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Sectors, Segments, and Circular Regions: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Two measures of a sector

A sector has radius $7$ cm and central angle $120^\circ$ . Find its exact area and the length of its selected arc.

The selected sector has the stated central angle.

Text description of this figure

A sector is shaded. A circle has center O, endpoints A and B, radius 7 cm, and the indicated central angle 120°. The circular incidence and indicated angle are drawn accurately.
1. Hint 1
  Compare the selected angle with a full turn.
2. Hint 2
  Apply that fraction separately to the disk area and the circumference.
Answer

$\begin{gathered}A=\frac{49\pi}{3}\text{ cm}^2\\s=\frac{14\pi}{3}\text{ cm}\end{gathered}$

Full solution

The selected angle is one third of a full turn, so the sector receives one third of the disk area.
$\begin{gathered}A=\frac13\pi(7)^2\\A=\frac{49\pi}{3}\text{ cm}^2\end{gathered}$

The arc receives the same fraction of the circumference.
$\begin{gathered}s=\frac13(2\pi\cdot7)\\s=\frac{14\pi}{3}\text{ cm}\end{gathered}$

Answer

$\begin{gathered}A=\frac{49\pi}{3}\text{ cm}^2\\s=\frac{14\pi}{3}\text{ cm}\end{gathered}$

Key idea

One angle fraction applies to area and arc length, but they remain different measurements.
Problem 2 A radian measure without pi

A sector has radius $9$ cm and central angle $2/9$ radian. Find its area and complete perimeter.

The small selected sector is drawn accurately.

Text description of this figure

A sector is shaded. A circle has center O, endpoints A and B, radius 9 cm, and the indicated central angle 2/9 rad. The circular incidence and indicated angle are drawn accurately.
1. Hint 1
  The angle is already in radians.
2. Hint 2
  Find the arc length as well as the area; the perimeter includes two straight edges.
Answer

$\begin{gathered}A=9\text{ cm}^2\\P=20\text{ cm}\end{gathered}$

Full solution

Use the radian area formula.
$\begin{gathered}A=\frac12(9)^2\left(\frac29\right)\\A=9\text{ cm}^2\end{gathered}$

Find the arc and add both radii.
$\begin{gathered}s=9\left(\frac29\right)\\s=2\text{ cm}\\P=9+9+2\\P=20\text{ cm}\end{gathered}$

Answer

$\begin{gathered}A=9\text{ cm}^2\\P=20\text{ cm}\end{gathered}$

Key idea

A rational radian measure can give exact answers with no pi in them.
Problem 3 Radii described by their sum and difference

An annulus has outer radius $R$ and positive inner radius $r$ . Their sum is $16$ cm.

Their difference is $6$ cm.

Find the annulus area and the total length of both boundary circles.

Schematic annulus with outer radius R and inner radius r.

Text description of this figure

A complete annulus is shaded between concentric circles centered at O. The outer radius is R and the inner radius is r. The drawing is schematic; use the stated measurements.
1. Hint 1
  The difference of squared radii can be factored.
2. Hint 2
  Use the stated sum and difference directly before deciding whether individual radii are needed.
Answer

$\begin{gathered}A=96\pi\text{ cm}^2\\P=32\pi\text{ cm}\end{gathered}$

Full solution

Factor the difference in the area calculation.
$\begin{gathered}A=\pi(R-r)(R+r)\\A=\pi(6)(16)\\A=96\pi\text{ cm}^2\end{gathered}$

Add the circumferences using the radius sum.
$\begin{gathered}P=2\pi(R+r)\\P=32\pi\text{ cm}\end{gathered}$

These givens describe radii 11 cm and 5 cm, so a positive inner circle fits inside the outer circle.

Answer

$\begin{gathered}A=96\pi\text{ cm}^2\\P=32\pi\text{ cm}\end{gathered}$

Key idea

A radius sum and difference can determine measurements without first finding each radius.
Problem 4 A small cap

Two radii of a radius- $9$ cm circle meet at $60^\circ$ . Find the area and complete perimeter of the minor circular segment cut off by the chord between their endpoints.

