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Circles: 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 The unfinished drawing

Complete the figure by drawing a diameter with $T$ as one endpoint and labeling its other endpoint $U$ .

A circle with center $K$ and a segment from $K$ to the point $T$ on the circle.

Text description of this figure

A circle with a dot at its center, labeled K. A point labeled T lies on the circle on its upper right, about halfway between the top of the circle and its rightmost point. A straight segment joins K to T. No other points, segments or lengths are drawn or labeled.
1. Hint 1
  A diameter passes through the center and reaches the circle at both ends.
2. Hint 2
  Segment $KT$ is one radius. A diameter needs a second radius from $K$ that makes a straight line with it.
Answer

Draw $TU$ through $K$ , with $U$ on the circle opposite $T$ .

Full solution

The center is $K$ .
A diameter starting at $T$ must pass through $K$ and continue to the other side of the circle.
Put $U$ at that second intersection and draw segment $TU$ .

The two parts $TK$ and $KU$ are radii of equal length, so the completed segment crosses the entire circle through its center.

Answer

Draw $TU$ through $K$ , with $U$ on the circle opposite $T$ .

Key idea

A diameter consists of two radii extending in opposite directions from the center.
Problem 1: Mark complete
Problem 2 A pair of lengths

A circle's diameter is 6 cm longer than its radius. Find the radius.
1. Hint 1
  The diameter contains two copies of the radius.
2. Hint 2
  If the radius is $r$ cm, compare $2r$ with $r+6$ .
Answer

6 cm.

Full solution

Let the radius be $r$ cm.
The diameter is both $2r$ cm and $(r+6)$ cm.
Equate the two descriptions and subtract $r$ .
$\begin{aligned}2r&=r+6\\r&=6\end{aligned}$

A radius of 6 cm gives a diameter of 12 cm, which is 6 cm longer than the radius.

Answer

6 cm.

Key idea

The difference between a diameter and its radius is one radius.
Problem 2: Mark complete
Problem 3 Pieces of the boundary

Three arcs make up an entire circle without overlap. Their lengths are $5\pi$ cm, $2\pi$ cm, and $\pi$ cm. Find the diameter.
1. Hint 1
  The lengths of all the arcs add to the circumference.
2. Hint 2
  The circumference is $\pi$ times the diameter, so divide the total arc length by $\pi$ .
Answer

8 cm.

Full solution

Add the three arc lengths to get the circumference in cm.
$\begin{aligned}C&=5\pi+2\pi+\pi\\C&=8\pi\end{aligned}$

Since $C=\pi d$ and $\pi$ is positive, divide by $\pi$ .
$\begin{aligned}d&=\frac{8\pi}{\pi}\\d&=8\end{aligned}$
A diameter of 8 cm has circumference $8\pi$ cm, equal to the three arcs together.

Answer

8 cm.

Key idea

Arcs that together make up a whole circle, without overlap, add to its circumference.
Problem 3: Mark complete
Problem 4 The badge outline

The figure shows a badge formed by a triangle and a semicircle. Find the area of the badge and its outer perimeter. Give exact answers in terms of $\pi$ , keeping $\pi$ as a symbol (for example, $(2+5\pi)$ cm).

A badge made of triangle $ABC$ and a semicircle that share the segment $AB$ .

Text description of this figure

A badge shape, lightly shaded. A horizontal segment AB is drawn dashed across the middle of the badge, with A at the left end, B at the right end, and a point M between them; matching tick marks show that AM and MB are equal. Above AB is triangle ABC, with C directly above M. Sides AC and BC are each labeled 5 cm. A dashed segment runs from C straight down to M, is labeled 4 cm, and carries a small square showing a right angle with AB. Thin lines drop straight down from A and B past the bottom of the semicircle, and a bracket between them, below the badge, marks the distance from A to B as 6 cm. Below AB is a semicircle that has AB as its diameter, so the triangle and the semicircle share AB. The outer edge of the badge is the two sides AC and BC and the curved edge of the semicircle.
1. Hint 1
  The shared straight edge separates the two regions but is inside the badge.
2. Hint 2
  The semicircle's diameter is $AB$ ; half of it is the radius.
3. Hint 3
  Add the triangle and semicircle areas, then trace just the two exposed triangle sides and the curved edge.
Answer

Area $(12+\frac{9\pi}{2})$ square cm, or $(12+4.5\pi)$ square cm; perimeter $(10+3\pi)$ cm.

Full solution

The semicircle has diameter $AB$ , which is 6 cm, so its radius is 3 cm.

The triangle has base 6 cm and perpendicular height 4 cm.
Its area $A_t$ , in square cm, is
$\begin{aligned}A_t&=\frac{1}{2}(6)(4)\\A_t&=12\end{aligned}$
The semicircle area $A_s$ , in square cm, is
$\begin{aligned}A_s&=\frac{1}{2}\pi(3)^2\\A_s&=\frac{9\pi}{2}\end{aligned}$
The total area is $(12+\frac{9\pi}{2})$ square cm.

