Question

1. The angle of a sector is 300 and the radius is 15 cm. Find its area and perimeter.
2. Obtain the area and perimeter of the segment of a circle whose radius is 11cm and central
angle of 75°?
3. The section of a rocket ship toy consists of a semicircle, a rectangle, and a triangle
as shown at the right. The altitude of a rectangle is three times the radius of the
semicircle, the altitude of the triangle is twice the same radius and the area of the
triangle is 20 sq. ft. Find the area of the section.

Answer 1

Answer on Question #57053 – Math – Geometry

Question

1. The angle of a sector is 300 and the radius is 15 cm. Find its area and perimeter

Solution

Assume that this question deals with the angle of 300 degrees.

The perimeter of the sector is the sum of the arc length $\left(2\pi r \frac{\theta^{*}}{360}\right)$ and the length of two radii $(2r)$.

The perimeter of the sector is

$$P = 2 \pi r \frac {\theta^ {*}}{3 6 0} + 2 r = 2 \pi \cdot 1 5 \cdot \frac {3 0 0}{3 6 0} + 2 \cdot 1 5 = 2 5 \pi + 3 0 \approx 1 0 8. 5 4 (c m).$$

The area of the sector is

$$A = \pi r ^ {2} \frac {\theta^ {*}}{3 6 0} = \pi * 1 5 ^ {2} * \frac {3 0 0}{3 6 0} = 1 8 7. 5 \pi \approx 5 8 9. 0 5 (c m ^ {2}).$$

Answer: 589.05 cm², 108.54 cm.

Question

2. Obtain the area and perimeter of the segment of a circle whose radius is 11 cm and central angle of $75{}^{\circ}$

Solution

The perimeter of the segment is the sum of the arc length $\left(2\pi r \frac{\theta^{*}}{360}\right)$ and the chord length $\left(2r \sin \left(\frac{\theta^{*}}{2}\right)\right)$.

The perimeter of the segment is

$$P = 2 \pi r \frac {\theta{}^ {\circ}}{3 6 0} + 2 r \sin \left(\frac {\theta{}^ {\circ}}{2}\right) = 2 \pi \cdot 1 1 \cdot \frac {7 5}{3 6 0} + 2 \cdot 1 1 \sin \frac {7 5 {}^ {\circ}}{2} = \frac {5 5}{1 2} \pi + 2 2 \sin (3 7. 5 ^ {0}) \approx 2 7. 7 9 c m.$$

The area of the segment is given by the area of the circular sector minus the area of the isosceles triangle.

The area of the segment is

$$A = \pi r ^ {2} \frac {\theta{}^ {\circ}}{3 6 0} - \frac {r ^ {2}}{2} \sin (\theta{}^ {\circ}) = \pi * 1 1 ^ {2} * \frac {7 5}{3 6 0} - \frac {1 1 ^ {2}}{2} * \sin (7 5 ^ {0}) = \frac {1 2 1}{2} \left(\frac {5}{1 2} \pi - \sin (7 5 ^ {0})\right) \approx 2 0. 7 6 (c m ^ {2}).$$

Answer: 20.76 cm², 27.79 cm.

Question

3. The section of a rocket ship toy consists of a semicircle, a rectangle, and a triangle as shown at the right. The altitude of a rectangle is three times the radius of the semicircle, the altitude of the triangle is twice the same radius and the area of the triangle is 20 sq. ft. Find the area of the section.

Solution

The area of triangle is

$\frac{1}{2} * 2r * 2r = 20 \rightarrow r = \sqrt{10}$ is the length of the radius of the semicircle.

The area of the section is the sum of areas of the semicircle $\left(\frac{\pi r^2}{2}\right)$ , the rectangle $(2r * 3r)$ and the triangle (20).

The area of the section is

$$A = \frac {\pi r ^ {2}}{2} + 2 r * 3 r + 2 0 = \left(\frac {\pi}{2} + 6\right) r ^ {2} + 2 0 = \left(\frac {\pi}{2} + 6\right) 1 0 + 2 0$$ $$\approx 95.71 (ft^{2}).$$

Answer: 95.71 ft².

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