The shaded minor segment is the requested region.

Text description of this figure

The minor circular segment is shaded. A circle has center O, endpoints A and B, radius 9 cm, and the indicated central angle 60°. The circular incidence and indicated angle are drawn accurately.
1. Hint 1
  Separate the curved cap from the sector that contains it.
2. Hint 2
  The two radii and the chord form an equilateral triangle.
3. Hint 3
  Its altitude comes from a special right triangle.
Answer

$\begin{gathered}A=\left(\frac{27\pi}{2}-\frac{81\sqrt3}{4}\right)\text{ cm}^2\\P=(9+3\pi)\text{ cm}\end{gathered}$

Full solution

The central triangle is equilateral, so its side is 9 cm and its altitude is half that side times the square root of three.
$\begin{gathered}h=\frac{9\sqrt3}{2}\text{ cm}\\A_{\triangle}=\frac{81\sqrt3}{4}\text{ cm}^2\end{gathered}$

The sector is one sixth of the disk.
Subtract the triangle.
$\begin{gathered}A_{\text{sector}}=\frac{27\pi}{2}\text{ cm}^2\\A=\left(\frac{27\pi}{2}-\frac{81\sqrt3}{4}\right)\text{ cm}^2\end{gathered}$

The chord is one side of the equilateral triangle, and the selected arc is one sixth of the circumference.
$\begin{gathered}s=3\pi\text{ cm}\\P=(9+3\pi)\text{ cm}\end{gathered}$

The solution adds the perpendicular height from O to the midpoint M of the chord.

Answer

$\begin{gathered}A=\left(\frac{27\pi}{2}-\frac{81\sqrt3}{4}\right)\text{ cm}^2\\P=(9+3\pi)\text{ cm}\end{gathered}$

Key idea

The triangle removed for area still supplies the chord used in the segment perimeter.
Problem 5 Keep the larger side of the chord

A circle has radius $8$ cm. The smaller central angle between the radii to two chord endpoints is $90^\circ$ .

Find the area and complete perimeter of the major circular segment.

The larger region bounded by the chord and major arc is shaded.

Text description of this figure

The major circular segment is shaded. A circle has center O, endpoints A and B, radius 8 cm, and the indicated central angle 90°. The circular incidence and indicated angle are drawn accurately.
1. Hint 1
  Decide which of the two regions on either side of the chord is requested.
2. Hint 2
  You can find the minor cap first and subtract it from the disk.
3. Hint 3
  The major arc and minor arc share the same chord.
Answer

$\begin{gathered}A=(48\pi+32)\text{ cm}^2\\P=(12\pi+8\sqrt2)\text{ cm}\end{gathered}$

Full solution

The minor sector is a quarter disk and its central triangle is right.
$\begin{gathered}A_{\text{minor}}=\frac14\pi(8)^2-\frac12(8)(8)\\A_{\text{minor}}=(16\pi-32)\text{ cm}^2\end{gathered}$

Subtract the minor cap from the disk to keep the major region.
$\begin{gathered}A=64\pi-(16\pi-32)\\A=(48\pi+32)\text{ cm}^2\end{gathered}$

The major arc is three quarters of the circumference.
The chord follows from the right triangle.
$\begin{gathered}s=12\pi\text{ cm}\\c=8\sqrt2\text{ cm}\\P=(12\pi+8\sqrt2)\text{ cm}\end{gathered}$

Answer

$\begin{gathered}A=(48\pi+32)\text{ cm}^2\\P=(12\pi+8\sqrt2)\text{ cm}\end{gathered}$

Key idea

Changing the chosen arc changes the segment area and curved boundary, but not the chord.
Problem 6 Three lines of work

A proper sector has radius $13$ cm and angle $2/5$ radian. The following work attempts to find its complete perimeter.

Line 1: $s=13\cdot\frac25$ cm.

Line 2: $P=s+13$ cm.

Line 3: $P=\frac{91}{5}$ cm.

Identify the first invalid line, explain why it fails, and give the correct perimeter.

