Arc, Sector, Segment & Chord
- In the previous chapter, you learned how right triangles help you calculate lengths and angles you cannot measure directly (Pythagoras and trig).
- Now we’ll use those same skills inside circles: chords create triangles, and sectors create curved lengths and “slice” areas.
Circle Vocabulary That Makes Geometry Precise
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.
- Don't mix up sector and segment.
- If you see two straight radii from the center, it’s a sector.
Arc Length
- A full circle is $360^{\circ}$ and has circumference $2 \pi r$.
- So an arc is the same fraction of the circumference as its angle is of $360^{\circ}$ : $$\ell=\frac{\theta}{360^{\circ}} \times 2 \pi r$$
- Useful rearrangement (solving for $\theta$ ): $$\theta=\frac{180 \ell}{\pi r}$$
- And solving for $r$ : $$r=\frac{180 \ell}{\pi \theta}$$
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
$$\theta=\frac{180 \ell}{\pi r}=\frac{180(5 \pi)}{\pi(6)}=\frac{900}{6}=150^{\circ}$$
Sector Area
- A full circle has area: $$A=\pi r^2$$
- A sector is a fraction of the full area:
- If $\theta$ in degrees: $$A_{\text {sector }}=\frac{\theta}{360} \times \pi r^2$$
- If $\theta$ in radians: $$A_{\text {sector }}=\frac{1}{2} r^2 \theta$$
A sector has radius 9 m and angle $40^{\circ}$. Find its area.
Solution
$$A_{\text {sector }}=\frac{40}{360} \times \pi\left(9^2\right)=\frac{1}{9} \times 81 \pi=9 \pi \approx 28.27 \mathrm{~m}^2$$
Perimeter of a Sector
- A sector’s perimeter includes:
- two radii, each length $r$
- one arc, length $s$
- So: $$P_{\text {sector }}=2 r+s$$
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: $$P=2 r+s=16+6 \pi \approx 34.85 \mathrm{~cm}$$
Chords Create Isosceles and Right Triangles
- Two radii and a chord form an isosceles triangle (because all radii are equal).
- If you draw a line from the center to the midpoint of the chord, it:
- bisects the chord (splits it into two equal halves)
- is perpendicular to the chord
- This creates two right-angled triangles.
- Once you have right triangles, you can use:
- Pythagoras, or
- trigonometry (especially sine)
Finding the Length of a Chord
- Let:
- radius $r$
- central angle $\theta$ (in degrees)
- chord length $c$
- Using the two right triangles formed by splitting the isosceles triangle: $$c=2 r \sin \left(\frac{\theta}{2}\right)$$
- To find the angle when you know the chord: $$\theta=2 \arcsin \left(\frac{c}{2 r}\right)$$
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
$$\theta=2 \arcsin \left(\frac{12}{2(10)}\right)=2 \arcsin (0.6) \approx 2\left(36.87^{\circ}\right) \approx 73.74^{\circ}$$
- State the difference between a sector and a segment.
- Write the arc length formula in degrees and explain where $\frac{\theta}{360}$ comes from.
- A sector has perimeter $2r+s$. What does each part represent?
- Explain why chord problems often lead to right triangles.
- If you know the arc length and radius, how can you find the central angle?