Maths-
General
Easy

Question


In the figure above, the circle has center O. What is the area of the shaded region?

  1. fraction numerator 8 pi over denominator 3 end fraction
  2. 1 third left parenthesis 12 square root of 3 minus 4 pi right parenthesis
  3. 1 third left parenthesis 8 pi minus 8 square root of 3 right parenthesis
  4. 1 third left parenthesis 8 pi minus 12 square root of 3 right parenthesis

The correct answer is: 1 third left parenthesis 8 pi minus 12 square root of 3 right parenthesis


    Step 1 :- find the area of triangle OAB
    As the angle BOA = 60° and OA = OB = 4 (radius of circle is equal)
    An isosceles triangle  with one angle 60° is an equilateral triangle .
    So the triangle  OAB  is an  equilateral triangle with side 4 .
    Area of an equilateral triangle with side a is fraction numerator square root of 3 a squared over denominator 4 end fraction
    Area of equilateral triangle  OAB = fraction numerator square root of 3 left parenthesis 4 right parenthesis squared over denominator 4 end fraction equals 4 square root of 3
    Step 2 :- Find area of sector OAB
    Area of sector making theta to the power of ring operator at centre of circle with radius r is theta to the power of ring operator over 360 to the power of ring operator cross times pi r squared
    Here r = 4 and theta to the power of ring operator equals 60 to the power of ring operator
    Area of sector OAB = 60 to the power of ring operator over 360 to the power of ring operator cross times pi left parenthesis 4 right parenthesis squared equals 16 pi over 6 equals 8 pi over 3
    area of shaded region = area of sector OAB - area of triangle OAB
    area of shaded region = fraction numerator 8 pi over denominator 3 end fraction minus 4 square root of 3 equals 1 third left parenthesis 8 pi minus 12 square root of 3 right parenthesis

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