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Circles and Construction Tools: 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 Place the points

A circle has center $K$ and radius $8$ cm.

Points $A,B,C,D$ all lie in the circle's plane, and their distances from $K$ are $8,5,10,0$ cm, respectively.

List the points on the circle, the points in the disk, and the points outside the disk.
1. Hint 1
  Separate exactly the radius from less than the radius.
2. Hint 2
  The disk includes its boundary and its center.
Answer

On the circle: $A$ . In the disk: $A,B,D$ . Outside the disk: $C$ .

Full solution

All tested points lie in the circle's plane.
Only $A$ is exactly $8$ cm from $K$ , so only $A$ is on the circle.

Distances $8,5,0$ are at most $8$ cm, so $A,B,D$ are in the disk.

Point $C$ is $10$ cm away, farther than the radius, so it is outside the disk.

Answer

On the circle: $A$ . In the disk: $A,B,D$ . Outside the disk: $C$ .

Key idea

In the circle's plane, a circle uses an exact distance condition and a disk includes smaller distances too.
Problem 2 Draw the named parts

On paper, draw a circle with center $O$ .

Mark opposite boundary points $A$ and $B$ and draw segment $AB$ through $O$ .

Choose another boundary point $C$ off line $AB$ , then draw $OC$ and $AC$ .

Identify a radius, a diameter, and a chord that is not a diameter among the segments you drew.
1. Hint 1
  Check the endpoints first, then whether a segment goes through the center.
2. Hint 2
  A radius begins at the center; a diameter joins boundary points through the center.
Answer

$OC$ is a radius, $AB$ is a diameter, and $AC$ is a chord that is not a diameter.

Full solution

Segment $OC$ joins the center to a boundary point, so it is a radius.

Segment $AB$ joins boundary points through $O$ , so it is a diameter.

Both endpoints of $AC$ lie on the circle, so it is a chord.

The positive radius makes $A$ different from $O$ , so the unique line through them is line $AB$ .

The line through $A$ and $C$ differs from line $AB$ because $C$ is off $AB$ .

If that line also contained $O$ , uniqueness through $A$ and $O$ would force it to be $AB$ , placing $C$ on $AB$ .

Thus $AC$ misses $O$ and is not a diameter.

The circle has center O and a positive radius. A and B are opposite boundary points, with the center on segment AB. C is another boundary point; AC is a chord and OC is a radius.

Answer

$OC$ is a radius, $AB$ is a diameter, and $AC$ is a chord that is not a diameter.

Key idea

Circle-segment names depend on endpoint and center conditions, not orientation.
Problem 3 Two uses of one formula

A circle has center $O$ and radius $7$ cm.

Chord $AB$ passes through $O$ , while chord $CD$ with distinct endpoints does not.

An argument says, "The radius is 7 cm, so both chords have length $2r=14$ cm." Assess the formula's use for each chord, find $AB$ , and decide whether the exact length $CD$ is determined by these givens.
1. Hint 1
  Which condition makes a chord a diameter?
2. Hint 2
  Check the proposed formula separately for $AB$ and $CD$ ; the latter chord's endpoint positions remain unspecified.
Answer

$AB=14$ cm, and the diameter formula is justified for $AB$ . Its use for $CD$ is unjustified because that chord misses the center; the exact length $CD$ is not determined.

Full solution

Chord $AB$ passes through $O$ , so it is a diameter.
Segment addition adds its two radius lengths in centimeters:
$7+7=14.$

Thus $AB=14$ cm, and the argument's first use of $2r$ is justified.
Its use for $CD$ lacks the diameter condition, because $CD$ does not pass through $O$ .

The givens do not specify the positions of $C$ and $D$ along the circle, so they do not determine that chord's exact length.
Applying the diameter formula to both chords is not justified.

Answer

$AB=14$ cm, and the diameter formula is justified for $AB$ . Its use for $CD$ is unjustified because that chord misses the center; the exact length $CD$ is not determined.

Key idea

Check the center condition separately before applying the diameter formula to a chord.
Problem 4 Make a separate copy

On paper, draw a source segment $AB$ of length $6$ cm and a separate ray $PX$ .

Put the ruler away.

