# A chord with a length of 6  runs from pi/12  to pi/2  radians on a circle. What is the area of the circle?

Jan 17, 2018

Area of the circle is $30.3$ sq.unit.

#### Explanation:

Formula for the length of a chord is ${L}_{c} = 2 r \sin \left(\frac{\theta}{2}\right)$

where $r$ is the radius of the circle and $\theta$ is the angle

subtended at the center by the chord.

$\theta = \frac{\pi}{2} - \frac{\pi}{12} = 90 - 15 = {75}^{0}$

:. L_c= 2 * r * sin (theta/2) ; L_c=6 , theta=75^0 unit or

$r = \frac{6}{2 \cdot \sin 37.5} = \frac{3}{\sin} 37.5 \approx 4.93$ unit.

Area of the circle is ${A}_{c} = \pi \cdot {r}^{2} = \pi \cdot {4.93}^{2} \approx 76.30 \left(2 \mathrm{dp}\right)$

sq.unit [Ans]

Jan 17, 2018

Area of circle ${A}_{c} \approx \textcolor{p u r p \le}{76.2942}$

#### Explanation:

Chord length $A B = C h = 2 \cdot r \cdot \sin \left(\frac{\theta}{2}\right)$ where $\theta$ is $\angle \left(A O M\right)$

$\theta = \angle \left(A O M\right) = \theta = \left(\frac{\pi}{2}\right) - \left(\frac{\pi}{12}\right) = \frac{5 \pi}{12}$

$\frac{\theta}{2} = \frac{\frac{5 \pi}{12}}{2} = \frac{5 \pi}{24}$

Given Chord length $A B = C h = 6$

r = 6 / (2 * sin ((5pi)/24) ~~color(blue)(4.928)

Area of the circle ${A}_{c} = \pi {r}^{2} = \pi \cdot {\left(4.928\right)}^{2}$

Area of circle ${A}_{c} \approx \textcolor{p u r p \le}{76.2942}$