3.4 Arc length and sector areaIB Maths: Applications and Interpretation HL: Revision notes

A circle with centre $O$ and radius $r$ has circumference $C=2\pi r$ and area $A=\pi r^2$.

• An arc is part of the circumference.
• A sector is the region between two radii and the arc joining them, like a slice of pizza. The angle between the radii is the angle at the centre, $\theta$.
• A chord is a straight line joining two points on the circle; it cuts off a segment. At SL, angles are in degrees. Radians are not required.

An arc is the same fraction of the circumference as the angle is of a full turn: $\text{arc length}=\frac{\theta}{360}\times2\pi r.$ Example: radius 12 cm, $\theta=75^\circ$: arc $=\frac{75}{360}\times2\pi\times12=5\pi=15.7$ cm. A point moving round a circle travels along an arc. A minute hand of length 9 cm turns $150^\circ$ in 25 minutes, so its tip travels $\frac{150}{360}\times2\pi\times9=23.6$ cm.

A sector has the same fraction of the circle's area as its angle is of a full turn: $\text{sector area}=\frac{\theta}{360}\times\pi r^2.$ Example: radius 12 cm, $\theta=75^\circ$: area $=\frac{75}{360}\times\pi\times12^2=30\pi=94.2$ cm$^2$. Arc length is in cm, sector area in cm$^2$: check that the units match what the question asks for.

The perimeter of a sector is the arc plus two radii: $P=\frac{\theta}{360}\times2\pi r+2r$. To find an unknown angle or radius, substitute the given values and solve (use your GDC solver if you prefer). Example: radius 8 cm and arc 14 cm. Then $\frac{\theta}{360}\times2\pi\times8=14$, so $\theta=\frac{14\times360}{16\pi}=100.3^\circ$. The sector area is then $\frac{100.27}{360}\times\pi\times8^2=56.0$ cm$^2$. Example: sector area 100 cm$^2$ with $\theta=90^\circ$: $\frac{90}{360}\pi r^2=100$, so $r^2=\frac{400}{\pi}$ and $r=11.3$ cm.

A segment is a sector with its triangle removed: $\text{segment area}=\text{sector area}-\tfrac12r^2\sin\theta.$ Here $\frac12r^2\sin\theta$ is the area of the triangle formed by the two radii. Example: $r=8$, $\theta=100.27^\circ$: sector $=56.0$ cm$^2$, triangle $=\frac12(8)(8)\sin100.27^\circ=31.5$ cm$^2$, so segment $=24.5$ cm$^2$. For the region between two sectors with the same centre and angle (a path round a flower bed), subtract the smaller sector from the larger: $\frac{\theta}{360}\pi(R^2-r^2)$. Percentages: divide the part by the whole, then multiply by 100.