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Parallelograms: 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 Opposite edges

Parallelogram $ABCD$ has side lengths in centimeters given by
$\begin{gathered}AB=4x-3,\\CD=29,\\BC=12.\end{gathered}$
Find $x$ and the length of $AD$ .
1. Hint 1
  Match each given side with the side opposite it.
2. Hint 2
  Use equal lengths for $AB$ and $CD$ before substituting into the expressions.
Answer

$\begin{gathered}x=8,\\AD=12\text{ cm}.\end{gathered}$

Full solution

Opposite sides of a parallelogram are congruent.
For $AB$ and $CD$ , this gives
$4x-3=29.$

Add $3$ to both sides and then divide by $4$ .
$\begin{aligned}4x&=32,\\x&=8.\end{aligned}$

The side opposite $BC$ is $AD$ , so $AD$ has length $12$ cm.

Answer

$\begin{gathered}x=8,\\AD=12\text{ cm}.\end{gathered}$

Key idea

Identify the opposite side before setting lengths equal.
Problem 2 Two intersecting diagonals

The diagonals of parallelogram $ABCD$ meet at $E$ . Lengths in centimeters satisfy
$\begin{gathered}AE=3u+1,\\EC=5u-9,\\BE=11.\end{gathered}$
Find $u$ , the full length $AC$ , and the full length $BD$ .
1. Hint 1
  Bisection compares the two parts of each diagonal separately.
2. Hint 2
  Set the expressions for $AE$ and $EC$ equal, then double a half length for each full diagonal.
Answer

$\begin{gathered}u=5,\\AC=32\text{ cm},\\BD=22\text{ cm}.\end{gathered}$

Full solution

The diagonals bisect each other, so $AE$ and $EC$ have equal lengths.
$3u+1=5u-9$

Solve this equation.
$\begin{aligned}10&=2u,\\u&=5.\end{aligned}$

Substitution makes each half of $AC$ equal to $16$ cm.
$\begin{aligned}AC&=2(16),\\AC&=32\text{ cm}.\end{aligned}$

Each half of $BD$ is $11$ cm.
$\begin{aligned}BD&=2(11),\\BD&=22\text{ cm}.\end{aligned}$

Answer

$\begin{gathered}u=5,\\AC=32\text{ cm},\\BD=22\text{ cm}.\end{gathered}$

Key idea

A full diagonal is twice either of its own halves.
Problem 3 A second route through the figure

In parallelogram $ABCD$ , draw diagonal $BD$ . Prove that $AB$ and $CD$ are congruent and that $AD$ and $CB$ are congruent, starting only with the parallel-side definition and previously taught triangle facts.

The given parallelogram with diagonal BD.

Text description of this figure

Parallelogram ABCD with diagonal BD. Single chevrons match AB with CD and double chevrons match AD with BC. No equal lengths are assumed.
1. Hint 1
  Look at the two triangles separated by the chosen diagonal.
2. Hint 2
  Use $BD$ as a transversal twice, once for each parallel-side pair.
3. Hint 3
  Find the common included side and write the triangle correspondence in order.
Answer

Triangles $ABD$ and $CDB$ are congruent by ASA, so corresponding opposite sides are congruent.

Full solution

Since $AB$ and $CD$ are parallel, diagonal $BD$ gives alternate interior angles.
$\angle ABD\cong\angle CDB$

Since $AD$ and $BC$ are parallel, the same diagonal gives a second pair.
$\angle ADB\cong\angle CBD$

Side $BD$ is common and lies between the matched angles.
ASA proves
$\triangle ABD\cong\triangle CDB.$

Corresponding sides of those congruent triangles now give the two requested conclusions.
$\begin{gathered}AB=CD,\\AD=CB.\end{gathered}$

Answer

Triangles $ABD$ and $CDB$ are congruent by ASA, so corresponding opposite sides are congruent.

Key idea

Parallel lines can supply the angle evidence needed before side equality is proved.
Problem 4 A paper diagram with useful marks

On paper, draw a convex parallelogram $ABCD$ that has unequal adjacent sides and no right angle. Draw its diagonals and name their intersection $E$ .

Add conventional markings for the two parallel-side pairs. Mark the half-diagonal lengths that must match, using different tick patterns for the two diagonals.

