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Rectangles, Rhombi, and Squares: 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 Two whole diagonals

Rectangle $ABCD$ has diagonals meeting at $O$ . Their lengths in centimeters are
$\begin{gathered}AC=7x-1\\BD=4x+14.\end{gathered}$
Find $x$ , the length of $BD$ , and the length of $AO$ .

The labels identify whole diagonals and their intersection.

Text description of this figure

Rectangle ABCD has its vertices in boundary order and both diagonals drawn, meeting at O. All vertex angles are marked right. The statement supplies the two diagonal expressions.
1. Hint 1
  Decide which fact compares the whole diagonals and which fact relates a diagonal to its half.
2. Hint 2
  Equate the two whole-diagonal expressions before finding a half-diagonal.
Answer

$\begin{gathered}x=5\\BD=34\text{ cm}\\AO=17\text{ cm}.\end{gathered}$

Full solution

A rectangle has congruent diagonals.
Equate the whole lengths and solve.
$\begin{aligned}7x-1&=4x+14\\3x&=15\\x&=5.\end{aligned}$

Substitution gives the whole diagonal length.
$\begin{aligned}BD&=4(5)+14\\BD&=34\text{ cm}.\end{aligned}$
The other expression gives the same length.

A rectangle is a parallelogram, so O bisects each diagonal.
$\begin{aligned}AO&=34\div2\\AO&=17\text{ cm}.\end{aligned}$

Answer

$\begin{gathered}x=5\\BD=34\text{ cm}\\AO=17\text{ cm}.\end{gathered}$

Key idea

Whole-diagonal congruence and diagonal bisection answer different parts of the problem.
Problem 2 Three angle locations

Rhombus $ABCD$ has diagonals meeting at $O$ . The whole angle $DAB$ measures $112^\circ$ .

Find the measures of angle $BAC$ , the whole angle $ABC$ , and angle $AOB$ .

The drawing labels the vertices and diagonal intersection.

Text description of this figure

Rhombus ABCD is drawn with its vertices in boundary order. Every side has a matching tick, and the diagonals meet at O. The statement specifies the whole angle at A as 112 degrees.
1. Hint 1
  Keep the vertex angles separate from the angles at the diagonal intersection.
2. Hint 2
  Use angle bisection at A, supplementary consecutive angles for B, and diagonal perpendicularity at O.
Answer

$\begin{gathered}m\angle BAC=56^\circ\\m\angle ABC=68^\circ\\m\angle AOB=90^\circ.\end{gathered}$

Full solution

The rhombus diagonal AC bisects the whole angle at A.
$\begin{aligned}m\angle BAC&=112^\circ\div2\\m\angle BAC&=56^\circ.\end{aligned}$

A rhombus is a parallelogram, so consecutive angles A and B are supplementary.
$\begin{aligned}m\angle ABC&=180^\circ-112^\circ\\m\angle ABC&=68^\circ.\end{aligned}$

The diagonals of a rhombus are perpendicular, so angle AOB is right.
Its measure is $90^\circ$ .

Answer

$\begin{gathered}m\angle BAC=56^\circ\\m\angle ABC=68^\circ\\m\angle AOB=90^\circ.\end{gathered}$

Key idea

Angle bisection, supplementary angles, and perpendicularity concern three different locations.
Problem 3 Repair the classification argument

An argument about a convex quadrilateral says, "Its diagonals are congruent, so it is a rectangle. Its diagonals are also perpendicular, so it is a square."

State one additional hypothesis about the diagonals that makes the entire argument valid. Rewrite the argument with the missing classification step, and explain why that step is needed.
1. Hint 1
  The special diagonal tests apply within a particular quadrilateral family.
2. Hint 2
  Look for diagonal evidence that first establishes a parallelogram.
Answer

Require that the diagonals bisect each other. Then diagonal bisection proves parallelogram, congruent diagonals prove rectangle, and perpendicular diagonals prove rhombus, so the quadrilateral is a square.

