A chord with a length of $9$ runs from $pi/8$ to $pi/6$ radians on a circle. What is the area of the circle?

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Answer 1 · 2016-10-03T22:15:57

$A ~~ 14872.29 units²$

Let $theta =$ the measure of the angle:

$theta = pi/6 - pi/8$

$theta = (8pi - 6pi)/48$

$theta = (2pi)/48$

$theta = (pi)/24$

This angle, the chord and two radii (one on either end of the chord) form a triangle. Therefore, we can use the law of cosines:

$c² = a² + b² - 2(a)(b)cos(theta)$

where $c = 9, a = b = r and theta = pi/24$ :

$9² = r² + r² -2(r)(r)cos(pi/24)$

$9² = 2r² -2(r²)cos(pi/24)$

$9² = r²(2 -2cos(pi/24))$

$r² = (pi9²)/(2 - 2cos(pi/24))$

The area of the circle, A, is the above multplied by $pi$ :

$A = (pi9²)/(2 - 2cos(pi/24))$

$A ~~ 14872.29 units²$

A chord with a length of 9 9 runs from pi/8 pi/8 to pi/6 pi/6 radians on a circle. What is the area of the circle?