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Circumference and Circle Area: 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 Three measurements from a diameter

A circle has diameter $17/3$ cm. Give exact results for each case.

(a) Its radius.

(b) Its circumference.

(c) Its disk area.

The diameter is given.

Text description of this figure

The circle diameter crosses center O and is labeled 17/3 cm. No radius or area value is printed.
1. Hint 1
  Check which measurement goes into each formula.
2. Hint 2
  The radius is half the diameter; square that radius for area.
Answer

$\begin{gathered}r=\frac{17}{6}\text{ cm}\\C=\frac{17\pi}{3}\text{ cm}\\A=\frac{289\pi}{36}\text{ cm}^2\end{gathered}$

Full solution

(a) Halve the diameter.
$\begin{aligned}r&=\frac12\left(\frac{17}{3}\right)\\r&=\frac{17}{6}\text{ cm}\end{aligned}$

(b) Multiply the diameter by pi.
$\begin{aligned}C&=\pi\left(\frac{17}{3}\right)\\C&=\frac{17\pi}{3}\text{ cm}\end{aligned}$

(c) Square the radius before multiplying by pi.
$\begin{aligned}A&=\pi\left(\frac{17}{6}\right)^2\\A&=\frac{289\pi}{36}\text{ cm}^2\end{aligned}$

Answer

$\begin{gathered}r=\frac{17}{6}\text{ cm}\\C=\frac{17\pi}{3}\text{ cm}\\A=\frac{289\pi}{36}\text{ cm}^2\end{gathered}$

Key idea

The diameter directly gives circumference, but it must be halved before calculating disk area.
Problem 2 Recover a circle from its area

A disk has exact area $300\pi$ square centimeters. Find the circle's radius, diameter, and circumference exactly.

The enclosed disk area is given.

Text description of this figure

A circle with an unmeasured radius r and the supplied disk area 300 pi square centimeters.
1. Hint 1
  Undo the multiplication by pi before undoing the square.
2. Hint 2
  Take the positive square root because the radius is a length.
Answer

$\begin{gathered}r=10\sqrt3\text{ cm}\\d=20\sqrt3\text{ cm}\\C=20\pi\sqrt3\text{ cm}\end{gathered}$

Full solution

Divide the area by pi.
$r^2=300$

The positive square root is
$r=10\sqrt3\text{ cm}$

Double the radius to obtain the diameter.
$d=20\sqrt3\text{ cm}$

Multiply the radius by twice pi.
$C=20\pi\sqrt3\text{ cm}$

Answer

$\begin{gathered}r=10\sqrt3\text{ cm}\\d=20\sqrt3\text{ cm}\\C=20\pi\sqrt3\text{ cm}\end{gathered}$

Key idea

A positive radius is recovered by dividing area by pi and taking the positive square root.
Problem 3 Two square enclosures

A circle has radius $3\sqrt2$ cm. One square is inscribed in it. Another square has parallel sides tangent to the circle.

(a) Find each square's perimeter and area.

(b) Give strict lower and upper bounds for the circle circumference and disk area from those squares.

A circle between inscribed and tangent squares.

Text description of this figure

A radius 3 square root 2 circle sits between two regular squares with parallel sides. The blue square is inscribed and the red dashed square is tangent; no computed side lengths or areas are shown.
1. Hint 1
  The inner square diagonal is a diameter; the outer square apothem is the radius.
2. Hint 2
  A square diagonal is its side multiplied by square root 2.
3. Hint 3
  Compare the circle separately with the straight perimeters and the polygon areas.
Answer

Inner: perimeter 24 cm, area 36 square cm. Outer: perimeter $24\sqrt2$ cm, area 72 square cm.
$\begin{gathered}24<C<24\sqrt2\text{ cm}\\36<A<72\text{ cm}^2\end{gathered}$

Full solution

(a) The diameter is $6\sqrt2$ cm, so the inner square has side 6 cm.
Its perimeter is 24 cm and its area is 36 square cm.