Construct its copy $PQ$ on that ray using a compass and unmarked straightedge.

Leave the compass circle visible and explain why the copied length equals $AB$ .
1. Hint 1
  Which tool can carry a distance without reading a number?
2. Hint 2
  The point on the target ray must lie on a circle centered at its starting point.
Answer

The constructed segment satisfies $PQ=AB=6$ cm; $PQ$ is a radius drawn with the retained opening $AB$ .

Full solution

Set the compass point at $A$ and the pencil at $B$ .

Without changing the opening, draw a circle centered at $P$ and mark its intersection with ray $PX$ as $Q$ .

That circle has radius $AB$ , so the definition of a circle gives $PQ=AB=6$ cm.

Source segment AB and target radius PQ have the same compass opening. The target ray starts at P, and Q is its marked circle intersection. X is a named point on the continuing target ray beyond Q.

Answer

The constructed segment satisfies $PQ=AB=6$ cm; $PQ$ is a radius drawn with the retained opening $AB$ .

Key idea

Keeping the compass opening transfers an exact distance in the ideal construction.
Problem 5 Reproduce an opening

On paper, use a protractor to draw a source angle of $65^\circ$ and draw a separate target ray $PX$ .

Put the protractor away.

Copy the source opening above the target ray with a compass and unmarked straightedge.

Leave the helper circles visible and identify the two different distance settings you used.
1. Hint 1
  The first setting marks a point on each source ray and is retained at the target vertex.
2. Hint 2
  The second setting is the distance between the two source-circle marks.
Answer

The target opening is $65^\circ$ in the ideal construction. The settings are the common source/target circle radius and the chord between the two source marks.

Full solution

Draw a circle centered at the source vertex, meeting the rays at $A$ and $B$ .

With the same radius, draw a circle centered at $P$ , meeting the target baseline at $D$ .

Set the compass to $AB$ , draw the circle centered at $D$ , and choose its intersection $E$ with the target circle above the target baseline.

Draw ray $PE$ to complete the copy.
The settings were the common radius and then chord $AB$ ; the full proof of the copied angle is deferred to lesson 4.5.

Answer

The target opening is $65^\circ$ in the ideal construction. The settings are the common source/target circle radius and the chord between the two source marks.

Key idea

An angle copy transfers a shared radius first and the distance between the source marks second.
Problem 6 Construct three matching sides

On paper, draw a segment $AB$ of length $4$ cm.

Construct an equilateral triangle on it using a compass and unmarked straightedge.

Retain the two circles and explain why each new side has the same length as $AB$ .
1. Hint 1
  The third vertex must be a distance $AB$ from both endpoints.
2. Hint 2
  Use one circle condition for each of the two required distances.
Answer

The construction gives $AB=AC=BC=4$ cm; $AC$ and $BC$ are radii of circles with the common radius $AB$ .

Full solution

Draw a circle centered at $A$ with radius $AB$ and a circle centered at $B$ with the same radius.

Choose either intersection as $C$ and join $AC$ and $BC$ .

The first circle gives $AC=AB=4$ cm; the second gives $BC=AB=4$ cm.

All three sides are equal, which is the definition of an equilateral triangle.

The two circles are centered at A and B, each with radius AB. Their upper intersection is C. Each new side is a radius of one circle.

Answer

The construction gives $AB=AC=BC=4$ cm; $AC$ and $BC$ are radii of circles with the common radius $AB$ .

Key idea

The two circle conditions separately justify the two newly drawn side lengths.
Problem 7 Different radius settings

Segment $AB$ has length $4$ cm.

A circle centered at $A$ has radius $4$ cm and a circle centered at $B$ has radius $6$ cm.

They meet at $C$ .

Find the three side lengths of triangle $ABC$ and decide whether it is equilateral.
1. Hint 1
  Read the distance to each center from that circle's own radius.
2. Hint 2
  Compare all three lengths rather than assuming different circles have equal radii.
Answer

$AB=4$ cm, $AC=4$ cm, and $BC=6$ cm; the triangle is not equilateral.

Full solution

Because $C$ is on the circle centered at $A$ , $AC=4$ cm.

