A chord with a length of #13 # runs from #pi/8 # to #pi/2 # radians on a circle. What is the area of the circle?

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Answer 1 · 2016-09-28T16:37:31

#pir^2 = (pi(13^2))/(2 - 2cos((3pi)/8)#

The angle, A, is:

#A = pi/2 - pi/8#

#A = (4pi)/8 - pi/8#

#A = 3pi/8#

Two radii and the chord form an isosceles triangle with sides, a = 13, b = r and c = r.

Using the law of cosines, #a² = b² + c² - 2(b)(c)cos(A)#, we make the substitutions:

#13² = r^2 + r^2 - 2(r)(r)cos((3pi)/8)#

#13² = 2r^2 - 2(r^2)cos((3pi)/8)#

#13² = r^2(2 - 2cos((3pi)/8))#

#r^2 = (13^2)/(2 - 2cos((3pi)/8)#

To obtain the area of the circle, multiply both sides by #pi#:

#pir^2 = (pi(13^2))/(2 - 2cos((3pi)/8)#

Answer 2 · 2016-09-28T16:37:46

#pir^2 = (pi(13^2))/(2 - 2cos((3pi)/8)#

The angle, A, is:

#A = pi/2 - pi/8#

#A = (4pi)/8 - pi/8#

#A = 3pi/8#

Two radii and the chord form an isosceles triangle with sides, a = 13, b = r and c = r.

Using the law of cosines, #a² = b² + c² - 2(b)(c)cos(A)#, we make the substitutions:

#13² = r^2 + r^2 - 2(r)(r)cos((3pi)/8)#

#13² = 2r^2 - 2(r^2)cos((3pi)/8)#

#13² = r^2(2 - 2cos((3pi)/8))#

#r^2 = (13^2)/(2 - 2cos((3pi)/8)#

To obtain the area of the circle, multiply both sides by #pi#:

#pir^2 = (pi(13^2))/(2 - 2cos((3pi)/8)#