A chord with a length of $2$ $2$ runs from $\frac{π}{12}$ $pi/12$ to $\frac{π}{2}$ $pi/2$ radians on a circle. What is the area of the circle?

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Answer 1 · 2017-07-20T14:21:40

The area is $= 8.48 u^{2}$ $=8.48u^2$

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The length of the chord is $A B = 2$ $AB=2$

$A C = 1$ $AC=1$

The angle subtended at the centre of the circle is

$ˆ A O B = \frac{π}{2} - \frac{π}{12} = \frac{5}{12} π$ $hat(AOB)=pi/2-pi/12=5/12pi$

$ˆ A O C = \frac{5}{24} π$ $hat(AOC)=5/24pi$

$r = \frac{A C}{sin (ˆ A O C)}$ $r=(AC)/sin(hat(AOC))$

$r = \frac{1}{sin (\frac{5}{24} π)} = \frac{1}{0.61} = 1.64$ $r=1/sin(5/24pi)=1/0.61=1.64$

The area of the circle is

$a r e a = π \cdot r^{2} = π \cdot {1.64}^{2} = 8.48$ $area=pi*r^2=pi*1.64^2=8.48$

A chord with a length of 2 22 runs from pi/12 π12pi/12 to pi/2 π2pi/2 radians on a circle. What is the area of the circle?