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?
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!
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).
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.