The shared diameter is inside the badge, so it is not part of the outer perimeter.
The curved edge is half the circumference of a circle of radius 3 cm, so its length is $\frac{1}{2}(2\pi)(3)=3\pi$ cm.
Add it to the two sides of length 5 cm.
$\begin{aligned}P&=5+5+3\pi\\P&=10+3\pi\end{aligned}$
The perimeter is $(10+3\pi)$ cm.

Answer

Area $(12+\frac{9\pi}{2})$ square cm, or $(12+4.5\pi)$ square cm; perimeter $(10+3\pi)$ cm.

Key idea

A shared edge contributes to the separate pieces but does not belong to the outer boundary of the combined shape.
Problem 4: Mark complete
Problem 5 The paper disk

A circular paper disk has diameter 22 cm. It is cut along one diameter into two semicircles, with no paper lost. Find the area of each piece and the increase in the total boundary length of the two separate pieces. Use $\pi\approx3.14$ .
1. Hint 1
  Cutting preserves area but creates new straight edges.
2. Hint 2
  Each piece has half the original area, while each gets one straight edge as long as the diameter.
3. Hint 3
  Compare the total of the two semicircle perimeters with the original circumference.
Answer

About 189.97 square cm per piece; the total boundary length increases by 44 cm.

Full solution

The radius is 11 cm.
Halving the disk area gives the area $A$ of each piece, in square cm.
$\begin{aligned}A&\approx\frac{1}{2}(3.14)(11)^2\\A&\approx189.97\end{aligned}$

The two curved edges together are the original circumference, about 69.08 cm.
Each semicircle also has a straight edge of 22 cm.
Thus the new total boundary length $P$ , in cm, is
$\begin{aligned}P&\approx69.08+2(22)\\P&\approx113.08\end{aligned}$
Subtracting the old circumference leaves an increase of exactly 44 cm, because the curved edges have the same total length as before.
The two areas add to about 379.94 square cm, the original disk area.

Answer

About 189.97 square cm per piece; the total boundary length increases by 44 cm.

Key idea

A cut can preserve the total area while adding boundary to the separate pieces.
Problem 5: Mark complete
Problem 6 Two sign designs

A maker can use at most 30 cm of edging and at most 60 square cm of material for one circular sign. Design A has radius 3.4 cm, and design B has radius 4.6 cm. Which design meets both limits? Give the edging length and material area for each design, using $\pi\approx3.14$ and rounding to the nearest hundredth.
1. Hint 1
  The edging follows the boundary, while the material fills the circular region.
2. Hint 2
  For each radius, calculate both $2\pi r$ and $\pi r^2$ .
3. Hint 3
  A design qualifies only if both of its measurements are within the stated limits.
Answer

Design A. A: about 21.35 cm and 36.30 square cm. B: about 28.89 cm and 66.44 square cm.

Full solution

For design A, use radius 3.4 cm.
The circumference $C_A$ is in cm and the area $A_A$ is in square cm.
$\begin{aligned}C_A&\approx2(3.14)(3.4)\\C_A&\approx21.352\end{aligned}$
This rounds to 21.35 cm.
$\begin{aligned}A_A&\approx3.14(3.4)^2\\A_A&\approx36.2984\end{aligned}$
This rounds to 36.30 square cm.
Both measurements are within the limits.

For design B, use radius 4.6 cm.
$\begin{aligned}C_B&\approx2(3.14)(4.6)\\C_B&\approx28.888\end{aligned}$
This rounds to 28.89 cm.
$\begin{aligned}A_B&\approx3.14(4.6)^2\\A_B&\approx66.4424\end{aligned}$
This rounds to 66.44 square cm.
Its edging fits the 30 cm limit, but its area exceeds 60 square cm.
Only design A meets both limits.

Answer

Design A. A: about 21.35 cm and 36.30 square cm. B: about 28.89 cm and 66.44 square cm.

Key idea

A circular design must satisfy its boundary and area requirements separately.
Problem 6: Mark complete
Problem 7 The circular frame

The shaded frame in the figure has area $(25\pi-24)$ square cm. The rectangular opening lies entirely inside the circle. Find its height $h$ .

A shaded circular frame around a rectangular opening. Not to scale.

Text description of this figure

A circle with its center marked O. A horizontal radius runs from O to the right edge of the circle and is labeled 5 cm. A rectangle with horizontal and vertical sides is centered at O and lies entirely inside the circle. A bracket below the rectangle marks its width as 4 cm, and a bracket to its left marks its height as h cm. The top right corner of the rectangle has a small square marking a right angle. The part of the disk outside the rectangle is shaded, forming the frame, and the rectangular opening is unshaded. The drawing is not to scale.
1. Hint 1
  The frame and the opening together fill the disk.
2. Hint 2
  Find the disk area, then subtract the stated frame area to get the opening area.
3. Hint 3
  Divide the opening area by its width to recover the height.
Answer

$h=6$ cm.

