A chord with a length of $5$ $5$ runs from $\frac{π}{12}$ $pi/12$ to $\frac{π}{8}$ $pi/8$ radians on a circle. What is the area of the circle?

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Answer 1 · 2018-05-18T15:18:11

$π {⎛ ⎜ ⎝ \frac{5 sin (\frac{23 π}{48})}{sin (\frac{π}{24})} ⎞ ⎟ ⎠}^{2}$ $color(blue)(pi((5sin((23pi)/48))/(sin(pi/24)))^2$

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From diagram:

$θ = \frac{π}{8} - \frac{π}{12} = \frac{π}{24}$ $theta=pi/8-(pi)/12=pi/24$

We need to find radius $r$ $bbr$ :

We have an isosceles triangle with angle at the apex of $\frac{π}{24}$ $pi/24$

Two remaining angles are:

$\frac{π - (\frac{π}{24})}{2} = \frac{23 π}{48}$ $(pi-(pi/24))/2=(23pi)/48$

Using sine rule:

$\frac{sin (\frac{π}{24})}{5} = \frac{sin (\frac{23 π}{48})}{r}$ $sin(pi/24)/5=sin((23pi)/48)/r$

$r = \frac{5 sin (\frac{23 π}{48})}{sin (\frac{π}{24})}$ $r=(5sin((23pi)/48))/(sin(pi/24))$

Area of circle is:

$π {⎛ ⎜ ⎝ \frac{5 sin (\frac{23 π}{48})}{sin (\frac{π}{24})} ⎞ ⎟ ⎠}^{2} \approx 4590.212958$ $pi((5sin((23pi)/48))/(sin(pi/24)))^2~~4590.212958$

A chord with a length of 55 runs from π12pi/12 to π8pi/8 radians on a circle. What is the area of the circle?