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Quadrilaterals and Polygons: Chapter Test

20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.

Core practice

10 problems from across the chapter. 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 Four equal angles

A convex quadrilateral has four congruent interior angles. Find their common measure.

Is the quadrilateral necessarily a regular polygon? If not, give a concrete convex counterexample with specified adjacent side lengths and explain which regularity condition fails.
1. Hint 1
  A quadrilateral’s total interior angle can be shared equally only because equality is given.
2. Hint 2
  After finding the angle measure, check whether the givens say anything about all four side lengths.
Answer

Each angle is $90^\circ$ . Regularity is not guaranteed; a rectangle with adjacent sides 5 cm and 9 cm has equal angles but unequal side lengths.

Full solution

The interior sum of a convex quadrilateral is
$(4-2)180^\circ=360^\circ$

All four angles are equal, so each is
$360^\circ\div4=90^\circ$

A 5-by-9 rectangle is convex and has four right angles.
Its unequal adjacent sides prevent it from being equilateral, so it is not regular.

Answer

Each angle is $90^\circ$ . Regularity is not guaranteed; a rectangle with adjacent sides 5 cm and 9 cm has equal angles but unequal side lengths.

Key idea

Equal angles do not supply the equal-side condition required for regularity.
Problem 2 Edges, a crossing, and an image

Parallelogram $ABCD$ has diagonals meeting at $O$ . Lengths in centimeters satisfy
$\begin{gathered}AB=5z+8\\CD=7z-18\\BO=4z-13\end{gathered}$
Find $z$ and the full length $BD$ . Name the image of vertex $B$ under the half-turn about $O$ .
1. Hint 1
  Separate the opposite-side equality from the relationship between a whole diagonal and its half.
2. Hint 2
  First equate AB and CD. Then evaluate BO, and follow the vertex opposite B along its diagonal.
Answer

$\begin{gathered}z=13\\BD=78\text{ cm}\\B\mapsto D\end{gathered}$

Full solution

Opposite sides of a parallelogram are congruent.
Thus
$5z+8=7z-18$

Solve the side equation.
$\begin{gathered}26=2z\\z=13\end{gathered}$

Substitute into the given half-diagonal expression.
$\begin{gathered}BO=4(13)-13\\BO=39\text{ cm}\end{gathered}$

The diagonals bisect each other, so
$\begin{gathered}BD=2(39)\\BD=78\text{ cm}\end{gathered}$

The point O is the midpoint of BD.
A half-turn about O therefore sends B to D.

Answer

$\begin{gathered}z=13\\BD=78\text{ cm}\\B\mapsto D\end{gathered}$

Key idea

Use each parallelogram property for the side, half-diagonal, or vertex image it actually relates.
Problem 3 Two unknown bases

A convex trapezoid has one base 18 cm longer than the other. The segment joining its leg midpoints is 32 cm long. Find both base lengths.

Would unequal leg lengths invalidate your calculation? Explain.
1. Hint 1
  The midpoint segment relates the sum of the chosen parallel bases to its own length.
2. Hint 2
  Let the shorter base be a. Write the longer base in terms of a, then use the midpoint-segment relationship.
Answer

$\begin{gathered}\text{shorter base}=23\text{ cm}\\\text{longer base}=41\text{ cm}\end{gathered}$
Unequal legs do not invalidate the calculation.

Full solution

Let the shorter base be a centimeters.
The longer base is a plus 18 centimeters.
The midpoint-segment relationship gives
$32=\frac{a+(a+18)}2$

Solve for the shorter base.
$\begin{gathered}64=2a+18\\46=2a\\a=23\end{gathered}$

The longer base is
$23+18=41\text{ cm}$

The theorem requires parallel bases and endpoints at the leg midpoints.
It does not require equal legs, so the calculation remains valid when the legs are unequal.

Answer

$\begin{gathered}\text{shorter base}=23\text{ cm}\\\text{longer base}=41\text{ cm}\end{gathered}$
Unequal legs do not invalidate the calculation.

Key idea

A trapezoid midpoint segment averages the bases without an isosceles hypothesis.
Problem 4 A regular boundary

An unmarked convex regular polygon has each interior angle 26 times its exterior turn. Find its number of sides, number and types of reflection axes, and least positive matching rotation angle.

