A chord with a length of $5$ runs from $pi/12$ to $pi/2$ radians on a circle. What is the area of the circle?

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Answer 1 · 2016-10-16T22:15:41

$A ~~ 52.98$

If one draws two radii from the center to each end of the chord, one has a triangle.

Compute the angle formed by the two radii:

$theta = pi/2 - pi/12 = (6pi)/12 - pi/12 = (5pi)/12$

We can use the Law of Cosines to find the length of $r^2$ :

$c^2 = a^2 + b^2 - 2(a)(b)cos(theta)$

where $c = 5, a = b = r$ and $theta = (5pi)/12$

$5^2 = r^2 + r^2 - 2(r)(r)cos((5pi)/12)$

$r^2 = 5^2/(2 - 2cos((5pi)/12)$

The area of the circle is the above multiplied by $pi$ :

$A = (5^2pi)/(2 - 2cos((5pi)/12)$

$A ~~ 52.98$

A chord with a length of 5 5 runs from pi/12 pi/12 to pi/2 pi/2 radians on a circle. What is the area of the circle?