Sectors, Segments, and Circular Regions

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Learning goals

Distinguish sectors, circular segments, and annuli by their regions and complete boundaries
Derive and use sector area formulas with degree or radian measures
Calculate circular segment areas using sectors and familiar right triangles
Determine annulus areas and include every circle in their total boundary length
Combine circular and straight pieces to find areas and complete perimeters

The same arc can bound different regions

A sector is the region bounded by two radii and a chosen arc. Its complete boundary includes the two straight radii as well as the curve.

The sector uses two radii and the selected arc.

A circular segment is the region bounded by a chord and a chosen arc with the same endpoints. A chord that is not a diameter determines a smaller, minor segment and a larger, major segment. State which one you want.

The segment uses the chord and the selected arc.

An annulus is the ring between two concentric circles, circles with the same center. The outer radius is larger than the positive inner radius. The ring has an outside edge and an edge around its hole.

A ring has an outer circle and an inner circle.

We use positive radii throughout. Unless a question asks for the outer boundary alone, the complete perimeter counts every boundary component, including the boundary of a hole. Guides drawn inside a region are not extra pieces of its perimeter.

Check your understanding

For the same minor arc, shorter than a semicircle, which has greater area: its sector or its circular segment?

A The circular segment B The sector C They have equal areas D There is not enough information to compare them

Sector area from a fraction of a turn

Equal central angles in one circle cut out congruent sectors, so they cut out equal areas. A disk of radius $6$ cm has area $36\pi$ square centimeters. Six equal $60^\circ$ sectors fill it without overlapping interiors, so each has area $6\pi$ square centimeters.

The circle-area argument was established in Circumference and Circle Area. Here the angle chooses a fraction of that area.

Why sector area follows its central angle#

Take a disk of radius $5$ cm and split a full turn into five equal $72^\circ$ angles. Rotating any one sector through these steps covers the disk with five congruent pieces. Each sector therefore receives one fifth of the disk’s area, which is $5\pi$ square centimeters. Two adjacent pieces receive twice that share.

More generally, divide a full turn into any whole number of equal angles, then take some of those sectors. Area addition gives the same fraction of the disk’s area as of the full turn. Nothing depends on using five pieces or on the radius chosen.

An angle need not be a rational fraction of a turn. You can trap its sector between sectors whose rational fractions are arbitrarily close. Their areas trap its area too. The exhaustion reasoning from the previous lesson extends the same rule to that angle.

If $\alpha$ is the numerical degree measure of the selected angle, its fraction of a full turn is $\alpha/360$ . Thus

A_{\text{sector}}=\frac{\alpha}{360}\pi r^2.

For a proper sector, the angle is strictly between zero and a full turn. The area rule also extends to zero area at zero degrees and the entire disk at $360^\circ$ .

Worked example 1 Measure the slice and its outline

Problem. A sector has radius $9$ cm and central angle $80^\circ$ . Find its exact area and complete perimeter.

The radius and central angle are the givens.

Solution. The sector occupies the same fraction of the disk as its angle occupies of a full turn.

\begin{aligned} A&=\frac{80}{360}\pi(9)^2,\\ A&=18\pi\text{ cm}^2. \end{aligned}

For the perimeter, first find the arc length using the same fraction of the circumference.

\begin{aligned} s&=\frac{80}{360}(2\pi\cdot9),\\ s&=4\pi\text{ cm}. \end{aligned}

The arc does not close the region by itself. Add both radii.

\begin{aligned} P&=9+9+4\pi,\\ P&=(18+4\pi)\text{ cm}. \end{aligned}

Check your understanding

A sector has radius $15$ cm and central angle $48^\circ$ . What is its exact area?

15\pi\text{ cm}^2

30\pi\text{ cm}^2

60\pi\text{ cm}^2

75\pi\text{ cm}^2

Radians simplify the same area rule

A sector of angle $2$ radians in a radius- $5$ circle occupies the fraction $2/(2\pi)$ of a turn. Multiplying that fraction by the disk area gives $25$ square units. The $\pi$ cancels; there is no approximation.