The outer square has apothem $3\sqrt2$ cm and side $6\sqrt2$ cm.
Its perimeter is $24\sqrt2$ cm and its area is 72 square cm.

(b) The circle lies between the two polygons.
Its circumference is greater than the inner perimeter and less than the outer perimeter.
$24<C<24\sqrt2$

Its disk area is greater than 36 and less than 72 square cm.
These are finite bounds, not an exact area calculation.

Answer

Inner: perimeter 24 cm, area 36 square cm. Outer: perimeter $24\sqrt2$ cm, area 72 square cm.
$\begin{gathered}24<C<24\sqrt2\text{ cm}\\36<A<72\text{ cm}^2\end{gathered}$

Key idea

A polygon enclosure supplies separate bounds for curved length and enclosed area.
Problem 4 Certify the rounded area

A circle has radius 5.6 cm. Use the certified bounds $3.14159<\pi<3.14160$ .

(a) State the exact disk area.

(b) Determine its area rounded to the nearest hundredth of a square centimeter, and explain why these bounds certify the rounding.
1. Hint 1
  Keep pi in the exact expression.
2. Hint 2
  Multiply both pi bounds by the positive squared radius.
3. Hint 3
  Check whether the resulting interval lies inside one rounding interval.
Answer

Exact area $31.36\pi$ square cm; rounded area 98.52 square cm.

Full solution

(a) The squared radius is 31.36, so the exact area is $31.36\pi$ square cm.

(b) Multiplication by 31.36 preserves the two inequalities.
$\begin{gathered}98.5202624<A\\A<98.520576\end{gathered}$

Both endpoints lie strictly between 98.515 and 98.525, the two rounding boundaries for 98.52.
Every possible area in the certified interval therefore rounds to 98.52 square cm.

Answer

Exact area $31.36\pi$ square cm; rounded area 98.52 square cm.

Key idea

A valid finite interval can certify rounding even though it does not provide an exact decimal value.
Problem 5 A length factor and an area change

A disk has area $75\pi$ square centimeters. A second circle has $7/5$ of the first circle's circumference.

Find the second disk's exact area and the percentage increase in area from the first disk.

The circumference ratio is 7 to 5; only the first disk area is supplied.

Text description of this figure

Two circles have the given circumference ratio. The smaller disk area is 75 pi square centimeters; the second area is unknown.
1. Hint 1
  Circumference and radius share a length scale factor.
2. Hint 2
  Area uses the square of that factor.
3. Hint 3
  For percentage increase, subtract the original area before dividing by it.
Answer

Area $147\pi$ square cm; increase 96 percent.

Full solution

The radius scale factor is $7/5$ , so the area scale factor is $49/25$ .

The new disk area is
$\begin{aligned}A_2&=75\pi\left(\frac{49}{25}\right)\\A_2&=147\pi\text{ cm}^2\end{aligned}$

The increase is $72\pi$ square cm.
Relative to $75\pi$ , it is
$\frac{72\pi}{75\pi}=0.96$
This is a 96 percent increase.

Answer

Area $147\pi$ square cm; increase 96 percent.

Key idea

A fractional circumference factor must be squared before it is applied to disk area.
Problem 6 Recover two radii

Two circles have positive radii. The larger circumference exceeds the smaller by $4\pi$ cm. The larger disk area exceeds the smaller by $52\pi$ square cm. Find both radii.
1. Hint 1
  Translate the two differences into equations involving the radii.
2. Hint 2
  The difference of two squares factors as the difference times the sum.
3. Hint 3
  Use the known difference and sum to recover the two positive lengths.
Answer

Larger radius 14 cm; smaller radius 12 cm.

Full solution

Let the larger and smaller radii, in centimeters, be R and r.
The circumference difference gives
$2\pi(R-r)=4\pi$
Thus $R-r=2$ .

The area difference gives
$R^2-r^2=52$
Factor it and use the difference already found.
$2(R+r)=52$
Thus $R+r=26$ .

