A chord with a length of #2 # runs from #pi/12 # to #pi/8 # radians on a circle. What is the area of the circle?

Question

A chord with a length of #2     # runs from #pi/12    # to #pi/8    # radians on a circle. What is the area of the circle?

Maverick Smith · Accepted Answer

# {2pi}/{1 - cos (pi/24)}#

Lost my answer in a tab crash twice.

I was plotting this, which is still in my clipboard:

(x^2+y^2-1)(y - x tan(pi/6) ) ( y(cos(pi/6) -1)-sin(pi/6)(x-1) )(x-cos(pi/6))=0

graph{(x^2+y^2-1)(y - x tan(pi/6) ) ( y(cos(pi/6) -1)-sin(pi/6)(x-1) )(x-cos(pi/6))=0 [-0.636, 1.469, -0.303, 0.75]}

I had another version with #pi/24# that didn't render very well.

Anyway, we have a sector of angle

#theta = pi/8 - pi/12 = pi/24 #.

The chord #c# forms an isosceles triangle with two radii of length #r#. The endpoints of the chord are # (r,0)# and #(r cos theta , r sin theta)# and the third vertex is the origin. The length of the chord #c# satisfies

#c^2 = (r cos theta - r )^2 + (r sin theta - 0)^2 #

#c^2 = r ^2 (cos^2 theta - 2 cos theta +1 + sin^ theta )#

#c^2 = r^2 (2 - 2 cos theta )#

We're interested in the area #A# of the circle,

# A = \pi r^2 = {pi c^2}/{2-2cos theta}#

Plugging in the numbers

# A = {pi(2^2)}/{2-2cos (pi/24)} = {2pi}/{1 - cos (pi/24)}#

We can actually get a nice radical form for #cos(pi/24)# but I'll spare you.

You'll have to work the calculator for yourself as well.

I should check but gotta go.

Jayden Jackson · Answer

undefined The length of a chord in a circle can be used to find the radius of the circle using the formula: $ r = \frac{l}{2\sin(\theta/2)} $, where $ r $ is the radius, $ l $ is the length of the chord, and $ \theta $ is the central angle subtended by the chord in radians. Once the radius is found, you can use the formula for the area of a circle, which is $ A = \pi r^2 $, to find the area of the circle.