For a radius $r$ and a selected angle $\theta$ measured in radians, the identical fraction argument gives

\begin{aligned} A_{\text{sector}}&=\frac{\theta}{2\pi}\pi r^2,\\ A_{\text{sector}}&=\frac12r^2\theta. \end{aligned}

This form requires radians. An angle given in degrees must be converted or used in the degree formula instead. Arc Length and Radians explains the angle convention and the corresponding arc-length rule.

For a proper sector, tracing its two radii and its selected arc gives

P=2r+s.

The full disk is an exception to that perimeter expression: its radii are internal guides, so its boundary is just the circle. Similarly, a zero-angle figure is not a proper sector with two distinct straight boundary edges.

Since the radian arc length is $r\theta$ , the area can also be written in terms of the arc length:

\begin{aligned} A_{\text{sector}}&=\frac12r(r\theta),\\ A_{\text{sector}}&=\frac12rs. \end{aligned}

Worked example 2 Keep the angle convention visible

Problem. A sector has radius $8$ cm and central angle $5/6$ radian. Find its exact area and complete perimeter.

The angle label is a radian measure.

Solution. Use the radian area form directly.

\begin{aligned} A&=\frac12(8)^2\left(\frac56\right),\\ A&=\frac{80}{3}\text{ cm}^2. \end{aligned}

The arc length is radius times radian measure.

\begin{aligned} s&=8\left(\frac56\right),\\ s&=\frac{20}{3}\text{ cm}. \end{aligned}

Add the straight edges to close the boundary.

\begin{aligned} P&=16+\frac{20}{3},\\ P&=\frac{68}{3}\text{ cm}. \end{aligned}

Check your understanding

A sector has radius $11$ cm and central angle $4/11$ radian. What is its area?

22\text{ cm}^2

44\text{ cm}^2

26\text{ cm}^2

2\text{ cm}^2

A segment is a sector with a triangle removed

For a minor central angle strictly between zero and $180^\circ$ , the two radii and chord form a triangle inside the minor sector. Subtracting this triangle leaves the minor circular segment.

For instance, a $90^\circ$ sector of radius $4$ has area $4\pi$ square units. Its two radii are perpendicular legs of length $4$ , so the triangle has area $8$ square units. The segment has area $4\pi-8$ square units.

For that configuration in general,

A_{\text{segment}}=A_{\text{sector}}-A_{\triangle OAB}.

Use triangle methods you already know. At $60^\circ$ , the triangle is equilateral: its two equal radii force the other angles to be $60^\circ$ as well. A side- $6$ instance has height $3\sqrt3$ and area $9\sqrt3$ . In general its height is $r\sqrt3/2$ , from the special right triangles formed by its altitude.

At $120^\circ$ , that altitude splits the triangle into two $30^\circ$ - $60^\circ$ - $90^\circ$ triangles. At $90^\circ$ , the two radii already are perpendicular legs. These cases need no general triangle-area formula involving sine.

Worked example 3 A cap has a chord instead of two radii

Problem. A circle has radius $10$ cm. Two radii meet at a right angle. Find the area and complete perimeter of the minor circular segment cut off by their endpoints’ chord.

The shaded minor segment lies between the chord and the quarter-circle arc.

Solution. Subtract the right triangle from the quarter disk.

\begin{aligned} A_{\text{sector}}&=\frac14\pi(10)^2,\\ A_{\text{sector}}&=25\pi,\\ A_{\triangle}&=\frac12(10)(10),\\ A_{\triangle}&=50. \end{aligned}

The segment area is $(25\pi-50)$ square centimeters.

The segment’s boundary uses the chord instead of the two radii. By Pythagoras, its chord length is $10\sqrt2$ cm. Its arc is one quarter of the circumference, or $5\pi$ cm. Thus

P=(5\pi+10\sqrt2)\text{ cm}.

The major segment is the rest of the disk after the minor segment is removed. Subtract the minor segment area from the whole disk area; do not subtract its triangle from the major sector. The major segment contains that triangle. Its perimeter uses the same chord but the longer, major arc.

