $\displaystyle \sqrt{27}$ from the integrand so we get only the area of the part of the circle you care about.

If you don't have integration consider this: The total area of the circle is $\displaystyle 36\pi$. We can inscribe a regular hexagon in the circle whose edges are each 6 units long. The hexagon is made up of 6 equilateral triangles each with base length 6 and height $\displaystyle \sqrt{27}$. This leads to the formula $\displaystyle \frac{(\text{AreaCircle - AreaHexagon})}{6}$ since the difference in the areas is the area of all 6 segements of the circle you are interested in (each has an equal area). - Mar 18th 2008, 03:01 PMVitamin
Well. This is my brothers 10th grade geometry homework. I used the equation Galactus told me above in the thread. Plugged in the numbers. Is my answer right?

$\displaystyle \frac{1}{2}r^{2}({\theta}-sin{\theta})$

$\displaystyle 6\;\pi - 9\sqrt{3}$ - Mar 18th 2008, 03:06 PMiknowone
Yes, that is the correct answer found by any of the above methods.

- Mar 18th 2008, 03:07 PMgalactus
Yes, now you got it. (Clapping)

- Mar 18th 2008, 03:26 PMgalactus
Something you may be interested in is the length of what is called the middle ordinate. The distance from the center of the chord to the circle.

This length can be found by the formula:

$\displaystyle R(1-cos(\frac{\theta}{2}))$

$\displaystyle 6(1-cos(\frac{\pi}{6}))=6-3\sqrt{3}$

So, we integrate from $\displaystyle 6-(6-3\sqrt{3})$ to 6.

Since this is 10th grade homework, integration is overkill but interesting and fun.

$\displaystyle 2\int_{3\sqrt{3}}^{6}\sqrt{36-y^{2}}dy=6{\pi}-9\sqrt{3}$

Gives you the same answer.