SOLUTION: (10) AB is a diameter with length 12 and ∠CBA = 30°. Find the area of the shaded region. Link to graph: https://ibb.co/RT8nPJf

Algebra ->  Surface-area -> SOLUTION: (10) AB is a diameter with length 12 and ∠CBA = 30°. Find the area of the shaded region. Link to graph: https://ibb.co/RT8nPJf      Log On


   

Question 1209481: (10) AB is a diameter with length 12 and ∠CBA = 30°. Find the area of the shaded region.
Link to graph: https://ibb.co/RT8nPJf
Answer by ikleyn(52750) About Me  (Show Source):
You can put this solution on YOUR website!
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(10) AB is a diameter with length 12 and ∠CBA = 30°. Find the area of the shaded region.
Link to graph: https://ibb.co/RT8nPJf
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Let point O be the center of the circle.

Connect points O and C by drawing the segment OC.


Then the shaded area is the union of the sector OAC and the isosceles triangle OCB.


The sector has the central angle of 60 degrees and its area is 1/6 of the area of the circle,
i.e. %281%2F6%29%2Api%2Ar%5E2.


The triangle OCB has the angle COB of 120 degrees, and its area is  


    %281%2F2%29%2Ar%2Ar%2Asin%28120%5Eo%29 = %281%2F2%29%2Ar%5E2%2A%28sqrt%283%29%2F2%29 = r%5E2%2A%28sqrt%283%29%2F4%29.


So, the shaded area is


    %281%2F6%29%2Api%2Ar%5E2 + %28sqrt%283%29%2F4%29%2Ar%5E2 = %28%28pi%2F6%29+%2B+%28sqrt%283%29%2F4%29%29%2Ar%5E2


To find the numerical value of the area, substitute r = 12/2 = 6.  You will get


    %28%28pi%2F6%29+%2B+%28sqrt%283%29%2F4%29%29%2A6%5E2 = %28%283.14159265%2F6%29+%2B+%28sqrt%283%29%2F4%29%29%2A6%5E2 = 34.43801317 square units (approx.)


You may round it as you want.

Solved.