Because $C$ is on the circle centered at $B$ , $BC=6$ cm.

The given base is $4$ cm, so the three sides are $4,4,6$ cm and are not all equal.

Answer

$AB=4$ cm, $AC=4$ cm, and $BC=6$ cm; the triangle is not equilateral.

Key idea

Radii from different circles agree only when their radius settings agree.
Problem 8 Choose a source mark

A source angle has vertex $O$ .

One mark $A$ lies on its first ray and on a circle centered at $O$ with radius $5$ cm.

A proposed second mark $B$ lies on the other ray, inside that circle, with $OB=3$ cm.

Decide whether these are the two source marks specified by the angle-copy procedure, and state where the second mark must be placed.
1. Hint 1
  The first circle is meant to give equal distances along the two source rays.
2. Hint 2
  The second mark must satisfy the same exact radius condition as the first.
Answer

They are not the specified source marks. Place the second mark on the other ray at its intersection with the same circle, $5$ cm from $O$ .

Full solution

The first source circle must meet both source rays, so both marked points must be $5$ cm from $O$ .

The proposed $B$ is only $3$ cm away and lies inside the disk rather than on the circle.

Replace it with the intersection of the other ray and the circle, where the distance from $O$ is $5$ cm.

Answer

They are not the specified source marks. Place the second mark on the other ray at its intersection with the same circle, $5$ cm from $O$ .

Key idea

The angle-copy marks must satisfy equal radius conditions before their chord is transferred.
Problem 9 Follow a construction chain

A source segment has length $(3t+1)$ cm.

It is copied exactly as the base of an equilateral triangle, whose base is also labeled $(5t-9)$ cm.

Find $t$ and the triangle's perimeter.
1. Hint 1
  The original segment and copied base have equal lengths.
2. Hint 2
  After solving, use the equal side lengths to find the perimeter.
Answer

$t=5$ ; each side is $16$ cm and the perimeter is $48$ cm.

Full solution

The segment copy gives $3t+1=5t-9$ .

Solving gives $10=2t$ , so $t=5$ .

The base has length $3(5)+1=16$ cm, and the equilateral construction makes each other side that length.

The perimeter is
$3\times16=48\text{ cm}.$

Answer

$t=5$ ; each side is $16$ cm and the perimeter is $48$ cm.

Key idea

Follow each preserved length through the construction before doing the final calculation.
Problem 10 Build from an unmeasured source

On paper, draw a nonzero source segment $UV$ and a separate ray $PX$ .

Construct an equilateral triangle $PQR$ with $Q$ on ray $PX$ and each side equal in length to $UV$ .

Use only a compass and unmarked straightedge after drawing the givens.

Leave all helper circles visible and mark the resulting equal lengths.
1. Hint 1
  First create the required base on the target ray.
2. Hint 2
  Then use that new base as the common radius of two endpoint-centered circles.
Answer

A valid construction has $Q$ on ray $PX$ and $PQ=PR=QR=UV$ , with the circle at $P$ reused from the segment copy and the circle at $Q$ both retained.

Full solution

Set the compass to $UV$ and draw a circle centered at $P$ , meeting ray $PX$ at $Q$ .
The radius condition gives $PQ=UV$ .

Keep that opening and draw a circle centered at $Q$ .
The first circle already supplies the required circle centered at $P$ .

Choose either intersection of these two circles as $R$ and join $PR$ and $QR$ .

Because $R$ lies on the circle centered at $P$ with radius $UV$ , $PR=UV$ .

Because $R$ lies on the circle centered at $Q$ with radius $UV$ , $QR=UV$ .

Mark the four lengths equal using the separately established equalities $PQ=UV$ , $PR=UV$ , and $QR=UV$ .

The two circles are centered at P and Q, each with radius PQ. Their upper intersection is R. Each new side is a radius of one circle. The separate source segment UV has the same compass length as the base. The target ray starts at P, passes through Q and X, and continues to the right.

Answer

A valid construction has $Q$ on ray $PX$ and $PQ=PR=QR=UV$ , with the circle at $P$ reused from the segment copy and the circle at $Q$ both retained.

Key idea

A copied segment can become the given side of a second construction without ever reading its numerical length.

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