113k views
13 votes
Please help! Area of part of a circle!!!

Please help! Area of part of a circle!!!-example-1

1 Answer

11 votes

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

User Paras Patidar
by
8.0k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories