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Arc Length and Radians: 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 A selected part of the circle

A circle has radius $18$ cm. Find the exact length of its selected minor arc with central angle $65^\circ$ .

The blue curve is the selected route.

Text description of this figure

Circle O has radius 18 centimeters. The selected blue minor arc AB has central angle 65 degrees. Its length is not supplied.
1. Hint 1
  Compare the selected turn with one full turn.
2. Hint 2
  Find the same fraction of the full circumference.
Answer

$\tfrac{13\pi}{2}$ cm.

Full solution

The selected turn is $65/360$ of a full turn.
The full circumference is $36\pi$ cm.

Apply that fraction to the full length.
$\begin{aligned}s&=\frac{65}{360}\cdot36\pi,\\s&=\frac{13\pi}{2}\text{ cm}.\end{aligned}$

Answer

$\tfrac{13\pi}{2}$ cm.

Key idea

An arc uses the same fraction of the circumference as its central turn uses of a full turn.
Problem 2 Two ways to name a turn

Convert each angle exactly.

Case A: $84^\circ$ to radians.

Case B: $17\pi/12$ radians to degrees.
1. Hint 1
  The two units agree on the size of a half-turn.
2. Hint 2
  Use pi radians per 180 degrees for Case A, and the reciprocal factor for Case B.
Answer

$\begin{gathered}\text{A: }\frac{7\pi}{15}\text{ radians},\\\text{B: }255^\circ.\end{gathered}$

Full solution

Case A: multiply the numerical degree measure by $\pi/180$ .
$\begin{aligned}\theta&=84\cdot\frac{\pi}{180},\\\theta&=\frac{7\pi}{15}.\end{aligned}$

Case B: multiply the radian measure by $180/\pi$ .
$\begin{aligned}\alpha&=\frac{17\pi}{12}\cdot\frac{180}{\pi},\\\alpha&=255^\circ.\end{aligned}$

Answer

$\begin{gathered}\text{A: }\frac{7\pi}{15}\text{ radians},\\\text{B: }255^\circ.\end{gathered}$

Key idea

Changing the angular unit changes its numerical measure, not the turn.
Problem 3 Recover both circle dimensions

A selected arc is $15\pi$ cm long and has central turn $5\pi/6$ radians. Find the circle's radius and diameter.

Use the stated arc length and central turn.

Text description of this figure

The selected blue arc AB has length 15 pi centimeters and central turn five pi over six radians. The radius is labeled r, with no computed radius or diameter shown.
1. Hint 1
  The radian measure is the arc length divided by the radius.
2. Hint 2
  Recover the radius from the ratio before doubling it for the diameter.
Answer

$\begin{gathered}r=18\text{ cm},\\d=36\text{ cm}.\end{gathered}$

Full solution

Reverse the arc-length product to recover the radius.
$\begin{aligned}r&=15\pi\div\frac{5\pi}{6},\\r&=18\text{ cm}.\end{aligned}$

The diameter contains two radii.
$d=36\text{ cm}.$

A radius of $18$ cm multiplied by $5\pi/6$ returns the given arc length of $15\pi$ cm.

Answer

$\begin{gathered}r=18\text{ cm},\\d=36\text{ cm}.\end{gathered}$

Key idea

Recover the radius from the arc ratio before finding a diameter.
Problem 4 The two routes between endpoints

A circle has center $O$ and radius $14$ cm. The minor turn from $OA$ to $OB$ is $5\pi/7$ radians. Point $C$ lies on the major arc from $A$ to $B$ .

Part A: find the minor arc length.

Part B: find the major arc's radian measure and length.

The two routes meet only at their endpoints.