The selected sector is proper and has two distinct radius edges.

Text description of this figure

A sector is shaded. A circle has center O, endpoints A and B, radius 13 cm, and the indicated central angle 2/5 rad. The circular incidence and indicated angle are drawn accurately.
1. Hint 1
  Trace a closed trip around the sector boundary.
2. Hint 2
  Check how many straight radius-length edges that trip includes.
Answer

Line 2 is the first invalid line.
$\begin{gathered}P=\frac{156}{5}\text{ cm}\end{gathered}$

Full solution

Line 1 correctly multiplies the radius by the radian angle.
$\begin{gathered}s=\frac{26}{5}\text{ cm}\end{gathered}$

Line 2 includes only one radius.
The sector boundary has two radii, so this is the first invalid line.
Line 3 is consistent with that mistaken setup, but it is not the complete perimeter.

Add both radii to the arc length.
$\begin{gathered}P=26+\frac{26}{5}\\P=\frac{156}{5}\text{ cm}\end{gathered}$

Answer

Line 2 is the first invalid line.
$\begin{gathered}P=\frac{156}{5}\text{ cm}\end{gathered}$

Key idea

Correct arithmetic cannot repair a missing boundary piece.
Problem 7 Four corner cutouts

A rectangular plate is $18$ cm by $14$ cm. At each corner, a quarter disk of radius $3$ cm, centered at that corner, is removed. The four removed regions do not overlap.

Find the remaining area and the complete perimeter after all four cuts. Give exact answers.

The shaded plate remains after the four specified quarter-disk cuts.

Text description of this figure

An 18 cm by 14 cm rectangle has a quarter disk of radius 3 cm removed at each corner. The four removed quarter disks do not overlap. The remaining region is shaded.
1. Hint 1
  Account for all four removed regions when finding area.
2. Hint 2
  For the perimeter, determine how much straight edge remains on each side.
3. Hint 3
  The four new quarter-circle arcs together have the length of one full circumference.
Answer

$\begin{gathered}A=(252-9\pi)\text{ cm}^2\\P=(40+6\pi)\text{ cm}\end{gathered}$

Full solution

The rectangle area is 252 square centimeters.
The four removed quarter disks together have the area of a radius-3 disk.
$\begin{gathered}A=18\cdot14-\pi(3)^2\\A=(252-9\pi)\text{ cm}^2\end{gathered}$

Each original side loses 3 cm at both ends.
The remaining straight pieces total
$\begin{gathered}L=2(18-6)+2(14-6)\\L=40\text{ cm}\end{gathered}$

The new arcs total one radius-3 circumference.
$\begin{gathered}s=6\pi\text{ cm}\\P=(40+6\pi)\text{ cm}\end{gathered}$

Answer

$\begin{gathered}A=(252-9\pi)\text{ cm}^2\\P=(40+6\pi)\text{ cm}\end{gathered}$

Key idea

Several curved cuts can simplify in area while still requiring a new boundary inventory.
Problem 8 A larger radius and a smaller turn

A proper sector has positive radius $r$ and angle $\theta$ radians, strictly between zero and $2\pi$ .

A second sector has radius $2r$ and angle $\theta/2$ .

Compare their areas, selected arc lengths, and complete perimeters. Justify your conclusions for every allowed radius and angle.
1. Hint 1
  Track the radius factor and angle factor separately.
2. Hint 2
  Compare the products in the area and arc-length formulas.
3. Hint 3
  The perimeter also contains the two radius-length edges.
Answer

$\begin{gathered}A_2=2A_1\\s_2=s_1\\P_2-P_1=2r\end{gathered}$

Full solution

For a concrete instance, take a radius of 6 units and an angle of 1 radian.
The second sector then has radius 12 units and angle one half radian.
$\begin{gathered}A_1=18\\A_2=36\\s_1=6\\s_2=6\\P_1=18\\P_2=30\end{gathered}$

For general positive radius and the stated radian angle, the squared-radius factor grows by four while the angle factor halves.
$\begin{gathered}A_2=\frac12(2r)^2\left(\frac\theta2\right)\\A_2=r^2\theta\\A_2=2A_1\end{gathered}$

