A chord with a length of #5 # runs from #pi/12 # to #pi/6 # radians on a circle. What is the area of the circle?

Question

1301 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-27T08:28:35

The area of the circle is #=1152.5u^2#

enter image source here

The angle subtended at the centre of the circle is

#hat(AOB)=theta=pi/6-pi/12=2/12pi-1/12pi=1/12pi#

The length of the chord is

#AB=5#

#AC=5/2=2.5#

#sin(theta/2)=(AC)/r#

The radius of the circle is

#r=(AC)/sin(theta/2)=2.5/sin(1/2*1/12pi)=2.5/sin(1/24pi)=19.2u#

The area of the circle is

#area=pir^2=pi*19.2^2=1152.5u^2#