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#