Full solution

The radius is 5 cm, so the full disk area is $25\pi$ square cm.
Subtract the frame area to get the rectangular opening area $A$ , in square cm.
$\begin{aligned}A&=25\pi-(25\pi-24)\\A&=24\end{aligned}$

The opening is 4 cm wide, so its area is $4h$ square cm.
$\begin{aligned}4h&=24\\h&=6\end{aligned}$
A 4 cm by 6 cm opening has area 24 square cm, leaving the stated frame area.

Answer

$h=6$ cm.

Key idea

Recover a missing dimension of a cutout by finding its area before reversing its area formula.
Problem 7: Mark complete
Problem 8 Two rows of pieces

The figure illustrates a disk of radius $r$ , with $r>0$ , cut into equal sectors and rearranged into two matching strips. Each strip uses half the sectors, pointing up and down in turn. No pieces overlap or are lost.

As the sectors get thinner, Lena says each strip approaches a parallelogram with base $\pi r$ and height $r$ , so the disk has area $2\pi r^2$ . Is her conclusion correct? Explain by finding the length that each strip's base gets closer to as the sectors get thinner, and deriving the disk area.

A disk cut into 16 equal sectors, and the same sectors rearranged into two strips, I and II.

Text description of this figure

At the top, a disk is divided into 16 equal pie-slice sectors by 16 radii drawn from its center, with the sectors shaded in two alternating tints. One radius is labeled r. Below the disk are two separate, matching strips, labeled I and II, one under the other. Each strip is a row of eight of the sectors, drawn the same size as in the disk, placed side by side and pointing up and down in turn, so that neighboring pieces meet along their straight edges without overlapping. Each piece keeps its curved edge: the pieces pointing up have their curved edges along the bottom of the strip, and the pieces pointing down have their curved edges along the top. No lengths other than r are labeled.
1. Hint 1
  The two strips must share all the curved edges from the original disk.
2. Hint 2
  Each strip contains half the sectors, and half of its curved edges lie along its lower boundary.
3. Hint 3
  Find the area of the parallelogram that one strip approaches, then account for both strips.
Answer

No. Each base approaches $\frac{\pi r}{2}$ , and the disk area is $\pi r^2$ .

Full solution

The original boundary has length $2\pi r$ .
Each strip receives half of those curved edges, and half of that share runs along its bottom.
The length $b$ that the base approaches is therefore
$\begin{aligned}b&=\frac{1}{4}(2\pi r)\\b&=\frac{\pi r}{2}\end{aligned}$

The height approaches $r$ .
Rearranging preserves area, so the disk area is the total area of the two parallelograms the strips approach.
$\begin{aligned}A&=2\left(\frac{\pi r}{2}\right)r\\A&=\pi r^2\end{aligned}$
Lena gave each strip the base belonging to one strip made from all the sectors, so her total is twice the disk area.

Answer

No. Each base approaches $\frac{\pi r}{2}$ , and the disk area is $\pi r^2$ .

Key idea

When pieces are divided among several strips, each strip receives only its share of the original boundary.
Problem 8: Mark complete
Problem 9 A circle in mixed units

A circle has circumference $15\pi$ cm and diameter 150 mm. There are 10 mm in 1 cm. Omar divides $15\pi$ by 150 and says the circle has a circumference-to-diameter ratio different from $\pi$ . Is Omar correct? Explain.
1. Hint 1
  The ratio defining $\pi$ compares two lengths measured in the same unit.
2. Hint 2
  Convert the diameter to cm before dividing the numerical measurements.
Answer

No. The circumference-to-diameter ratio is $\pi$ .

Full solution

Convert the diameter from mm to cm.
$150\div10=15$
The diameter is 15 cm, so the ratio of the two lengths is
$\frac{15\pi}{15}=\pi$

Omar divided numbers recorded in different units.
His calculation does not show a different value of $\pi$ ; using one unit gives the same ratio that every circle has.

Answer

No. The circumference-to-diameter ratio is $\pi$ .

Key idea

The fixed circumference-to-diameter ratio uses lengths expressed in the same unit.
Problem 9: Mark complete
Problem 10 The pair of circles

Two circles have positive radii whose sum is 10 cm. Kai says that the sum of their circumferences is fixed even though neither individual radius is known. Is Kai correct? Justify your decision and give that sum if it is fixed.
1. Hint 1
  Write each circumference in terms of its own radius.
2. Hint 2
  Look for a common factor in the sum of the two circumference expressions.
Answer

Yes; the sum is $20\pi$ cm.

Full solution

Let the radii in cm be $a$ and $b$ , so $a+b=10$ .
The circumference sum $C$ , in cm, is
$\begin{aligned}C&=2\pi a+2\pi b\\C&=2\pi(a+b)\\C&=20\pi\end{aligned}$
This depends on the radius sum, not on the separate radii, so Kai is correct.

For example, radii 4 cm and 6 cm give circumferences $8\pi$ cm and $12\pi$ cm.
Their sum is $20\pi$ cm, consistent with the result.

Answer

Yes; the sum is $20\pi$ cm.

Key idea

Adding circle circumferences is equivalent to multiplying the sum of the radii by $2\pi$ .
Problem 10: Mark complete

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