Decide whether a 50-degree rotation about its symmetry center matches the polygon. Give an exact reason.
1. Hint 1
  The interior angle and exterior turn together make a straight angle.
2. Hint 2
  Express both angles using the exterior turn, then recover the side count from a full turn.
3. Hint 3
  A matching rotation must be a whole number of least matching steps.
Answer

There are 54 sides and 54 reflection axes: 27 of each type. The least positive rotation is $\frac{20}{3}^\circ$ ; a $50^\circ$ turn does not match.

Full solution

Write e for the exterior turn, measured in degrees.
Its neighboring interior angle has measure 26e.
Thus
$\begin{gathered}26e+e=180\\27e=180\\e=\frac{20}{3}\end{gathered}$

The side count is
$\begin{gathered}n=360\div\frac{20}{3}\\n=54\end{gathered}$

A convex regular 54-gon has 54 reflection axes.
Its even side count gives 27 vertex-to-opposite-vertex axes and 27 opposite-side-midpoint axes.

Its least positive matching rotation is
$\frac{360^\circ}{54}=\frac{20}{3}^\circ$

The proposed rotation is not a whole number of those steps.
$50\div\frac{20}{3}=7.5$
Therefore it does not match.

Answer

There are 54 sides and 54 reflection axes: 27 of each type. The least positive rotation is $\frac{20}{3}^\circ$ ; a $50^\circ$ turn does not match.

Key idea

Regular side count connects angle calculations with whole-figure symmetry, but a matching turn still needs an integer step count.
Problem 5 Two recorded equalities

Convex quadrilateral $ABCD$ satisfies $AB=CD$ and $\angle BAC\cong\angle DCA$ . Prove that both pairs of opposite sides are parallel. Give a complete argument with reasons.

The given configuration.

Text description of this figure

Convex quadrilateral ABCD has vertices in boundary order. AB and CD are given congruent and marked with matching ticks. Diagonal AC joins opposite vertices. One arc at A between AB and AC matches one arc at C between CD and CA, representing the given angle congruence. The diagram supplies no other length or angle measure.
1. Hint 1
  Locate the given equal angles on their actual lines before assigning a shape name.
2. Hint 2
  The segment AC acts as a transversal. It also belongs to two triangles with a supplied side and included-angle match.
3. Hint 3
  Compare triangles BAC and DCA, then identify the corresponding angles concerning AD and BC.
Answer

$\begin{gathered}AB\parallel CD\\AD\parallel BC\end{gathered}$

Full solution

Angles BAC and DCA are alternate interior angles for lines AB and CD cut by AC.
Their given congruence proves
$AB\parallel CD$

Triangles BAC and DCA have the supplied match BA with DC, the common side AC, and the given congruent included angles.
SAS gives
$\triangle BAC\cong\triangle DCA$

Corresponding angles give
$\angle BCA\cong\angle DAC$

These are alternate interior angles for BC and AD cut by AC.
Their converse therefore proves
$AD\parallel BC$

Both opposite-side pairs are parallel.
Every parallelism conclusion follows from an angle converse rather than from an assumed parallelogram.

Answer

$\begin{gathered}AB\parallel CD\\AD\parallel BC\end{gathered}$

Key idea

Match the given angle and side evidence before using corresponding angles to prove the remaining directions.
Problem 6 An equality at the crossing

Parallelogram $ABCD$ has diagonals meeting at $O$ , and $AO=BO$ . Prove that each of the four triangles $AOB$ , $BOC$ , $COD$ , and $DOA$ is isosceles. Prove also that $ABCD$ is a rectangle.

Is a square guaranteed? Justify your answer with a possible nonsquare example.

The given parallelogram and equal segments.

Text description of this figure

Parallelogram ABCD has diagonals AC and BD meeting at O. A matching single tick is placed on AO and BO to represent their given equality. The remaining two half-diagonals have no matching ticks supplied. No vertex right angles or boundary-side equalities are marked.
1. Hint 1
  Combine the given equality with the midpoint facts already available for a parallelogram.
2. Hint 2
  Relate the recovered half-diagonal equalities to whole diagonal lengths, then check the hypotheses of a special-shape test.
3. Hint 3
  A rectangle can have unequal adjacent sides while retaining equal, bisected diagonals.
Answer

All four triangles are isosceles, and ABCD is a rectangle. A square is not guaranteed; a rectangle with adjacent sides 7 cm and 10 cm is a valid nonsquare example.

