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Maths-

General

Easy

Question

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

$\frac{8 π}{3}$
$\frac{1}{3} (12 \sqrt{3} - 4 π)$
$\frac{1}{3} (8 π - 8 \sqrt{3})$
$\frac{1}{3} (8 π - 12 \sqrt{3})$

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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