A chord with a length of $15$ $15$ runs from $\frac{π}{3}$ $pi/3$ to $\frac{π}{2}$ $pi/2$ radians on a circle. What is the area of the circle?

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Answer 1 · 2017-07-07T14:44:07

The area of the circle is $= 2638 u^{2}$ $=2638u^2$

The angle subtended at the center of the circle is

$θ = \frac{π}{2} - \frac{π}{3} = \frac{1}{6} π$ $theta=pi/2-pi/3=1/6pi$

The length of the chord is

$l = 2 \cdot r \cdot sin (\frac{θ}{2})$ $l=2*r*sin(theta/2)$

So,

$r = \frac{l}{2 sin (\frac{θ}{2})}$ $r=l/(2sin(theta/2))$

$= \frac{15}{2 sin (\frac{1}{12} π)}$ $=15/(2sin(1/12pi))$

$= 29$ $=29$

The area of the circle is

$A = π \cdot r^{2} = π \cdot 29^{2} = 2638 u^{2}$ $A=pi*r^2=pi*29^2=2638u^2$

A chord with a length of 15 1515 runs from pi/3 π3pi/3 to pi/2 π2pi/2 radians on a circle. What is the area of the circle?