Full solution

Add the hypothesis that the diagonals bisect each other.
The diagonal-bisection converse proves that the convex quadrilateral is a parallelogram.

Now the congruent-diagonal test applies and proves that it is a rectangle.
The perpendicular-diagonal test applies and proves that it is a rhombus.

A quadrilateral that is both a rectangle and a rhombus is a square.
The missing step matters because equal or perpendicular diagonals alone do not establish these special classifications for an arbitrary convex quadrilateral.

Answer

Require that the diagonals bisect each other. Then diagonal bisection proves parallelogram, congruent diagonals prove rectangle, and perpendicular diagonals prove rhombus, so the quadrilateral is a square.

Key idea

A sufficient test can be used only after its hypotheses have been established.
Problem 4 From diagonal evidence to angles

Let $ABCD$ be a parallelogram whose diagonals $AC$ and $BD$ are congruent.

Prove that angle $ABC$ is right by comparing two triangles, and conclude that $ABCD$ is a rectangle. Name the congruence criterion and the angle relationship that finishes the proof.

The drawing records the givens without marking a right angle.

Text description of this figure

Parallelogram ABCD has both diagonals drawn. Opposite sides are marked in distinct congruent pairs, and the whole diagonals have matching triple ticks. No vertex angles are marked right.
1. Hint 1
  Use triangles that each contain a whole diagonal and share a boundary side.
2. Hint 2
  Compare triangles DAB and CBA by SSS. Then combine congruent consecutive angles with their sum.
Answer

Triangles $DAB$ and $CBA$ are congruent by SSS. Corresponding consecutive angles are congruent and supplementary, so angle $ABC$ is right and $ABCD$ is a rectangle.

Full solution

Opposite parallelogram sides AD and BC are congruent, AB is common, and the diagonals DB and CA are given congruent.
Therefore
$\triangle DAB\cong\triangle CBA$
by SSS.

The corresponding angles DAB and CBA are congruent.
They are consecutive angles of a parallelogram, so they are also supplementary.
Two equal angles summing to a straight angle each measure $90^\circ$ .

Angle CBA is the same angle as ABC, so angle ABC is right.
The opposite-angle and consecutive-angle properties make all four vertex angles right.
Thus ABCD is a rectangle.

Answer

Triangles $DAB$ and $CBA$ are congruent by SSS. Corresponding consecutive angles are congruent and supplementary, so angle $ABC$ is right and $ABCD$ is a rectangle.

Key idea

Congruence supplies angle equality; the parallelogram supplies supplementary angles.
Problem 5 The other diagonal

Let $ABCD$ be a rhombus. Prove that diagonal $BD$ bisects the angles at both $B$ and $D$ .

Give the ordered triangle correspondence and the two resulting angle congruences.

The rhombus with its diagonals.

Text description of this figure

Rhombus ABCD is named in boundary order. Its four sides have matching ticks, and both diagonals are drawn. No angle-bisection marks are given.
1. Hint 1
  Draw the two triangles on opposite sides of BD.
2. Hint 2
  The four congruent boundary sides supply two corresponding side pairs; the diagonal supplies the third.
Answer

$\begin{gathered}\triangle ABD\cong\triangle CBD\\\angle ABD\cong\angle CBD\\\angle ADB\cong\angle CDB.\end{gathered}$
Therefore BD bisects the angles at B and D.

Full solution

All four rhombus sides are congruent.
In triangles ABD and CBD, AB corresponds to CB, AD corresponds to CD, and BD is common.
SSS gives
$\triangle ABD\cong\triangle CBD.$

Corresponding angles at B are congruent.
$\angle ABD\cong\angle CBD$
Since BD lies inside the convex rhombus, these are the two parts of its whole angle at B.

Corresponding angles at D are also congruent.
$\angle ADB\cong\angle CDB$
These are the two parts of its whole angle at D.
Thus BD bisects both endpoint angles.

Answer

$\begin{gathered}\triangle ABD\cong\triangle CBD\\\angle ABD\cong\angle CBD\\\angle ADB\cong\angle CDB.\end{gathered}$
Therefore BD bisects the angles at B and D.

