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Segments, Length, and Distance: 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 Separate the two notations

A segment with endpoints $A$ and $B$ has length $12.5$ centimeters.

Write a statement using $AB$ for its length.

Another segment $\overline{CD}$ is congruent to it.

Write a congruence statement and state $CD$ .
1. Hint 1
  The segment and its numerical measurement use different notation.
2. Hint 2
  Use bars and the congruence symbol for the objects; use an equals sign for the lengths.
Answer

$AB=12.5$ cm; $\overline{AB}\cong\overline{CD}$ ; $CD=12.5$ cm.

Full solution

The unmarked pair $AB$ denotes length, so $AB=12.5$ centimeters.

The bars identify the two segments: $\overline{AB}\cong\overline{CD}$ .
By the definition of segment congruence, their lengths are equal, so $CD=12.5$ centimeters too.

Answer

$AB=12.5$ cm; $\overline{AB}\cong\overline{CD}$ ; $CD=12.5$ cm.

Key idea

Congruence describes geometric objects, while equality compares their lengths.
Problem 2 Why the midpoint average works

Points $P$ and $Q$ on one number line have coordinates $71$ and $-54$ , respectively.

Calculate the average of their coordinates.

Show that the point $M$ at this coordinate lies between the endpoints and has equal distances to them.

Then explain why swapping the endpoint names changes neither the average nor the midpoint's location.
1. Hint 1
  A midpoint must lie on the segment and have equal distances to its endpoints.
2. Hint 2
  Check the coordinate order and both endpoint gaps. For the last part, compare $a+b$ with $b+a$ before dividing by two.
Answer

The average is $8.5$ . It lies between $-54$ and $71$ , and $PM=MQ=62.5$ units, so $M$ is the midpoint. Swapping the endpoints gives the same average and therefore the same point.

Full solution

Compute the average:
$\frac{71+(-54)}{2}=8.5.$

The strict order $-54<8.5<71$ places $M$ between $Q$ and $P$ .

The distance from $P$ is
$PM=71-8.5=62.5.$

The distance from $Q$ is
$QM=8.5-(-54)=62.5.$

The location and the equal distances together meet the midpoint definition.

Swapping the endpoint coordinates gives
$\frac{-54+71}{2}=8.5.$

In general, $a+b=b+a$ , so dividing either sum by two gives the same average.
Under the accepted coordinate correspondence, the same coordinate identifies the same point.

Answer

The average is $8.5$ . It lies between $-54$ and $71$ , and $PM=MQ=62.5$ units, so $M$ is the midpoint. Swapping the endpoints gives the same average and therefore the same point.

Key idea

The midpoint average gives an equal-distance point between the endpoints, independent of which endpoint is named first.
Problem 3 The missing adjacent length

Distinct points $J$ , $K$ , and $L$ occur in that order on a line.

If $JL=37.5$ millimeters and $JK=14.8$ millimeters, find $KL$ .

State why subtraction applies.

The named points occur from left to right in the stated order. Length labels belong to the indicated parts; the lower bracket identifies the whole.

Text description of this figure

The named points occur from left to right in the stated order. Length labels belong to the indicated parts; the lower bracket identifies the whole.
1. Hint 1
  Identify the whole segment and the two adjacent parts.
2. Hint 2
  Use $JK+KL=JL$ and isolate the missing term.
Answer

$KL=22.7$ mm; $K$ is between $J$ and $L$ , so segment addition applies.

Full solution

The given order makes $K$ lie between $J$ and $L$ .
Segment addition gives $JK+KL=JL$ .

Subtract the known part from the whole to get $KL=37.5-14.8=22.7$ millimeters.

The lengths check because
$14.8+22.7=37.5.$

Answer

$KL=22.7$ mm; $K$ is between $J$ and $L$ , so segment addition applies.

Key idea

Betweenness tells you which lengths are parts and which length is the whole.
Problem 4 Equal lengths in different places

Segments $\overline{UV}$ and $\overline{WX}$ are congruent. Their lengths in the same unit are $UV=7t-5$ and $WX=4t+16$ . Find $t$ and their common length.
1. Hint 1
  Translate congruence into equality of lengths.
2. Hint 2
  Set the two expressions equal, solve for $t$ , then evaluate both lengths.
Answer

$t=7$ ; both segments have length $44$ units.

