Core practice ← Back to lesson

Trapezoids and Kites: 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)

0 of 10 done

Progress saved in this browser.

Problem 1 of 10

Problem 1 A pair of parallel bases

A convex isosceles trapezoid $ABCD$ has chosen bases $AB$ and $CD$ . Its angle at $D$ measures $117^\circ$ .

Its diagonal lengths, in centimeters, are
$\begin{gathered}AC=7x-8\\BD=4x+13.\end{gathered}$
Find all four vertex angles, $x$ , and both diagonal lengths. Explain the relationships you use.

The named bases and diagonals; measurements come from the statement.

Text description of this figure

Convex ABCD with AB the bottom base and CD the top base. The bases are parallel. The dashed segments, when present, are auxiliaries described in the adjacent text. The drawing is schematic.
1. Hint 1
  Which angles share a base, and which angles lie along one leg?
2. Hint 2
  The two diagonals are congruent. Equate their given expressions, then substitute.
Answer

Angles:
$\begin{gathered}m\angle A=63^\circ\\m\angle B=63^\circ\\m\angle C=117^\circ\\m\angle D=117^\circ.\end{gathered}$
Length results:
$\begin{gathered}x=7\\AC=41\text{ cm}\\BD=41\text{ cm}.\end{gathered}$

Full solution

Angles at base CD are congruent, so the angle at C is 117 degrees.
Each angle at AB supplements the angle at the other end of its leg, so the angles at A and B are both 63 degrees.

Congruent diagonals give
$\begin{aligned}7x-8&=4x+13\\3x&=21\\x&=7.\end{aligned}$

Substituting into the two expressions gives
$\begin{gathered}AC=41\text{ cm}\\BD=41\text{ cm}.\end{gathered}$

Answer

Angles:
$\begin{gathered}m\angle A=63^\circ\\m\angle B=63^\circ\\m\angle C=117^\circ\\m\angle D=117^\circ.\end{gathered}$
Length results:
$\begin{gathered}x=7\\AC=41\text{ cm}\\BD=41\text{ cm}.\end{gathered}$

Key idea

Use base-angle equality, leg supplements, and diagonal congruence for their separate jobs.
Problem 2 Lengths at a diagonal crossing

Convex kite $ABCD$ has equal adjacent pairs $AB,AD$ and $BC,CD$ . Its diagonals meet at $O$ . The given lengths $AO,OC,BO$ are $8$ cm, $20$ cm, and $15$ cm, respectively.

Find $DO$ , $BD$ , $AB$ , and $BC$ exactly. Justify which right triangles and equal halves you use.

The three given lengths are drawn to one scale.

Text description of this figure

Convex kite ABCD has adjacent equal pairs AB with AD, and BC with CD. AC joins the vertices where those pairs meet. O is the intersection of AC and BD. Dashed lines show the diagonals. Numeric AO, OC, and BO labels, when all present, are drawn to one scale. Derived right-angle and half-diagonal markings occur only in a solution figure.
1. Hint 1
  Name the diagonal through the two equal-side meeting vertices.
2. Hint 2
  After locating the perpendicular bisection, use AO with BO for AB, and OC with BO for BC.
Answer

$\begin{gathered}DO=15\text{ cm}\\BD=30\text{ cm}\\AB=17\text{ cm}\\BC=25\text{ cm}.\end{gathered}$

Full solution

AC joins the equal-side meeting vertices, so it is the perpendicular bisector of BD.
Thus DO is 15 cm, and BD is 30 cm.

Triangle AOB is right at O.
Pythagoras gives
$\begin{gathered}AB^2=8^2+15^2\\AB^2=289\\AB=17\text{ cm}.\end{gathered}$

Triangle COB is also right at O, but uses the other part of AC.
$\begin{gathered}BC^2=20^2+15^2\\BC^2=625\\BC=25\text{ cm}.\end{gathered}$

Perpendicularity and equal halves are established in the solution.