Key idea

A diagonal angle-bisection proof identifies the endpoint angles through a triangle correspondence.
Problem 6 Two sets of evidence

Case A. A parallelogram has congruent diagonals and two adjacent sides of different lengths. State its strongest guaranteed classification and whether it can be a square.

Case B. A parallelogram has perpendicular diagonals of different lengths. State its strongest guaranteed classification and whether it can be a square.
1. Hint 1
  Treat the two cases separately, applying the diagonal test appropriate to each.
2. Hint 2
  Use the unequal measurements to rule out the extra classification needed for a square.
Answer

Case A is a rectangle that is not a square. Case B is a rhombus that is not a square.

Full solution

Case A.
Congruent diagonals make the parallelogram a rectangle.
Its unequal adjacent sides prevent it from being a rhombus or a square.

Case B.
Perpendicular diagonals make the parallelogram a rhombus.
Its unequal whole diagonals prevent it from being a rectangle or a square.

Answer

Case A is a rectangle that is not a square. Case B is a rhombus that is not a square.

Key idea

A test can guarantee one family while another given excludes the intersection.
Problem 7 One bisected angle

Construct or describe a convex quadrilateral in boundary order with a diagonal that bisects one of its endpoint vertex angles, but that is not a rhombus.

Give exact lengths or construction instructions, justify the angle bisection, and prove that your quadrilateral is not a rhombus. A sketch alone is insufficient.
1. Hint 1
  Two triangles sharing a diagonal can be congruent without all four boundary sides being equal.
2. Hint 2
  You can place unequal lengths on two opposite rays and equal lengths on the two perpendicular rays.
Answer

One example has four perpendicular rays at O, with AO 7 units, OC 2 units, and BO and DO each 4 units. Diagonal AC bisects angle DAB, but the quadrilateral is not a rhombus.

Full solution

Take four rays from O, with consecutive rays perpendicular.
Place A and C on opposite rays at distances 7 and 2 units from O.
Place B and D on the other two rays at 4 units each, then join A, B, C, D in that boundary order.
This gives a convex quadrilateral.

Triangles AOB and AOD have common AO, equal legs OB and OD, and equal included right angles.
They are congruent by SAS, so angle BAO is congruent to angle DAO.
Thus AC bisects the whole angle DAB.

If the quadrilateral were a rhombus, it would be a parallelogram and its diagonals would bisect each other.
But AO is 7 units and OC is 2 units, so O is not the midpoint of AC.
The quadrilateral is therefore not a rhombus.

One construction satisfying the requested angle condition.

Answer

One example has four perpendicular rays at O, with AO 7 units, OC 2 units, and BO and DO each 4 units. Diagonal AC bisects angle DAB, but the quadrilateral is not a rhombus.

Key idea

A counterexample must satisfy the angle condition and separately fail a necessary rhombus condition.
Problem 8 Make the markings precise

On paper, draw a nonsquare rhombus $PQRS$ in boundary order and draw its diagonals meeting at $T$ . Choose an orientation with no side horizontal.

Use conventional markings to show all congruent sides, the two separate midpoint equalities, a right angle where the diagonals meet, and the two equal parts of the angle at $P$ . Use different tick patterns for the two half-diagonal pairs.

Write one sentence explaining why the angle at the intersection is right. Do not use a measurement or the appearance of the sketch as evidence.
1. Hint 1
  The midpoint pairs compare halves of the same diagonal, not halves of different diagonals.
2. Hint 2
  A pair of congruent triangles on opposite sides of PT creates congruent adjacent angles on a straight line.
Answer

A valid drawing marks all four sides congruent, PT congruent to TR, QT congruent to TS, PR perpendicular to QS, and angle QPT congruent to angle SPT. Congruent adjacent angles at T sum to a straight angle, so each is right.

Full solution

Draw a slanted nonsquare rhombus and label its vertices P, Q, R, S around the boundary.
Mark all four boundary sides with one common pattern.
Label the diagonal intersection T.

