# Math Help - URGENT HELP! cirlcess

1. ## URGENT HELP! cirlcess

im not sure how i can draw the specific image i have for the problem to be solved, but i'll try to explain it and then maybe someone can help me out with this. so there is a circle with radius r and center O. The points A, B, and C are on the circle and AOC = the angle θ, situated across the arc ABC. The area of the sector OABC is 4/3π and the length of the arc ABC is 2/3π.

Find the value of r and θ.

2. The area of the sector is $\frac{1}{2}r^{2}{\theta}=\frac{4\pi}{3}$

The length of the curve is: $r{\theta}=\frac{2\pi}{3}$

Using the two equations, solve for r and theta.

BTW, use proper grouping symbols. The way you have them written would suggest $\frac{4}{{\pi}3}$. I assume you mean what I have above?.

3. Originally Posted by miley_22
im not sure how i can draw the specific image i have for the problem to be solved, but i'll try to explain it and then maybe someone can help me out with this. so there is a circle with radius r and center O. The points A, B, and C are on the circle and AOC = the angle θ, situated across the arc ABC. The area of the sector OABC is 4/3π and the length of the arc ABC is 2/3π.

Find the value of r and θ.

Area of Sector OABC = [(2pi -theta)/(2pi)]*pi(r^2) = (4/3)pi
[(2pi)/(2pi) -theta/(2pi)]*pi(r^2) = 4pi/3
[1 -theta/(2pi)](r^2) = 4/3 -----------------(1)

Arc ABC = r(theta) = (2/3)pi
theta = 2pi/(3r) -----------------------(2)
Substitute that into (1),
[1 -(2pi)/(3r)/(2pi)]r^2 = 4/3
[1 -1/(3r)]r^2 = 4/3
r^2 -r/3 = 4/3
Clear the fractions, multiply both sides by 3,
3r^2 -r = 4
3r^2 -r -4 = 0
Factor that, ------- or use the Quadratic Formula
(3r -4)(r +1) = 0
r = 4/3 or -1

Reject r = -1 because there are no negative dimensions.

So, r = 4/3 units -------------------answer.

theta = 2pi/(3r) -------------(2)
theta = 2pi/[3(4/3)]

4. Hello, miley_22!

Another approach . . .

There is a circle with radius r and center O.
The points A, B, and C are on the circle
and $\theta = \angle AOC$ subtends arc $\widehat{ABC}.$

The area of the sector $\overline{OABC}\:=\:\frac{4}{3}\pi$
. . and the length of the arc $\widehat{ABC} \:=\:\frac{2}{3}\pi$

Find the value of $r$ and $\theta$.
Code:
              * * *
*           *   A
*               o
*              /  o B
/
*           / θ     *
*         * - - - - o C
*         O         *

*                 *
*               *
*           *
* * *

The area of the sector is: . $A \:=\:\frac{1}{2}r^2\theta \:=\:\frac{4}{3}\pi$ .[1]

The arc length of $\widehat{ABC}$ is: . $s \:=\:r\theta\:=\:\frac{2\pi}{3}$ .[2]

Divide [1] by [2]: . $\frac{\frac{1}{2}r^2\theta}{r\theta} \:=\:\frac{\frac{4\pi}{3}}{\frac{2\pi}{3}}\quad\Ri ghtarrow\quad\frac{1}{2}r\:=\:2\quad\Rightarrow\qu ad\boxed{ r \:=\:4}$

Substitute into [2]: . $4\theta\:=\:\frac{2\pi}{3}\quad\Rightarrow\boxed{\ theta \:=\:\frac{\pi}{6}}$