Answer

$\begin{gathered}DO=15\text{ cm}\\BD=30\text{ cm}\\AB=17\text{ cm}\\BC=25\text{ cm}.\end{gathered}$

Key idea

The two parts of the axis diagonal can differ even when the other diagonal is bisected.
Problem 3 The two diagonals

A convex trapezoid $ABCD$ has chosen parallel bases $AB$ and $CD$ and congruent angles at $A$ and $B$ . Prove that its diagonals $AC$ and $BD$ are congruent. State the matching that makes your argument work.

The two diagonals are drawn without equality markings.

Text description of this figure

Convex ABCD with AB the bottom base and CD the top base. The bases are parallel. The dashed segments, when present, are auxiliaries described in the adjacent text. The drawing is schematic.
1. Hint 1
  The base-angle condition gives an isosceles trapezoid and therefore congruent legs.
2. Hint 2
  Compare triangles DAB and CBA using the common base, the equal legs, and the included base angles.
Answer

Diagonals AC and BD are congruent. A valid matching is
$\triangle DAB\cong\triangle CBA.$
Reflection in the common base bisector is also valid.

Full solution

The base-angle condition makes the trapezoid isosceles, so the proved leg property gives congruent sides AD and BC.

Triangles DAB and CBA share AB.
Their included angles DAB and CBA are congruent by the given base-angle condition.
SAS therefore proves the matching
$\triangle DAB\cong\triangle CBA.$

In that matching, DB corresponds to CA.
Corresponding sides of congruent triangles give the required diagonal congruence.

Alternatively, the isosceles reflection swaps A with B and D with C, mapping diagonal AC onto BD and preserving its length.

Answer

Diagonals AC and BD are congruent. A valid matching is
$\triangle DAB\cong\triangle CBA.$
Reflection in the common base bisector is also valid.

Key idea

A congruence proof needs a correct vertex matching before corresponding sides can be used.
Problem 4 Angles across a diagonal

Convex kite $ABCD$ has adjacent equal pairs $AB,AD$ and $BC,CD$ . Prove that the whole angles at $B$ and $D$ are congruent and that $AC$ bisects the angles at $A$ and $C$ . Do not assume a reflection axis before proving the angle facts.

Only the side-pair givens are marked.

Text description of this figure

Convex kite ABCD has adjacent equal pairs AB with AD, and BC with CD. AC joins the vertices where those pairs meet. O is the intersection of AC and BD. Dashed lines show the diagonals. Numeric AO, OC, and BO labels, when all present, are drawn to one scale. Derived right-angle and half-diagonal markings occur only in a solution figure.
1. Hint 1
  One diagonal divides the kite into triangles with three known matching side pairs.
2. Hint 2
  Write the correspondence for triangles ABC and ADC. Separate the whole opposite angles from the two angles cut in half.
Answer

$\begin{gathered}\angle ABC\cong\angle ADC\\\angle BAC\cong\angle DAC\\\angle BCA\cong\angle DCA.\end{gathered}$
Thus AC bisects the angles at A and C.

Full solution

Triangles ABC and ADC have the given adjacent side pairs, together with the common side AC.
SSS gives
$\triangle ABC\cong\triangle ADC.$

The whole angle at B corresponds to the whole angle at D, giving their congruence.

At A, the corresponding angles BAC and DAC are congruent.
Convexity puts AC inside the vertex angle, so these equal parts show that AC bisects it.

At C, the corresponding angles BCA and DCA are congruent.
Again the diagonal lies inside the vertex angle, so it bisects the angle at C.

Answer

$\begin{gathered}\angle ABC\cong\angle ADC\\\angle BAC\cong\angle DAC\\\angle BCA\cong\angle DCA.\end{gathered}$
Thus AC bisects the angles at A and C.

Key idea

The kite side pairs fix a correspondence that distinguishes whole equal angles from bisected angles.
Problem 5 Three symmetry records

For each case, give the number of reflection axes and the number of rotational symmetries, including the identity. Describe the axes and the nonidentity rotations.

