A chord with a length of $6$ runs from $pi/8$ to $pi/6$ radians on a circle. What is the area of the circle?

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Answer 1 · 2017-11-21T13:24:50

$2103pi$

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From diagram. chord = c = 6.

Angle C subtended at the centre is:

$pi/6-pi/8=pi/24$

The sum of the angles of triangle ABC = $pi$

Triangle ABC is isosceles so angles A and B are equal.

Angles A and B are:

$pi-pi/24=1/2*(23pi)/24=(23pi)/48$

a and b are radii, so we can solve for either of these.

Solving for a using the Sine Rule:

$sinA/a=sinB/b=sinC/c$

We know angle C and side c

$:.$

$sin((23pi)/48)/a=sin(pi/24)/6=>a=(6sin((23pi)/48))/(sin(pi/24))~~45.869$

So radius is $~~45.869$

Area of a circle:

$pir^2$

$pi(45.869)^2=2103pi$

A chord with a length of 6 6 runs from pi/8 pi/8 to pi/6 pi/6 radians on a circle. What is the area of the circle?