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#