Adding the sum and difference gives $2R=28$ , so R is 14 cm.
Then r is 12 cm.
Their circumference difference is $4\pi$ cm, and their area difference is $52\pi$ square cm.

Answer

Larger radius 14 cm; smaller radius 12 cm.

Key idea

The two measurement differences reveal the difference and the sum of the radii.
Problem 7 Three enclosures

An unknown positive length x, measured in centimeters, has each of these certified enclosures.

First: $6.28<x<6.29$ .

Second: $6.283<x<6.284$ .

Third: $6.2831<x<6.2832$ .

Do these three rows alone force the exact equality $x=6.28318$ ? Give two different numerical possibilities that the rows do not distinguish, and explain what an exact exhaustion conclusion requires.
1. Hint 1
  Test whether more than one number fits even the narrowest interval.
2. Hint 2
  Choose two different numbers strictly between 6.2831 and 6.2832.
3. Hint 3
  Distinguish an arbitrarily refinable enclosure from a fixed finite list.
Answer

No. For example, 6.28315 and 6.28319 both satisfy every row. Exact exhaustion requires valid bounds with arbitrarily small gaps around the quantity and the proposed exact value.

Full solution

The values 6.28315 and 6.28319 both lie strictly between 6.2831 and 6.2832, and therefore lie in the two wider intervals too.
The three rows cannot distinguish them.

Thus the table does not force the particular decimal 6.28318.
It gives a finite precision enclosure.

An exact exhaustion argument needs valid bounds that can be tightened past any prescribed positive error, with both the actual quantity and the proposed exact value inside.
A fixed positive difference would then be ruled out by a narrower enclosure.

Answer

No. For example, 6.28315 and 6.28319 both satisfy every row. Exact exhaustion requires valid bounds with arbitrarily small gaps around the quantity and the proposed exact value.

Key idea

A finite table can narrow possibilities without selecting one exact number.
Problem 8 An exact area from shrinking bounds

A circle has positive radius $r$ and circumference $C$ . Regular inscribed polygons have perimeters $P$ and apothems $h$ . Parallel-sided tangent polygons have perimeters $Q$ .

As their arcs are repeatedly bisected, the perimeter gap can be made arbitrarily small, both perimeters approach C, and the apothems h approach r.

Derive an exact expression for the disk area A using only C and r. Explain why a nonzero difference from your expression is impossible.
1. Hint 1
  The disk area lies between the two polygon areas.
2. Hint 2
  Express each polygon area through its perimeter and its perpendicular center-to-side distance.
3. Hint 3
  Show that both polygon areas approach one value, and use the shrinking gap.
Answer

$A=Cr/2$ .

Full solution

The inner polygon area is $I=Ph/2$ , and the outer polygon area is $U=Qr/2$ .
The outer apothem is exactly r because its sides are tangent.
The disk area lies between I and U.

The inequalities $P<C<Q$ and $h<r$ also place the value $Cr/2$ between I and U.

As refinement continues, both polygon areas approach $Cr/2$ .
Their gap can be made as small as desired, since
$\begin{aligned}E_p&=\frac{r(Q-P)}2,\\E_h&=\frac{P(r-h)}2,\\U-I&=E_p+E_h.\end{aligned}$
Here P remains bounded, and both differences shrink as far as required.

If A differed from $Cr/2$ by a fixed positive amount, choose an enclosure narrower than that amount.
Both numbers remain inside it, which is impossible for numbers that far apart.
Therefore $A=Cr/2$ exactly.

Answer

$A=Cr/2$ .

Key idea

Exact equality follows when every possible positive discrepancy is ruled out by a valid shrinking enclosure.
Problem 9 A range of panel sizes

A circular panel must cover at least $18\pi$ square meters. Its edge can use at most $10\pi$ meters of trim.

Find all positive radii that meet both requirements. Give the range exactly and approximate its lower endpoint to the nearest hundredth of a meter.
1. Hint 1
  Translate the area and boundary requirements separately.
2. Hint 2
  For a positive radius, the area lower bound becomes a lower bound on r.
3. Hint 3
  Use the intersection of the two allowed ranges.
Answer

$3\sqrt2\le r\le5$ meters; the lower endpoint is about 4.24 meters.