When the chord is a diameter, each segment is a semicircle. The triangle has zero area, and the boundary is the semicircular arc plus the diameter.

Check your understanding

A minor circular segment comes from a $60^\circ$ sector of radius $12$ cm. What is its area?

(24\pi-72)\text{ cm}^2

(48\pi-36\sqrt3)\text{ cm}^2

(24\pi+36\sqrt3)\text{ cm}^2

(24\pi-36\sqrt3)\text{ cm}^2

Rings have two circular boundaries

An annulus with outer radius $9$ and inner radius $4$ contains the outer disk except for the inner disk. Their areas differ by $65\pi$ square units. Its two circular boundaries have lengths $18\pi$ and $8\pi$ , for a total of $26\pi$ units.

Write $R$ for the outer radius and $r$ for the inner radius, with $0<r<R$ . Subtraction of areas and addition of boundary lengths give

\begin{aligned} A_{\text{annulus}}&=\pi(R^2-r^2),\\ P_{\text{all boundaries}}&=2\pi(R+r). \end{aligned}

The difference of squares can also be factored:

A_{\text{annulus}}=\pi(R-r)(R+r).

The width of the ring is $R-r$ . Squaring that width would find a disk’s area, not the area of this ring.

Worked example 4 Measure a ring's material and edging

Problem. An annulus has outer radius $12$ cm and inner radius $7$ cm. Find its exact area and the total length of all its boundary circles.

The shaded material lies between the two concentric circles.

Solution. Subtract the inner disk from the outer disk.

\begin{aligned} A&=\pi(12^2-7^2),\\ A&=95\pi\text{ cm}^2. \end{aligned}

Both circles touch the material and count toward its complete boundary.

\begin{aligned} P&=2\pi(12)+2\pi(7),\\ P&=38\pi\text{ cm}. \end{aligned}

Check your understanding

An annulus has outer radius $11$ cm and inner radius $3$ cm. What is the total length of both boundary circles?

16\pi\text{ cm}

22\pi\text{ cm}

28\pi\text{ cm}

112\pi\text{ cm}

Composite regions need a fresh boundary trace

Add areas of pieces with non-overlapping interiors, or subtract an explicitly removed region. Then start the perimeter calculation afresh. A shared internal join contributes no perimeter; a hole contributes its own boundary. Composite and Shaded Figures uses the same area and boundary distinction for earlier shapes.

Worked example 5 A quarter-disk cut removes straight edges

Problem. A square has side length $14$ cm. A quarter disk, centered at one corner with radius $14$ cm, is removed from the square. Find the area and complete perimeter of the remaining region.

The shaded corner remains after the quarter disk is removed.

Solution. Subtract the quarter disk area from the square area.

\begin{aligned} A&=14^2-\frac14\pi(14)^2,\\ A&=(196-49\pi)\text{ cm}^2. \end{aligned}

The two square sides from the cut’s center are removed. The other two sides remain. The new curved edge is a quarter circumference of radius $14$ cm.

\begin{aligned} s&=\frac14(2\pi\cdot14),\\ s&=7\pi\text{ cm},\\ P&=14+14+7\pi,\\ P&=(28+7\pi)\text{ cm}. \end{aligned}

A partial ring is another useful composite region. If you keep one quarter of an annulus with radii $10$ and $6$ , subtract the two quarter-disk areas. The result is $16\pi$ square units.

Both arcs and both connectors bound the shaded quarter ring.

Its two arcs have combined length $8\pi$ units. The two straight connectors each have length $4$ , so the complete perimeter is $(8+8\pi)$ units. Taking a fraction of the annulus’s circular boundaries alone would miss those new straight edges. For any selected angle strictly between zero and a full turn, the same reasoning uses that angle’s fraction of both disk areas, then counts both arcs and both connectors.

Sectors, Segments, and Circular Regions

Learning goals

The same arc can bound different regions

Sector area from a fraction of a turn

Radians simplify the same area rule

A segment is a sector with a triangle removed

Rings have two circular boundaries

Composite regions need a fresh boundary trace

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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