What is the area of the shaded segment shown in 0 below?

First, let's find the area of the wedge between MO and NO.

If we find the area of the circle we can then use proportions to find the area of the 30° segment knowing that the full circle is 360°:

The radius is 12 in, so:

A = πr²
A = π * 12²
A = 144π

That's for the full circle, so for the wedge we get:

144π / 360 = x / 30
30 * 144π = 360x
4320π = 360x
4320π / 360 = x
12π = x

So that's the area of the wedge.

Now we need to find the area of the triangle that we subtract out of the wedge. We know it's an isosceles triangle as it has two sides of 12 in each. We know the inner angle and need the opposite side. We can use law of cosines to find the length of the third side:

c² = a² + b² - 2ab cos(C)
c² = 12² + 12² - 2(12)(12) cos(30)
c² = 144 + 144 - 288 (√3 / 2)
c² = 288 - 144√3
c = √(288 - 144√3)
c = √[144 * (2 - √3)]
c = 12√(2 - √3)

That's the length of the base of the triangle. Since this is an isosceles triangle, if we drop a line from the tip to the base that is perpendicular to the base, we cut the base in half. Since we now have two right triangles we can solve for the height using 12 as the hypotenuse and half of the base calculated as one of the two legs:

a² + b² = c²
[6√(2 - √3)]² + b² = 12²
36(2 - √3) + b² = 144
b² = 144 - 36(2 - √3)
b = √[144 - 36(2 - √3)]
b = √{36[4 - (2 - √3)]}
b = 6√[4 - (2 - √3)]
b = 6√(4 - 2 + √3)
b = 6√(2 + √3)

Now that we have the height and the base of the triangle we can find the area:

A = bh/2
A = 12√(2 - √3) * 6√(2 + √3) / 2
A = 6√(2 - √3) * 6√(2 + √3)
A = 36√(2 - √3) * √(2 + √3)
A = 36√[(2 - √3)(2 + √3)]
A = 36√(4 - 3)
A = 36√1
A = 36(1)
A = 36

Now that we have the area of the triangle we can subtract it from the wedge to get the area of the shaded section:

(12π - 36) in²

Which is approx 1.699 in²

Area of sector OMN - Area of triangle OMN is the area of the shaded segment shown in Circle O.
12 pi - 36 = 1.699 in^2

Total area of circle = πr^2 = π(12^2) = 144π
Area of sector = 144π x 30/360 = 12π

Using A = 0.5absinC where A = area of triangle, a = 12, b = 12 and C = 30
A = 0.5(12)(12)(sin30) = 72x0.5 = 36

Area of sector - area of triangle = 12π - 36 = 1.7in. squared (2s.f)