Question

Please help! Area of part of a circle!!!

Answer 1

The sector (shaded segment + triangle) makes up 1/3 of the circle (which is evident from the fact that the labeled arc measures 120° and a full circle measures 360°). The circle has radius 96 cm, so its total area is π (96 cm)² = 9216π cm². The area of the sector is then 1/3 • 9216π cm² = 3072π cm².

The triangle is isosceles since two of its legs coincide with the radius of the circle, and the angle between these sides measures 120°, same as the arc it subtends. If b is the length of the third side in the triangle, then by the law of cosines

b² = 2 • (96 cm)² - 2 (96 cm)² cos(120°)   ⇒   b = 96√3 cm

Call b the base of this triangle.

The vertex angle is 120°, so the other two angles have measure θ such that

120° + 2θ = 180°

since the interior angles of any triangle sum to 180°. Solve for θ :

2θ = 60°

θ = 30°

Draw an altitude for the triangle that connects the vertex to the base. This cuts the triangle into two smaller right triangles. Let h be the height of all these triangles. Using some trig, we find

tan(30°) = h / (b/2)   ⇒   h = 48 cm

Then the area of the triangle is

1/2 bh = 1/2 • (96√3 cm) • (48 cm) = 2304√3 cm²

and the area of the shaded segment is the difference between the area of the sector and the area of the triangle:

3072π cm² - 2304√3 cm² ≈ 5660.3 cm²