Case A: a non-parallelogram isosceles trapezoid.

Case B: a nonsquare rectangle.

Case C: a square.

Explain why Case A has no half-turn even though Cases B and C do.

Three cases with no candidate reflection axes drawn.

Text description of this figure

Three separate convex quadrilaterals. Case A is a non-parallelogram isosceles trapezoid. Case B is a nonsquare rectangle. Case C is a square. No candidate axes are drawn and no matching ticks compare the panels.
1. Hint 1
  The inclusive definition allows these cases to share a family name without sharing a symmetry count.
2. Hint 2
  A half-turn interchanges opposite sides. Compare the base lengths in Case A.
Answer

Case A: one reflection axis and one rotation. Case B: two reflection axes and two rotations. Case C: four reflection axes and four rotations. Case A has unequal bases, which a half-turn cannot interchange.

Full solution

Case A: the common perpendicular bisector of the bases is the only reflection axis.
Its sole rotation is the identity.
The bases have unequal lengths, since parallel congruent opposite bases would make it a parallelogram; a half-turn cannot swap unequal sides.

Case B: the two lines through opposite side midpoints are reflection axes.
The rotations are the identity and a half-turn.

Case C: the two midpoint lines and the two diagonal lines are reflection axes.
The rotations are the identity, a quarter-turn, a half-turn, and a three-quarter-turn.

Answer

Case A: one reflection axis and one rotation. Case B: two reflection axes and two rotations. Case C: four reflection axes and four rotations. Case A has unequal bases, which a half-turn cannot interchange.

Key idea

State the special case and the identity convention before comparing symmetry counts.
Problem 6 An annotated drawing

On paper, draw a convex kite $ABCD$ that is not a rhombus. Let its equal-side pairs meet at $A$ and $C$ . Draw both diagonals and label their intersection $O$ .

Mark the two adjacent equal-side pairs, the guaranteed right angle at $O$ , the guaranteed equal half-diagonals, and both vertex-angle bisections. State which pair of half-diagonals you have no right to mark congruent in this kite.
1. Hint 1
  Use different tick patterns for the two side pairs, and a third pattern for the bisected diagonal.
2. Hint 2
  The diagonal through A and C bisects BD; reverse bisection would force the rhombus special case.
Answer

A valid convex drawing marks the two adjacent side pairs, both vertex-angle bisections, a right angle at O, and congruent halves of BD. It leaves the two parts of AC unequal.

Full solution

Place A and C on a line on opposite sides of an interior point O, with unequal distances from O.
Put B and D at equal distances from O on the perpendicular through O, then join the vertices in boundary order.
This produces a convex non-rhombus kite.

Mark AB with AD using one tick, and BC with CD using two ticks.
Mark BO and DO with a third tick pattern.
Add a right-angle square at O.

Mark the two parts of angle A with matching arcs, and the two parts of angle C with a different matching arc pattern.
These are separate bisection claims, not equality between the whole angles at A and C.

Do not mark AO and OC congruent.
In a kite that extra equality would make both diagonals bisect each other and force a rhombus.

One possible drawing, with distinct equality patterns for the side pairs, half-diagonals, and angle halves.

Answer

A valid convex drawing marks the two adjacent side pairs, both vertex-angle bisections, a right angle at O, and congruent halves of BD. It leaves the two parts of AC unequal.

Key idea

Use distinct mark patterns to separate different equalities, and mark only facts the hypotheses justify.
Problem 7 Testing the equal-leg condition

Give a convex quadrilateral $ABCD$ with chosen parallel bases $AB$ and $CD$ and congruent legs $AD,BC$ , but with unequal angles at base $AB$ . Specify angle measures and justify every required property.