Full solution

Parallelogram diagonals bisect each other, giving
$\begin{gathered}AO=OC\\BO=OD\end{gathered}$

Together with the given equality AO equals BO, these make all four half-diagonal lengths equal.
Each named triangle has two of these as sides, so each is isosceles.

The two whole diagonal lengths therefore agree.
$\begin{gathered}AC=2AO\\BD=2BO\\AC=BD\end{gathered}$

The known parallelogram with congruent diagonals is a rectangle.

A rectangle with adjacent sides 7 cm and 10 cm has congruent whole diagonals that bisect each other, so AO equals BO.
Its unequal adjacent sides prevent it from being a square.
This verifies the counterexample against the original condition.

Answer

All four triangles are isosceles, and ABCD is a rectangle. A square is not guaranteed; a rectangle with adjacent sides 7 cm and 10 cm is a valid nonsquare example.

Key idea

Equal half-diagonals can establish a rectangle without establishing four equal boundary sides.
Problem 7 A whole family of tiles

Let n be any integer at least seven. Can congruent copies of one convex regular n-gon make a regular tessellation meeting whole side to whole side? Give one argument that covers every such n, rather than testing a few individual polygons.
1. Hint 1
  Compare an interior corner with the angle available when at least three tiles meet at a point.
2. Hint 2
  An exterior turn of a regular polygon is a full turn divided by its side count. Bound that turn when n is at least seven.
3. Hint 3
  Explain both why two corners cannot fill a turn and why three or more then fail.
Answer

No such regular tessellation exists for any integer n at least seven.

Full solution

Let e and i be the exterior and interior angle measures in degrees.
Since n is at least seven, the exterior turn is no larger than 360 divided by seven, which is less than 60.
Thus
$\begin{gathered}e<60\\i=180-e\\i>120\end{gathered}$

Convexity also gives i less than 180.
Two corners therefore total less than a full turn, so at least three tiles must meet at a regular tiling vertex.

But three of these corners already exceed a full turn.
$3i>360$
Any larger number of corners exceeds it as well.

Consequently no integer corner count fills a full turn, so the necessary local condition fails for every n at least seven.
This conclusion concerns congruent regular n-gons meeting whole side to whole side; it does not rule out mixed or nonregular tilings.

Answer

No such regular tessellation exists for any integer n at least seven.

Key idea

A bound can rule out an entire polygon family when every possible vertex count fails the angle condition.
Problem 8 A proposed argument

Convex kite $UVWX$ has $UV=UX$ and $VW=WX$ . Its diagonals $UW$ and $VX$ meet at $T$ . Consider this argument.

Line 1: $VT=TX$ by the kite diagonal property.

Line 2: $UT=TW$ because the diagonals are perpendicular.

Line 3: Both diagonals bisect each other, so the quadrilateral is a parallelogram.

Line 4: Its adjacent congruent sides then make it a rhombus.

Identify the first unsupported line and explain the gap. State an additional hypothesis that repairs the argument. Then give an exact convex counterexample to the original rhombus conclusion and verify its required side pairs.
1. Hint 1
  Check which diagonal is guaranteed to bisect which other diagonal in this kite.
2. Hint 2
  Perpendicularity concerns an angle at the crossing, rather than the distances along both lines.
3. Hint 3
  For a counterexample, use perpendicular lines through T, equal distances toward V and X, and unequal positive distances toward U and W.
Answer

Line 2 is unsupported. Add $UT=TW$ . One convex counterexample on perpendicular rays has $UT=8$ , $TW=15$ , and $VT=TX=6$ , in common units.

Full solution

Line 1 is valid: UW joins the equal-side meeting vertices U and W, so it bisects VX.

Line 2 is unsupported.
A right angle at T does not imply that T is the midpoint of UW.

Repair: add the hypothesis
$UT=TW$
Then both diagonal midpoint conditions hold.
The parallelogram test is valid, and its adjacent kite pairs make all four sides congruent, proving rhombus.

Counterexample: draw perpendicular lines through T.
Put U and W on opposite rays at distances 8 and 15 units.
Put V and X on the other two rays at 6 units each.
Join U, V, W, X in order.
Each edge lies within one right-angle sector, and every vertex angle combines two acute angles, so the boundary is simple and strictly convex.