Write the vertex mapping for the half-turn about $E$ . Explain why your markings do not assert that the full diagonals are equal or perpendicular.
1. Hint 1
  Draw two pairs of parallel lines with different directions, avoiding a right angle. Their intersections form the four corners.
2. Hint 2
  A half-turn exchanges opposite vertices. Diagonal bisection gives two distinct equal-length pairs.
Answer

Drawings vary. Parallel pairs: $AB$ with $CD$ , and $AD$ with $BC$ . Tick pairs: $AE$ with $EC$ , and $BE$ with $ED$ . The half-turn exchanges $A$ with $C$ and $B$ with $D$ . Do not mark equal or perpendicular full diagonals.

Full solution

The drawing uses separate chevron patterns for the two given pairs of parallel sides.

The diagonal theorem gives equal lengths within each diagonal.
Use one tick on $AE$ and $EC$ , and two ticks on $BE$ and $ED$ .
Different patterns avoid asserting equality between the two diagonals.

Because $E$ is the midpoint of both diagonals, the half-turn exchanges $A$ with $C$ and $B$ with $D$ .
Bisection does not prove that the crossing is perpendicular, so no square corner mark is added.

One possible drawing and annotation.

Answer

Drawings vary. Parallel pairs: $AB$ with $CD$ , and $AD$ with $BC$ . Tick pairs: $AE$ with $EC$ , and $BE$ with $ED$ . The half-turn exchanges $A$ with $C$ and $B$ with $D$ . Do not mark equal or perpendicular full diagonals.

Key idea

A marking should assert exactly the relationship you have evidence for.
Problem 5 Two angle totals

Let $ABCD$ be a parallelogram. Prove that the angle measures at $A$ and $B$ add to $180^\circ$ , and that the measures at $B$ and $C$ also add to $180^\circ$ . Name the parallel lines and transversal for each conclusion.

The two marked pairs of parallel sides.

Text description of this figure

Parallelogram ABCD with the two given parallel-side pairs marked. No angle measure is given.
1. Hint 1
  Consecutive angles can be same-side interior angles of parallel lines.
2. Hint 2
  For the first pair, use $AB$ as a transversal. For the second pair, use $BC$ .
Answer

Both sums are $180^\circ$ . For $A$ and $B$ , use parallel $AD$ and $BC$ cut by $AB$ ; for $B$ and $C$ , use parallel $AB$ and $CD$ cut by $BC$ .

Full solution

Lines $AD$ and $BC$ are parallel by the parallelogram definition.
Side $AB$ cuts them as a transversal.
The interior angles on the same side of that transversal are angles $DAB$ and $ABC$ , so
$m\angle A+m\angle B=180^\circ.$

For the other pair, lines $AB$ and $CD$ are parallel and $BC$ is the transversal.
The same-side interior angles are $ABC$ and $BCD$ , so
$m\angle B+m\angle C=180^\circ.$

Answer

Both sums are $180^\circ$ . For $A$ and $B$ , use parallel $AD$ and $BC$ cut by $AB$ ; for $B$ and $C$ , use parallel $AB$ and $CD$ cut by $BC$ .

Key idea

A supplementary-angle claim should name the parallel lines and the transversal that support it.
Problem 6 Repair the bisection claim

The diagonals of parallelogram $ABCD$ meet at $E$ . Segment $AE$ is $7$ cm and segment $BE$ is $4$ cm.

A written solution says, “The diagonals bisect each other, so $AE$ and $BE$ must be equal. The given measurements are impossible.” Repair this reasoning. State the actual values of $EC$ and $ED$ , and decide whether these measurements contradict the parallelogram hypothesis.

The two labeled segments meet at E.

Text description of this figure

Parallelogram ABCD with diagonals meeting at E, AE labeled 7 cm and BE labeled 4 cm. No equality tick joins pieces of different diagonals.
1. Hint 1
  Ask which whole segment each of the named parts belongs to.
2. Hint 2
  Match $AE$ to the other part of $AC$ , and $BE$ to the other part of $BD$ .
Answer

The measurements are consistent.
$\begin{gathered}EC=7\text{ cm},\\ED=4\text{ cm}.\end{gathered}$

Full solution

The claim compares parts of different diagonals.
Bisection says that each diagonal has its own two equal halves.