Text description of this figure

The circle has center O and radius 14 centimeters. The blue minor arc AB has turn five pi over seven radians. C marks a point on the dashed red major arc ACB. Neither arc length is stated.
1. Hint 1
  The two turns together make one full turn.
2. Hint 2
  Find the remaining radian measure before applying the arc-length product.
Answer

$\begin{gathered}\text{A: }10\pi\text{ cm},\\\text{B: }\frac{9\pi}{7}\text{ radians},\\s_{\mathrm{major}}=18\pi\text{ cm}.\end{gathered}$

Full solution

Part A: multiply the radius by the minor radian measure.
$\begin{aligned}s_{\mathrm{minor}}&=14\cdot\frac{5\pi}{7},\\s_{\mathrm{minor}}&=10\pi\text{ cm}.\end{aligned}$

Part B: subtract the minor turn from a full turn.
$\begin{aligned}\theta_{\mathrm{major}}&=2\pi-\frac{5\pi}{7},\\\theta_{\mathrm{major}}&=\frac{9\pi}{7}.\end{aligned}$

Use that turn for the longer route.
$\begin{aligned}s_{\mathrm{major}}&=14\cdot\frac{9\pi}{7},\\s_{\mathrm{major}}&=18\pi\text{ cm}.\end{aligned}$

The lengths add to $28\pi$ cm, which is the full circumference.

Answer

$\begin{gathered}\text{A: }10\pi\text{ cm},\\\text{B: }\frac{9\pi}{7}\text{ radians},\\s_{\mathrm{major}}=18\pi\text{ cm}.\end{gathered}$

Key idea

A major arc uses the full turn left after the minor turn, not the smaller angle again.
Problem 5 A probe follows a curved guide

A probe follows a circular guide of radius $45$ mm through a central turn of $4\pi/9$ radians. Find its exact travel distance in millimeters and centimeters. Then give the centimeter distance to the nearest hundredth, using a calculator value of $\pi$ .

The distance follows the curve, not the straight separation.

Text description of this figure

The circular guide has radius 45 millimeters, and the selected blue arc has central turn four pi over nine radians. The drawing identifies the traveled curve, not its chord.
1. Hint 1
  The stated turn is already in radians.
2. Hint 2
  Measure the curved travel in millimeters first, then use 10 millimeters per centimeter.
Answer

$\begin{gathered}20\pi\text{ mm},\\2\pi\text{ cm},\\\text{about }6.28\text{ cm}.\end{gathered}$

Full solution

Multiply the radius by the radian measure.
$\begin{aligned}s&=45\cdot\frac{4\pi}{9},\\s&=20\pi\text{ mm}.\end{aligned}$

Divide the millimeter distance by $10$ to express it in centimeters.
$s=2\pi\text{ cm}.$

Keep the exact value until the requested rounding.
$s\approx6.28\text{ cm}.$

Answer

$\begin{gathered}20\pi\text{ mm},\\2\pi\text{ cm},\\\text{about }6.28\text{ cm}.\end{gathered}$

Key idea

Changing the length unit changes both the exact expression and its numerical approximation consistently.
Problem 6 Matching turns in unequal circles

The same central turn selects corresponding arcs in two circles. In the smaller circle, the radius is $9$ cm and the selected arc is $12\pi/5$ cm long. In the larger circle, the radius is $15$ cm.

Find the common radian measure and the larger selected arc length.

A common turn gives corresponding arcs under a dilation.

Text description of this figure

Two circles show matching selected central turns. The first has radius 9 centimeters and selected arc length twelve pi over five centimeters. The second has radius 15 centimeters and unknown corresponding arc length s.
1. Hint 1
  The turn is measured by the arc-length to radius ratio in either circle.
2. Hint 2
  Use the known smaller circle to find that ratio, then use it in the larger circle.
Answer

$\begin{gathered}\theta=\frac{4\pi}{15}\text{ radians},\\s_{\mathrm{large}}=4\pi\text{ cm}.\end{gathered}$

Full solution

Compute the angle from the smaller circle using matching length units.
$\begin{aligned}\theta&=\frac{12\pi/5}{9},\\\theta&=\frac{4\pi}{15}.\end{aligned}$

The same angle selects the larger arc.
$\begin{aligned}s_{\mathrm{large}}&=15\cdot\frac{4\pi}{15},\\s_{\mathrm{large}}&=4\pi\text{ cm}.\end{aligned}$

The arc lengths also have the radius ratio $15/9$ , which checks the result.