Triangles UTV and UTX share UT, have equal sides TV and TX, and have right included angles.
SAS gives UV equal to UX.
The same SAS comparison of WTV and WTX gives WV equal to WX.
Thus both disjoint adjacent kite pairs hold.

The diagonal parts UT and TW are unequal.
A rhombus would be a parallelogram with mutual diagonal bisection, so this exact kite is not a rhombus.

One constructed counterexample.

Answer

Line 2 is unsupported. Add $UT=TW$ . One convex counterexample on perpendicular rays has $UT=8$ , $TW=15$ , and $VT=TX=6$ , in common units.

Key idea

A correct kite bisection does not supply the reverse bisection needed for the parallelogram converse.
Problem 9 Same counts, different lengths

On paper, draw two different convex quadrilaterals. Each must have exactly two reflection axes and exactly two rotational symmetries, including the identity.

The first must have congruent whole diagonals. The second must have different whole diagonal lengths. Name the family of each example, mark all its reflection axes, and describe all its matching rotations.

Justify why your drawings meet the two different diagonal requirements.
1. Hint 1
  Equal symmetry counts do not require the axes to occupy the same positions relative to the sides.
2. Hint 2
  Compare two nonsquare special parallelogram families, paying attention to their diagonal properties.
3. Hint 3
  A rectangle’s axes join opposite side midpoints; a nonsquare rhombus’s axes are its unequal diagonals.
Answer

One valid pair is a nonsquare rectangle and a nonsquare rhombus. Each has two axes and rotations of zero degrees and 180 degrees. Rectangle diagonals are congruent; choose unequal rhombus diagonals.

Full solution

First example: draw a rectangle with adjacent sides 14 cm and 9 cm.
Its two axes join opposite side midpoints.
Its rotations are identity and a half-turn; unequal side lengths exclude quarter-turns.

A rectangle has congruent whole diagonals, so the first diagonal requirement holds.

Second example: draw perpendicular segments of 22 cm and 28 cm crossing at their shared midpoint.
Join their endpoints in alternating order.
Mutual diagonal bisection establishes a parallelogram, and perpendicular diagonals establish a rhombus.

The unequal diagonal lengths exclude a square.
This nonsquare rhombus has exactly its two diagonal reflection axes and the rotations identity and a half-turn.

The specified whole lengths
$22\ne28$
verify the second requirement directly.
Other correctly justified pairs of drawings are accepted.

One valid pair of drawings.

Answer

One valid pair is a nonsquare rectangle and a nonsquare rhombus. Each has two axes and rotations of zero degrees and 180 degrees. Rectangle diagonals are congruent; choose unequal rhombus diagonals.

Key idea

The number of symmetries does not by itself distinguish a nonsquare rectangle from a nonsquare rhombus.
Problem 10 A configuration with extra conditions

Convex isosceles trapezoid $ABCD$ has chosen parallel bases $AB$ and $CD$ . The bases have equal length, and diagonal $AC$ bisects the whole angle $DAB$ . Prove that $ABCD$ is a square.
1. Hint 1
  Keep track of the two facts about the chosen bases before using properties of a narrower family.
2. Hint 2
  The isosceles condition concerns the two angles at base AB. Once a parallelogram is established, those same consecutive angles also have a known sum.
3. Hint 3
  Use the endpoint-angle bisection only after establishing the family required by its converse test.
Answer

ABCD is a square.

Full solution

The chosen bases are parallel and congruent.
They form the same opposite-side pair, so the one-pair converse establishes that ABCD is a parallelogram.

Because the trapezoid is isosceles relative to AB and CD, its angles at A and B are congruent.
As consecutive parallelogram angles, their measures also satisfy
$m\angle A+m\angle B=180^\circ$

The two congruent angles each measure 90 degrees.
Opposite-angle congruence then supplies the other two right angles, so ABCD is a rectangle.

The given diagonal AC bisects a whole endpoint angle of the known parallelogram.
The angle-bisector converse therefore also makes ABCD a rhombus.

The quadrilateral is both a rectangle and a rhombus.
It has four right angles and four congruent sides, so it is a square.

Answer

ABCD is a square.

Key idea

Establish each test’s hypotheses in order when combining inclusive trapezoid, parallelogram, and special-shape facts.

Multiple choice

Core practice