Full solution

The area condition is
$\pi r^2\ge18\pi$
Since r is positive, this requires
$r\ge3\sqrt2.$

The trim condition is
$2\pi r\le10\pi$
It requires $r\le5$ .

Both conditions hold exactly when $3\sqrt2\le r\le5$ , in meters.
The lower endpoint rounds to 4.24 meters; the exact endpoint remains $3\sqrt2$ , so 4.24 itself is not substituted for that exact bound.

Answer

$3\sqrt2\le r\le5$ meters; the lower endpoint is about 4.24 meters.

Key idea

A feasible radius must satisfy the area and boundary inequalities at the same time.
Problem 10 A circle measured through a chord

Circle O has chord AB of length 14 cm. The perpendicular distance from O to AB is 4 cm.

(a) Find its radius and exact circumference.

(b) Find the disk's exact area.

(c) A second disk has area $260\pi$ square centimeters. How many times the first circle's circumference is the second circle's circumference?

(d) Certified bounds are $3.14159<\pi<3.14160$ . Give a numerical enclosure for the first disk's area and its area rounded to the nearest hundredth of a square centimeter.

Explain what this finite enclosure guarantees and how that differs from the limiting argument behind the exact circle-area formula.

Chord AB and its perpendicular center distance.

Text description of this figure

Circle O has chord AB of length 14 cm. Its perpendicular foot is M with OM 4 cm. Radius OA is unmeasured. No chord half-length or computed radius is supplied.
1. Hint 1
  The perpendicular from the center bisects the chord.
2. Hint 2
  Use a right triangle with half the chord and the center distance.
3. Hint 3
  The ratio of positive radii is the square root of the disk-area ratio.
4. Hint 4
  Multiply both positive pi bounds by the squared radius. Check which single rounding bin contains the whole enclosure.
Answer

$\begin{gathered}r=\sqrt{65}\text{ cm}\\C=2\pi\sqrt{65}\text{ cm}\\A=65\pi\text{ cm}^2\\204.20335<A\\A<204.204\end{gathered}$
Second circumference: twice first. Bounds certify 204.20 square cm rounded; exhaustion gives exactness.

Full solution

(a) Let M be the perpendicular foot.
The center-perpendicular chord theorem gives AM as 7 cm.
The radius is the hypotenuse of right triangle OMA.
$\begin{aligned}r^2&=7^2+4^2\\r^2&=65\end{aligned}$
The positive radius is $\sqrt{65}$ cm, and the circumference is $2\pi\sqrt{65}$ cm.

(b) The area uses the squared radius directly.
$A=65\pi\text{ cm}^2$

(c) The second-to-first area ratio is 4.
The positive radius ratio, and therefore the circumference ratio, is 2.
Equivalently, the second radius is $\sqrt{260}=2\sqrt{65}$ cm.

(d) Multiplying the certified bounds by the positive squared radius gives
$\begin{aligned}65(3.14159)&=204.20335\\65(3.14160)&=204.204\end{aligned}$

The area, in square centimeters, satisfies
$\begin{gathered}204.20335<A\\A<204.204.\end{gathered}$
Every value in this interval rounds to $204.20\text{ cm}^2$ .

This fixed enclosure has positive width, so it certifies the rounding, not the exact area by itself.
The exact formula comes from indefinitely refinable inner and outer polygon areas approaching the same limit, here $65\pi$ square centimeters.
The disk area is trapped between them, so it shares that limit.

Answer

$\begin{gathered}r=\sqrt{65}\text{ cm}\\C=2\pi\sqrt{65}\text{ cm}\\A=65\pi\text{ cm}^2\\204.20335<A\\A<204.204\end{gathered}$
Second circumference: twice first. Bounds certify 204.20 square cm rounded; exhaustion gives exactness.

Key idea

Exact circle measures, certified numerical bounds and circumference scaling answer different measurement questions.

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