Answer

$\begin{gathered}\theta=\frac{4\pi}{15}\text{ radians},\\s_{\mathrm{large}}=4\pi\text{ cm}.\end{gathered}$

Key idea

One corresponding arc can determine the angle and therefore the arc in another circle.
Problem 7 What a dilation preserves

A circle has radius $11$ cm and a selected arc of length $22$ cm. A dilation by factor $4$ produces a second circle and its corresponding arc. Find the new radius, arc length, and common radian measure.

Then let a positive dilation factor $k$ act on a circle of positive radius $r$ with selected arc length $s$ . Justify why the radian measure stays unchanged.
1. Hint 1
  A dilation scales every corresponding length by the same positive factor.
2. Hint 2
  In the general quotient, both the numerator and denominator contain the same factor.
Answer

$\begin{gathered}R=44\text{ cm},\\S=88\text{ cm},\\\theta=2\text{ radians}.\end{gathered}$
The radian measure is unchanged under every positive dilation.

Full solution

In the numerical example, both lengths are multiplied by $4$ .
$\begin{gathered}R=44\text{ cm},\\S=88\text{ cm}.\end{gathered}$

Their ratios are equal.
$\begin{gathered}\frac{22}{11}=2,\\\frac{88}{44}=2.\end{gathered}$
The common turn therefore measures $2$ radians.

In general, the new radius is $kr$ and the new corresponding arc length is $ks$ .
The curved arc scales because its arbitrarily close polygonal approximations scale by the same factor.

Both factors in the denominator are positive, so cancellation is valid.
$\frac{ks}{kr}=\frac{s}{r}.$
This quotient defines the radian measure in the two circles.

Both measured lengths receive the same scale factor.

Answer

$\begin{gathered}R=44\text{ cm},\\S=88\text{ cm},\\\theta=2\text{ radians}.\end{gathered}$
The radian measure is unchanged under every positive dilation.

Key idea

A common positive scale factor cancels from the arc-length to radius ratio.
Problem 8 Which records describe a single arc?

A circle has radius $6$ cm. For each record below, decide whether it describes a positive-length arc traced once without repeating any part of the circle. If it does, give its length.

Case A: a central turn of $3\pi/2$ radians.

Case B: a central turn of $7$ radians.

Case C: a zero central turn.
1. Hint 1
  Compare each turn with zero and one full turn.
2. Hint 2
  An angle that exceeds a full turn repeats part of the circle; a zero turn travels no positive length.
Answer

A: valid, $9\pi$ cm. B: invalid for one nonrepeating arc. C: zero distance, not a positive-length arc.

Full solution

Case A: the turn is positive and less than $2\pi$ .
$\begin{aligned}s_A&=6\cdot\frac{3\pi}{2},\\s_A&=9\pi\text{ cm}.\end{aligned}$
It selects a major arc.

Case B: $7$ radians exceeds a full turn of $2\pi$ radians.
Its travel distance would be $42$ cm, which is greater than the circumference, so it repeats part of the circle.

Case C: the zero turn travels zero distance.
It does not meet the stated positive-length condition.

Case A is a valid single arc; the other records fail the stated conditions.

Answer

A: valid, $9\pi$ cm. B: invalid for one nonrepeating arc. C: zero distance, not a positive-length arc.

Key idea

An algebraic distance can describe a repeated journey without describing a single nonrepeating arc.
Problem 9 Two adjacent arc pieces

Point $B$ lies on the selected minor arc from $A$ to $C$ in a circle of radius $18$ cm. That whole arc has central turn $4/3$ radians.

The arc from $A$ to $B$ is $(x+4)$ cm long.

The arc from $B$ to $C$ is $(2x-1)$ cm long.

Find $x$ and the two arc lengths. Check that both lengths are positive.