Use one matching tick pattern on PT and TR, and a different pattern on QT and TS.
Add a square at T and matching angle arcs on the two parts of the whole angle at P.

For the justification, triangles PTQ and PTS are congruent by SSS: PQ and PS are rhombus sides, TQ and TS are equal diagonal halves, and PT is common.
Their corresponding angles PTQ and PTS are equal adjacent angles whose sum is a straight angle, so each is right.

One possible fully annotated drawing.

Answer

A valid drawing marks all four sides congruent, PT congruent to TR, QT congruent to TS, PR perpendicular to QS, and angle QPT congruent to angle SPT. Congruent adjacent angles at T sum to a straight angle, so each is right.

Key idea

Separate marking patterns keep side equality, midpoint information, and angle equality unambiguous.
Problem 9 Assemble a classification

Convex quadrilateral $EFGH$ has diagonals meeting at $O$ . The diagonals bisect each other and are congruent. In addition, adjacent sides $EF$ and $FG$ are congruent.

Prove the strongest classification forced by the evidence. Explain the role of all three pieces of information.
1. Hint 1
  First decide which given establishes the family in which the special tests work.
2. Hint 2
  Congruent whole diagonals and congruent adjacent sides provide different special classifications.
Answer

The quadrilateral is a square. Bisection establishes parallelogram, congruent diagonals establish rectangle, and congruent adjacent sides establish rhombus.

Full solution

The diagonals bisect each other, so the converse parallelogram test establishes that EFGH is a parallelogram.

Its congruent whole diagonals make that parallelogram a rectangle.

Its congruent adjacent sides EF and FG, together with congruent opposite sides of a parallelogram, make all four sides congruent.
It is therefore also a rhombus.

Being both a rectangle and a rhombus makes EFGH a square.
Each given supplies a distinct needed step.

Answer

The quadrilateral is a square. Bisection establishes parallelogram, congruent diagonals establish rectangle, and congruent adjacent sides establish rhombus.

Key idea

Build classification in an order that establishes every test hypothesis before using it.
Problem 10 Both directions and one missing hypothesis

Let $ABCD$ be a convex quadrilateral whose diagonals bisect each other. Prove both directions of this statement: $ABCD$ is a square if and only if its diagonals are congruent and perpendicular.

Then decide whether the statement remains true after removing the assumption that the diagonals bisect each other. Justify your decision with exact construction data.
1. Hint 1
  In the forward direction, use both families to which a square belongs.
2. Hint 2
  In the reverse direction, establish parallelogram first.
3. Hint 3
  For the final question, equal total diagonal lengths do not force equal parts on each diagonal.
Answer

With diagonal bisection assumed, both directions are true. Without it, the reverse direction fails; perpendicular rays with AO 5, OC 1, BO 3, and DO 3 give a convex nonsquare with congruent, perpendicular diagonals.

Full solution

Forward direction.
A square is a rectangle, so its diagonals are congruent.
It is also a rhombus, so its diagonals are perpendicular.

Reverse direction.
The assumed diagonal bisection establishes a parallelogram.
Congruent diagonals then establish a rectangle, and perpendicular diagonals establish a rhombus.
A quadrilateral in both families is a square.

Removing the assumption.
Take perpendicular rays through O.
Put A and C on opposite rays at distances 5 and 1 units, and B and D on the other two rays at 3 units each.
Joining A, B, C, D gives a convex quadrilateral.

Both whole diagonals have length 6 units, and they are perpendicular by construction.
But AO and OC are unequal, so the diagonals do not bisect each other.
It is not a parallelogram and cannot be a square.
Thus the reverse direction fails without the bisection assumption, while every square still has the two stated properties.

The diagonal facts used in the valid reverse direction.

Answer

With diagonal bisection assumed, both directions are true. Without it, the reverse direction fails; perpendicular rays with AO 5, OC 1, BO 3, and DO 3 give a convex nonsquare with congruent, perpendicular diagonals.

Key idea

A biconditional needs two proofs, and removing a hypothesis requires checking each direction again.

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