State which extra hypothesis would turn the equal-leg condition into a sufficient isosceles-trapezoid test.
1. Hint 1
  The inclusive trapezoid definition includes figures with two opposite parallel pairs.
2. Hint 2
  An oblique parallelogram has congruent opposite sides and supplementary unequal consecutive angles.
Answer

One valid example is an oblique parallelogram with angles A and B of 50 degrees and 130 degrees. The sufficient test needs exactly one opposite parallel pair.

Full solution

Take an oblique parallelogram whose consecutive angles at A and B are 50 degrees and 130 degrees.
Its opposite sides AB and CD are parallel, so it satisfies the trapezoid definition for those chosen bases.

Its other opposite sides AD and BC are congruent by the parallelogram property, so the equal-leg hypothesis holds.
Its two angles at AB are unequal, so it fails the isosceles definition.

The extra hypothesis is that exactly one opposite-side pair is parallel.
It excludes this counterexample and allows the auxiliary-parallel proof of the equal-legs converse.

The marked equal legs coexist with different angles at the chosen base.

Answer

One valid example is an oblique parallelogram with angles A and B of 50 degrees and 130 degrees. The sufficient test needs exactly one opposite parallel pair.

Key idea

A counterexample must satisfy all the proposed hypotheses while failing the proposed conclusion.
Problem 8 Evidence from the crossing

Two segments $AC$ and $BD$ meet internally at $O$ . They are perpendicular, and $O$ is the midpoint of $BD$ . Joining $A,B,C,D$ in that order gives a convex quadrilateral.

Prove that $ABCD$ is a kite under this course's definition. State the two disjoint adjacent congruent pairs.

The right angle and equal halves of BD are givens; the side equalities must be proved.

Text description of this figure

Convex ABCD has diagonals AC and BD meeting internally at O. The givens mark AC perpendicular to BD and BO congruent to DO. No boundary-side equality or vertex-angle bisection is marked.
1. Hint 1
  The right triangles on either side of AC share a leg and have equal legs along BD.
2. Hint 2
  Prove the pair at A first, and then use the same structure for the pair at C.
Answer

The disjoint adjacent pairs are
$\begin{gathered}AB=AD\\BC=CD.\end{gathered}$
With the stated convexity, ABCD is a kite.

Full solution

Triangles AOB and AOD share AO.
The midpoint hypothesis gives congruent sides BO and DO, and their included angles at O are both right angles.
SAS proves the triangles congruent, so AB and AD are congruent.

Triangles COB and COD share CO and use the same congruent halves BO and DO and right included angles.
SAS therefore gives congruent sides CB and CD.

These two adjacent pairs are disjoint: the first uses AB and AD, while the second uses BC and CD.
Every boundary side is used once, and convexity is given, so the kite definition is satisfied.

Answer

The disjoint adjacent pairs are
$\begin{gathered}AB=AD\\BC=CD.\end{gathered}$
With the stated convexity, ABCD is a kite.

Key idea

Perpendicular bisection supplies two SAS arguments, one for each adjacent side pair.
Problem 9 Successive cuts

Convex trapezoid $ABCD$ has parallel bases $AB$ and $CD$ of lengths $13$ cm and $37$ cm, respectively. Let $M,N$ be the midpoints of $AD,BC$ .

Let $P,Q$ be the midpoints of $AM,BN$ , and let $R,S$ be the midpoints of $MD,NC$ . Find the lengths $PQ,MN,RS$ , justifying why the same rule applies at each stage.

Explain why the five widths AB, PQ, MN, RS, CD have equal successive increases, and say whether the argument requires an isosceles trapezoid.

The successive midpoint locations, with the three requested lengths left unlabeled.

Text description of this figure

Convex ABCD has parallel bases AB at the top and CD at the bottom, of lengths 13 cm and 37 cm. M and N are leg midpoints. P and Q are midpoints of AM and BN; R and S are midpoints of MD and NC. The three cross-segments PQ, MN, and RS are shown without their lengths. The base ratio and midpoint locations are drawn accurately.
1. Hint 1
  The original midsegment becomes a base of each smaller trapezoid.
2. Hint 2
  First average 13 and 37. Then average that new width separately with each original base.
Answer

$\begin{gathered}PQ=19\text{ cm}\\MN=25\text{ cm}\\RS=31\text{ cm}.\end{gathered}$
The widths increase by 6 cm each time; isosceles is not required.

