# A chord with a length of 24  runs from pi/3  to (5 pi )/6  radians on a circle. What is the area of the circle?

Oct 3, 2017

The area of the circle is $\frac{2304}{\pi}$ or $733.386$

#### Explanation:

So, a circle has an internal angle of 2$\pi$ radians. First we need to figure out the angle of the chord we have.

$\frac{5 \pi}{6} - \frac{\pi}{3} = \frac{5 \pi}{6} - \frac{2 \pi}{6} = \frac{3 \pi}{6} = \frac{\pi}{2}$

This menas that $\frac{\pi}{2}$ represents a length of 24. In order to know the circumference of the circle we need to know the length represented by $2 \pi$. We can get this by multiplying $\frac{\pi}{2}$ by 4, so the circumference is 96. Remember the formula for the circumference is:

$C = 2 \pi \cdot r$
$96 = 2 \pi \cdot r$
$r = \frac{96}{2 \pi}$
$r = \frac{48}{\pi}$

Finally, we want the area of the cicle, which is given by the following:
$A = \pi \cdot {r}^{2}$

substituting r into the equation, we get:

$A = \pi {\left(\frac{48}{\pi}\right)}^{2}$
$A = {48}^{2} / \pi = \frac{2304}{\pi} = 733.386$

Oct 3, 2017

Area of circle$= 905.14$

#### Explanation:

$\theta = \left(\frac{5 \pi}{6}\right) - \left(\frac{\pi}{3}\right) = \frac{5 \pi - 2 \pi}{6} = 3 \frac{\pi}{6} = \frac{\pi}{2} = 90$degrees
$\frac{\theta}{2} = 45$ deg
$\sin \left(\frac{\theta}{2}\right) =$ opp. side / hypotenuse = (chord/2)/radius
$\sin 45 = \frac{\frac{24}{2}}{r}$
$r = \frac{12}{\sin} 45 = \frac{12}{\frac{1}{\sqrt{2}}} = 12 \sqrt{2}$
Area of the circle $= \pi {r}^{2} = \frac{22 \cdot 12 \sqrt{2} \cdot 12 \sqrt{2}}{7}$
$= \frac{22 \cdot 288}{7} = 905.14$