Full solution

Congruence gives
$7t-5=4t+16.$
Subtracting $4t$ and adding $5$ gives $3t=21$ , so $t=7$ .

Substitution gives $UV=7(7)-5=44$ and
$WX=4(7)+16=44.$
The equal positive lengths satisfy the geometric condition.

Answer

$t=7$ ; both segments have length $44$ units.

Key idea

A geometric condition supplies the equation; substitution checks the resulting lengths.
Problem 5 Locate the halfway point

The endpoints of a number-line segment are at $-18$ and $9$ . Find the midpoint coordinate and the length of each half.
1. Hint 1
  A midpoint has equal distances to the endpoints.
2. Hint 2
  Average the coordinates for the location and halve the full distance for the half length.
Answer

The midpoint coordinate is $-4.5$ ; each half has length $13.5$ units.

Full solution

The midpoint coordinate is
$(-18+9)/2=-4.5.$

The whole length is $9-(-18)=27$ units, so each half has length $27/2=13.5$ units.

The distances from $-4.5$ to $-18$ and to $9$ both equal $13.5$ .

Answer

The midpoint coordinate is $-4.5$ ; each half has length $13.5$ units.

Key idea

A midpoint's coordinate is a location, while a half length is a distance.
Problem 6 A crossing with enough information

A line crosses $\overline{EF}$ at $N$ and meets the segment nowhere else.

The part lengths are $EN=2u+7$ and $NF=5u-8$ in the same unit.

For what value of $u$ is the line a segment bisector?

Find $EF$ for that value.

No crossing angle is specified.

The named points occur from left to right in the stated order. Length labels belong to the indicated parts; the lower bracket identifies the whole. One possible crossing at N is shown; its angle is not specified.

Text description of this figure

The named points occur from left to right in the stated order. Length labels belong to the indicated parts; the lower bracket identifies the whole. One possible crossing at N is shown; its angle is not specified.
1. Hint 1
  Bisection requires equal lengths on the two sides of the crossing.
2. Hint 2
  Equate the expressions for $EN$ and $NF$ , then add their positive values.
Answer

$u=5$ ; $EF=34$ units.

Full solution

The line bisects the segment precisely when the crossing point is its midpoint, which requires $2u+7=5u-8$ .

Rearranging gives $15=3u$ , so $u=5$ .
Then $EN=17$ and $NF=17$ , both positive.

Since $N$ lies on the segment, segment addition gives
$EF=17+17=34.$
The crossing angle is irrelevant to this bisection condition.

Answer

$u=5$ ; $EF=34$ units.

Key idea

Equal halves establish bisection without any perpendicularity assumption.
Problem 7 Measure a misplaced ruler

A paper strip starts at the $27$ mm mark and ends at the $86$ mm mark of a ruler.

Find its measured length in millimeters and centimeters.

A second strip is measured as $6.2$ cm with the same precision.

Which is longer, and by how much?
1. Hint 1
  The first strip's length is the difference between two readings.
2. Hint 2
  Convert the second length to millimeters before comparing.
Answer

The first strip measures $59$ mm, or $5.9$ cm. The second is longer by $3$ mm, or $0.3$ cm.

Full solution

Subtracting endpoint readings gives $86-27=59$ millimeters.
This is $5.9$ centimeters.

The second strip measures $6.2$ centimeters, or $62$ millimeters.
The difference is $62-59=3$ millimeters, so the second strip is longer by that measured amount.

These are physical measurements at the ruler's precision, rather than exact geometric congruence claims.

Answer

The first strip measures $59$ mm, or $5.9$ cm. The second is longer by $3$ mm, or $0.3$ cm.

Key idea

A ruler's zero position is optional, but subtraction and consistent units are essential.
Problem 8 Two possible locations

Point $A$ has number-line coordinate $3$ .

Point $B$ on the same number line is $10$ units from $A$ .