The two pieces meet at B without overlapping.

Text description of this figure

A, B, and C occur in that order along the selected blue minor arc. Its whole central turn is four thirds of a radian and its radius is 18 centimeters. The adjacent arc lengths are x plus 4 and 2x minus 1 centimeters.
1. Hint 1
  First find the whole selected arc length from its radius and turn.
2. Hint 2
  The two adjacent arc lengths add to that total, producing a linear equation.
Answer

$\begin{gathered}x=7,\\s_{AB}=11\text{ cm},\\s_{BC}=13\text{ cm}.\end{gathered}$
Both lengths are positive.

Full solution

The whole selected arc length is
$\begin{aligned}s_{AC}&=18\cdot\frac43,\\s_{AC}&=24\text{ cm}.\end{aligned}$

The two pieces meet only at B, so add their lengths.
$\begin{aligned}(x+4)+(2x-1)&=24,\\3x+3&=24,\\x&=7.\end{aligned}$

Substitute into the two stated lengths.
$\begin{gathered}s_{AB}=11\text{ cm},\\s_{BC}=13\text{ cm}.\end{gathered}$
Both are positive and their sum is $24$ cm.

Answer

$\begin{gathered}x=7,\\s_{AB}=11\text{ cm},\\s_{BC}=13\text{ cm}.\end{gathered}$
Both lengths are positive.

Key idea

Adjacent arc lengths add before the arc-length relationship is reversed or checked.
Problem 10 Reconstruct the circle and the shorter turn

Two distinct points $A$ and $B$ divide a circle into a minor arc and a major arc. The major arc is $45$ cm long. The ratio of major arc length to minor arc length is $5:2$ .

Part A: find the minor arc length and the circle's radius exactly.

Part B: find the minor arc's central turn in radians and in degrees exactly.

The given ratio compares curved lengths on the same circle.

Text description of this figure

The two arcs share endpoints A and B. The major route through C is dashed red and has given length 45 centimeters. The minor route is blue. The radius is unknown, and the given major-to-minor length ratio is five to two.
1. Hint 1
  The known major length is five equal shares of the combined circumference.
2. Hint 2
  Find the remaining two shares, then use the full circumference to recover the radius.
3. Hint 3
  The minor turn occupies the same fraction of a full turn as the minor arc occupies of the circumference.
Answer

$\begin{gathered}s_{\mathrm{minor}}=18\text{ cm},\\r=\frac{63}{2\pi}\text{ cm},\\\theta_{\mathrm{minor}}=\frac{4\pi}{7}\text{ radians},\\\alpha_{\mathrm{minor}}=\frac{720}{7}^\circ.\end{gathered}$

Full solution

Part A: five shares have length $45$ cm, so one share is $9$ cm and the minor arc's two shares total $18$ cm.
The circumference is
$C=63\text{ cm}.$

Recover the radius from the full circumference.
$r=\frac{63}{2\pi}\text{ cm}.$
This is positive.

Part B: the minor arc occupies two of the seven total shares.
Its central turn occupies that fraction of a full turn.
$\begin{aligned}\theta_{\mathrm{minor}}&=\frac27\cdot2\pi,\\\theta_{\mathrm{minor}}&=\frac{4\pi}{7}.\end{aligned}$

Convert the same turn to degrees.
$\alpha_{\mathrm{minor}}=\frac{720}{7}^\circ.$

The turn is less than a half-turn, as a minor arc requires.
Multiplying the recovered radius by $4\pi/7$ returns the minor length of $18$ cm.

Answer

$\begin{gathered}s_{\mathrm{minor}}=18\text{ cm},\\r=\frac{63}{2\pi}\text{ cm},\\\theta_{\mathrm{minor}}=\frac{4\pi}{7}\text{ radians},\\\alpha_{\mathrm{minor}}=\frac{720}{7}^\circ.\end{gathered}$

Key idea

An arc-length ratio determines fractions of both the circumference and the full central turn.

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