Loading subject…
Chord
A line segment joining two points on the circumference of a circle.
Arc
A part of the circumference of a circle between two points.
Sector
The region bounded by two radii and the arc between them (a “pizza slice”).
Segment
The region bounded by a chord and the arc between the chord’s endpoints.
Radius $r=7 \mathrm{~cm}$, angle $\theta=120^{\circ}$. Find arc length $\ell$.
Solution
$$\ell=\frac{120}{360} \times 2 \pi(7)=\frac{1}{3} \times 14 \pi=\frac{14 \pi}{3} \approx 14.66 \mathrm{~cm}$$
An arc has length $\ell=5 \pi \mathrm{~cm}$ in a circle of radius $r=6 \mathrm{~cm}$. Find $\theta$.
Solution
$$\theta=\frac{180 \ell}{\pi r}=\frac{180(5 \pi)}{\pi(6)}=\frac{900}{6}=150^{\circ}$$
A sector has radius 9 m and angle $40^{\circ}$. Find its area.
Solution
$$A_{\text {sector }}=\frac{40}{360} \times \pi\left(9^2\right)=\frac{1}{9} \times 81 \pi=9 \pi \approx 28.27 \mathrm{~m}^2$$
A sector has radius 8 cm and central angle $135^{\circ}$. Find its perimeter.
Solution
First find arc length: $$s=\frac{135}{360} \times 2 \pi(8)=\frac{3}{8} \times 16 \pi=6 \pi$$
Now perimeter: $$P=2 r+s=16+6 \pi \approx 34.85 \mathrm{~cm}$$
A circle has radius $r=9 \mathrm{~cm}$. A chord subtends $\theta=80^{\circ}$ at the center. Find $c$.
Solution
$$c=2 r \sin (\frac{\theta}{2})=2(9) \sin \left(40^{\circ}\right)=18 \sin 40^{\circ} \approx 11.57 \mathrm{~cm}$$
A circle has radius $r=10 \mathrm{~cm}$ and chord length $c=12 \mathrm{~cm}$. Find $\theta$.
Solution
$$\theta=2 \arcsin \left(\frac{12}{2(10)}\right)=2 \arcsin (0.6) \approx 2\left(36.87^{\circ}\right) \approx 73.74^{\circ}$$