Segment $AE$ is paired with $EC$ , and $BE$ is paired with $ED$ .
$\begin{gathered}EC=7\text{ cm},\\ED=4\text{ cm}.\end{gathered}$

The two full diagonals are therefore $14$ cm and $8$ cm.
They need not be equal, so the given measurements do not contradict a parallelogram.

Answer

The measurements are consistent.
$\begin{gathered}EC=7\text{ cm},\\ED=4\text{ cm}.\end{gathered}$

Key idea

Equal halves of each diagonal do not mean all four half-diagonal lengths match.
Problem 7 Compare two motions

Figure $ABCD$ is an oblique parallelogram with adjacent sides $9$ cm and $6$ cm and an angle of $70^\circ$ . Figure $WXYZ$ is a rectangle with adjacent sides $9$ cm and $6$ cm. Both are drawn with their longer opposite sides horizontal.

Case A: Does each shape map onto itself under a half-turn about its own diagonal intersection? Explain.

Case B: Reflect each shape in the horizontal line halfway between its horizontal sides. For each shape, decide whether the outlines match and explain using vertex images.

Test the indicated motion separately on each figure.

Text description of this figure

Top: solid blue oblique parallelogram ABCD, with a horizontal line halfway between its horizontal sides. Bottom: dashed red rectangle WXYZ, with its own horizontal middle line. Both shapes have adjacent side ratio 9 to 6.
1. Hint 1
  The half-turn theorem applies to every parallelogram, including rectangles.
2. Hint 2
  For the reflection, follow the lower-left vertex straight up across the proposed axis. Does it reach the upper-left vertex?
Answer

Case A: Both have the half-turn symmetry. Case B: The oblique parallelogram does not match under this reflection; the rectangle does, exchanging $W$ with $Z$ and $X$ with $Y$ .

Full solution

Case A: Both figures are parallelograms.
Their diagonals bisect each other, so the half-turn exchanges opposite vertices and maps every side to its opposite side in each figure.

Case B: Reflection in a horizontal line preserves the horizontal position of each point.
In the oblique parallelogram, the image of $A$ lies directly above $A$ , while $D$ lies farther right.
The image is not an original vertex, so that reflection fails.

Case B: In the rectangle, the upper and lower vertices are vertically aligned at equal distances from the middle line.
Reflection sends $W$ to $Z$ and $X$ to $Y$ , and the sides match.
This is one of the extra symmetries possible for a special parallelogram.

Answer

Case A: Both have the half-turn symmetry. Case B: The oblique parallelogram does not match under this reflection; the rectangle does, exchanging $W$ with $Z$ and $X$ with $Y$ .

Key idea

The general half-turn property persists when a special parallelogram gains reflection symmetry.
Problem 8 Across the diagonal intersection

Diagonals $AC$ and $BD$ of parallelogram $ABCD$ meet at $E$ . Prove that $AE$ and $CE$ have equal lengths and that $DE$ and $BE$ have equal lengths. Use triangles $ADE$ and $CBE$ without assuming that $E$ is already a midpoint.

Use the specified triangles at the crossing.

Text description of this figure

Parallelogram ABCD with diagonals AC and BD intersecting at E. Parallel markings are given; no midpoint markings are shown.
1. Hint 1
  The opposite-side theorem is available before the diagonal theorem.
2. Hint 2
  Use parallel $AD$ and $BC$ to find two alternate interior-angle pairs.
3. Hint 3
  The equal opposite sides lie between the matched angles.
Answer

Triangles $ADE$ and $CBE$ are congruent by ASA, so the two parts of each diagonal are equal.

Full solution

The already-proved opposite-side property gives
$AD=CB.$

Using $AC$ as a transversal of parallel $AD$ and $BC$ gives
$\angle DAE\cong\angle BCE.$

Using $BD$ as a transversal of the same parallel pair gives
$\angle ADE\cong\angle CBE.$

The equal included sides give ASA congruence in the required order.
$\triangle ADE\cong\triangle CBE$

Corresponding sides then supply both midpoint equalities.
$\begin{gathered}AE=CE,\\DE=BE.\end{gathered}$

Answer

Triangles $ADE$ and $CBE$ are congruent by ASA, so the two parts of each diagonal are equal.