Find every possible coordinate of $B$ and the midpoint coordinate for each resulting segment.
1. Hint 1
  Distance gives a gap but does not select a direction.
2. Hint 2
  Place $B$ ten units to the left or right of $3$ , then average each endpoint pair.
Answer

If $B$ has coordinate $-7$ , the midpoint coordinate is $-2$ . If $B$ has coordinate $13$ , the midpoint coordinate is $8$ .

Full solution

Moving ten units left from $3$ gives coordinate $3-10=-7$ for $B$ .
The midpoint coordinate is
$(3+(-7))/2=-2.$

Moving ten units right gives coordinate $3+10=13$ for $B$ .
The midpoint coordinate is
$(3+13)/2=8.$

The two choices exhaust the possible directions on the number line, and each midpoint is five units from $A$ .

Answer

If $B$ has coordinate $-7$ , the midpoint coordinate is $-2$ . If $B$ has coordinate $13$ , the midpoint coordinate is $8$ .

Key idea

A distance condition on a line can leave two possible locations.
Problem 9 What each marking tells you

Use only these verbal descriptions of marks; no drawing is supplied before the solution.

Situation A: a line crosses a segment at a marked square corner, with no other marks.

Situation B: a line crosses a segment whose two resulting parts carry matching congruence ticks, with no angle mark.

For each situation, state whether perpendicularity, bisection, both, or neither is established. The descriptions are the complete givens.
1. Hint 1
  Treat each described mark as information. No other crossing information is given.
2. Hint 2
  A square corner concerns the angle; matching length ticks concern equal halves.
Answer

Situation A establishes perpendicularity only. Situation B establishes bisection only.

Full solution

The square-corner mark described in Situation A declares perpendicularity.
It gives no equality of part lengths, so bisection is not established.

The matching ticks described in Situation B declare equal part lengths at a crossing point on the segment.
That point is the midpoint, so the line bisects the segment.
No given angle information establishes perpendicularity.

Situation A marks a right angle but no equality of the parts. Situation B marks equal parts but no right angle. The slanted crossing is deliberately different from a perpendicular crossing.

Answer

Situation A establishes perpendicularity only. Situation B establishes bisection only.

Key idea

Perpendicularity and bisection require different geometric information.
Problem 10 Combine order and a bisector

Distinct points occur in order $A$ , $B$ , $M$ , $C$ on a line.

A second line bisects $\overline{AC}$ at $M$ .

In one unit, $AB=2x+1$ , $BM=x+2$ , and $MC=4x-5$ .

Find $x$ , $AC$ , and $BC$ .

The named points occur from left to right in the stated order. Length labels belong to the indicated parts; the lower bracket identifies the whole.

Text description of this figure

The named points occur from left to right in the stated order. Length labels belong to the indicated parts; the lower bracket identifies the whole.
1. Hint 1
  The bisector supplies equal halves of the larger segment.
2. Hint 2
  First express $AM$ as $AB+BM$ , then set it equal to $MC$ .
3. Hint 3
  After solving, use the stated order to add the parts belonging to each requested segment.
Answer

$x=8$ , $AC=54$ units, and $BC=37$ units.

Full solution

The given order makes $AM=AB+BM$ .

Substitute the two adjacent lengths to get
$AM=(2x+1)+(x+2).$
Combining terms gives $AM=3x+3$ .

The bisector makes $AM=MC$ , so $3x+3=4x-5$ .

Solving gives $x=8$ .
The three adjacent lengths are $AB=17$ , $BM=10$ , and $MC=27$ , all positive.

Apply segment addition again to get $AC=AM+MC$ .
Substituting the first part sum gives
$AC=17+10+27.$
Thus $AC=54$ units.

Since $B$ , $M$ , and $C$ occur in that order, $BC=BM+MC$ .
Substituting gives $BC=10+27$ , so $BC=37$ units.

For the midpoint check, $AM=17+10$ .
This gives $AM=27$ , equal to the stated $MC$ .

Answer

$x=8$ , $AC=54$ units, and $BC=37$ units.

Key idea

A midpoint equation and the order of points let you reconstruct several segment lengths.

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