Arc length uses only the first power of radius, so its two scale factors cancel.
$\begin{gathered}s_2=(2r)\left(\frac\theta2\right)\\s_2=r\theta\\s_2=s_1\end{gathered}$

Both angles remain proper.
Each sector keeps the same arc length, but the second has longer straight edges.
$\begin{gathered}P_1=2r+s_1\\P_2=4r+s_1\\P_2-P_1=2r\end{gathered}$

Answer

$\begin{gathered}A_2=2A_1\\s_2=s_1\\P_2-P_1=2r\end{gathered}$

Key idea

Changing radius and angle together affects area, arc length, and complete perimeter in different ways.
Problem 9 Material around two holes

A disk has radius $15$ cm. Two equal circular holes lie entirely inside it, do not touch each other, and have common positive radius $r$ .

The material left after removing both holes has area $207\pi\text{ cm}^2$ . Find $r$ and the total length of every boundary of the material.

The two equal holes are wholly inside the outer disk. Schematic.

Text description of this figure

A disk of radius 15 cm contains two equal circular holes, entirely inside the disk and not touching each other. Their common radius r is unknown. The material outside the holes is shaded. Schematic; use the stated measurements.
1. Hint 1
  The removed area belongs to two disks, not one.
2. Hint 2
  Use the material area to find the common hole radius.
3. Hint 3
  Count the outside circle and each hole circle separately.
Answer

$\begin{gathered}r=3\text{ cm}\\P=42\pi\text{ cm}\end{gathered}$

Full solution

Subtract both equal hole areas from the outer disk area.
$\begin{gathered}225\pi-2\pi r^2=207\pi\\2r^2=18\\r=3\text{ cm}\end{gathered}$

The positive root is required.
Two holes of this radius can fit strictly inside the radius-15 disk without touching, as the statement requires.

The total boundary consists of three circles.
$\begin{gathered}P=2\pi(15)+2(2\pi\cdot3)\\P=42\pi\text{ cm}\end{gathered}$

Answer

$\begin{gathered}r=3\text{ cm}\\P=42\pi\text{ cm}\end{gathered}$

Key idea

Separate holes contribute separate boundary circles even when their areas were combined in one subtraction.
Problem 10 Recover the inside of a quarter ring

A quarter of an annulus has outer radius $13$ cm and unknown positive inner radius $r$ , smaller than the outer radius.

Its area is $36\pi\text{ cm}^2$ . Find its inner radius and its complete perimeter, including both straight connectors.

A quarter ring with the stated outer radius and unknown inner radius. Schematic.

Text description of this figure

A selected part of an annulus is shaded between concentric circles centered at O. The outer radius is 13 cm and the inner radius is r. The selected central angle is 90°. The drawing is schematic; use the stated measurements.
1. Hint 1
  The region is the difference of two quarter disks.
2. Hint 2
  Recover the inner radius from the area before calculating boundary lengths.
3. Hint 3
  There are two quarter-circle arcs and two connectors, each spanning the difference of radii.
Answer

$\begin{gathered}r=5\text{ cm}\\P=(16+9\pi)\text{ cm}\end{gathered}$

Full solution

Use the stated quarter-ring area.
$\begin{gathered}\frac\pi4(13^2-r^2)=36\pi\\169-r^2=144\\r^2=25\\r=5\text{ cm}\end{gathered}$

The positive radius is less than 13 cm, so it satisfies the geometry.
The two arcs together have length
$\begin{gathered}s=\frac14(2\pi)(13+5)\\s=9\pi\text{ cm}\end{gathered}$

Each straight connector has length 8 cm.
Add both connectors.
$\begin{gathered}P=9\pi+2(13-5)\\P=(16+9\pi)\text{ cm}\end{gathered}$

Answer

$\begin{gathered}r=5\text{ cm}\\P=(16+9\pi)\text{ cm}\end{gathered}$

Key idea

Inverse area information can determine the radii needed for a complete boundary calculation.

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