Key idea

A proof of midpoint equality must establish congruence before using the equality as evidence.
Problem 9 Check the available conditions

A convex quadrilateral $ABCD$ has $AB$ parallel to $CD$ . Its angle at $A$ is $60^\circ$ and its angle at $C$ is $120^\circ$ , as shown. No other parallelism is given.

Can you apply the parallelogram opposite-angle theorem to conclude that those two angles are equal? State the missing hypothesis and explain how this example shows why the known parallel pair alone is insufficient.

Only one opposite-side pair is known to be parallel.

Text description of this figure

A convex quadrilateral ABCD with AB parallel to CD. Angle A is 60 degrees and opposite angle C is 120 degrees. The nonparallel legs tilt in opposite directions.
1. Hint 1
  Read the hypothesis of the theorem before its conclusion.
2. Hint 2
  A parallelogram requires a second parallel opposite-side pair. Compare that requirement with the displayed unequal opposite angles.
Answer

No. The second pair, $AD$ and $BC$ , must also be parallel before that theorem applies. This convex example has the known parallel pair and unequal opposite angles.

Full solution

The opposite-angle theorem begins with a parallelogram, which means both opposite-side pairs are parallel.
Only the pair $AB$ and $CD$ is supplied here.

The missing hypothesis is that $AD$ is parallel to $BC$ .
In this example the angles at $A$ and $C$ are $60^\circ$ and $120^\circ$ , so they are unequal despite the one given parallel pair.

The example satisfies the stated weaker conditions and fails the proposed conclusion.
It therefore shows why a single parallel pair does not justify importing all parallelogram properties.

Answer

No. The second pair, $AD$ and $BC$ , must also be parallel before that theorem applies. This convex example has the known parallel pair and unequal opposite angles.

Key idea

A direct theorem can be used only after all its hypotheses are known.
Problem 10 A fold and a turn

A convex quadrilateral has a reflection axis along one diagonal. Must a half-turn about its diagonal intersection also map it onto itself?

Decide the claim by giving either a proof or a specific convex counterexample. Explain which length condition in the parallelogram half-turn proof your argument needs to check.
1. Hint 1
  Reflection pairs points across an axis, while a half-turn pairs points across a center. These need different distance equalities.
2. Hint 2
  For a possible counterexample, place two opposite vertices on a vertical axis and put the other two equally far to either side of it. Try unequal distances from the intersection to the two vertices on the axis.
3. Hint 3
  Verify convexity and the reflection, then follow one vertex under the proposed half-turn.
Answer

No. Use perpendicular diagonals meeting at $O$ , with $AO$ of $4$ units, $OC$ of $2$ units, and $BO$ and $OD$ each $3$ units. Reflection in $AC$ works, but $O$ is not the midpoint of $AC$ , so the half-turn fails.

Full solution

Place $A$ above $O$ at distance $4$ and $C$ below $O$ at distance $2$ .
Place $B$ to the left and $D$ to the right at distance $3$ from $O$ , on a line perpendicular to $AC$ .
Join the vertices in order $A,B,C,D$ .
This gives a convex quadrilateral, since the two diagonals cross inside it and each boundary turn goes the same way.

Reflection in line $AC$ fixes $A$ and $C$ and exchanges $B$ with $D$ , so the whole boundary matches.
It is indeed a reflection axis.

The half-turn about $O$ sends $A$ to a point $4$ units below $O$ .
That point is not $C$ , which is only $2$ units below $O$ , and it is not any other vertex.
The shape therefore fails this half-turn.

The parallelogram argument required the intersection to be the midpoint of both diagonals.
A reflection in one diagonal does not supply the missing equality on that diagonal.

Reflection in AC and the image of A under a half-turn.

Answer

No. Use perpendicular diagonals meeting at $O$ , with $AO$ of $4$ units, $OC$ of $2$ units, and $BO$ and $OD$ each $3$ units. Reflection in $AC$ works, but $O$ is not the midpoint of $AC$ , so the half-turn fails.

Key idea

A symmetry claim needs its own point mapping; a successful reflection alone does not prove a half-turn.

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