Full solution

MN joins the original leg midpoints, so it is parallel to the bases.
Its length is
$MN=\frac{13+37}{2}.$
Thus MN is 25 cm long.

ABNM is a convex trapezoid with bases AB and MN.
P and Q are its leg midpoints, so
$PQ=\frac{13+25}{2}.$
Its length is 19 cm.

MNCD is a convex trapezoid with bases MN and CD.
R and S are its leg midpoints, so
$RS=\frac{25+37}{2}.$
Its length is 31 cm.

The five widths are 13, 19, 25, 31, and 37 cm.
Every successive increase is 6 cm.
Averaging creates the middle width halfway between the originals, then the two remaining half-intervals are each halved again.

For general base lengths a and b, these quarter-step increases each have length one fourth of the difference between b and a, with its sign if b is smaller.
The argument uses parallel bases and leg midpoints throughout; it does not require congruent legs or angles.

Answer

$\begin{gathered}PQ=19\text{ cm}\\MN=25\text{ cm}\\RS=31\text{ cm}.\end{gathered}$
The widths increase by 6 cm each time; isosceles is not required.

Key idea

A proved relationship can be reapplied to smaller figures once their hypotheses are checked.
Problem 10 Two evidence records

In both records, $ABCD$ is a convex kite with equal-side pairs meeting at $A$ and $C$ . Its diagonals meet at $O$ . Decide whether being a rhombus is guaranteed, possible but not guaranteed, or impossible, and justify your answer.

Record A: $AO$ is $3$ cm and $OC$ is $10$ cm.

Record B: $O$ is also the midpoint of $AC$ .

For Record B, decide whether a square is forced. If it is not forced, give possible unequal diagonal lengths that distinguish a nonsquare rhombus from a square, and explain why they work.
1. Hint 1
  Separate what every kite guarantees from the bisection direction added in Record B.
2. Hint 2
  A rhombus is a parallelogram, so its diagonals bisect each other. Conversely, mutual bisection gives a parallelogram; combine this with the kite side pairs.
3. Hint 3
  A square has congruent diagonals. Can perpendicular diagonals bisect each other without having equal total lengths?
Answer

Record A: a rhombus is impossible. Record B: a rhombus is guaranteed, but a square is not forced. Diagonal lengths of 10 cm and 24 cm give one valid nonsquare rhombus.

Full solution

Record A: if the kite were a rhombus, it would be a parallelogram and its diagonals would bisect each other.
The given 3 cm and 10 cm parts show that BD does not bisect AC.
A rhombus is therefore impossible.

Record B: the kite property makes O the midpoint of BD, and the extra record makes it the midpoint of AC.
Both diagonals bisect each other, so the parallelogram test applies.
The kite adjacent congruent pairs then force all four boundary sides to be congruent, proving a rhombus.

Record B does not force a square.
Draw perpendicular segments of 10 cm and 24 cm crossing at their shared midpoint.
Their four half-segments form four congruent right triangles with legs of 5 cm and 12 cm, so all four boundary sides are 13 cm.

The resulting convex quadrilateral is a rhombus.
Its unequal diagonal lengths prevent it from being a rectangle, since every rectangle has congruent diagonals; it is therefore not a square.

The example uses perpendicular diagonals of 10 cm and 24 cm with a shared midpoint.

Answer

Record A: a rhombus is impossible. Record B: a rhombus is guaranteed, but a square is not forced. Diagonal lengths of 10 cm and 24 cm give one valid nonsquare rhombus.

Key idea

A kite guarantees one diagonal bisection; the second bisection distinguishes its rhombus special case.

← Back to lesson