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9. Polygons
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Geometry

9.03 Special parallelograms

Lesson
Practice

Rectangles, rhombuses and squares

The following are special types of parallelograms, with specific properties about their sides, angles, and/or diagonals that help identify them:

Rectangle

A quadrilateral containing four right angles.

A quadrilateral with four right angles.

These are three examples of rectangles:

Three rectangles with different orientations, lengths, and width.
Square

A quadrilateral with four right angles and four congruent sides.

A quadrilateral with four congruent sides and four right angles.
Rhombus

A quadrilateral with four congruent sides.

A quadrilateral with four congruent sides.

These are two examples of rhombi:

Two rhombi with different orientations, lengths, and widths.

Exploration

Explore the applet by dragging the vertices of the polygons.

Loading interactive...

  1. Which polygon(s) are always rectangles? Can you create a rectangle with any of the polygons?

  2. How do you know which polygon(s) form a rhombus versus a square?

  3. Which polygon(s) are always parallelograms? Can you create a parallelogram with any of the polygons?

The following theorems relate to the special parallelograms:

Rectangle diagonals theorem

A parallelogram is a rectangle if and only if its diagonals are congruent.

Rectangle A B C D with diagonals A C and B D. The four segments formed from each of the vertex to the point of intersection of the diagonals are all congruent.
Rhombus diagonals theorem

A parallelogram is a rhombus if and only if its diagonals are perpendicular.

Rhombus P Q R S with diagonals P R and Q S drawn. Q S is perpendicular to P R.
Rhombus opposite angles theorem

A parallelogram is a rhombus if and only if each diagonal bisects a pair of opposite angles.

Rhombus P Q R S with diagonals P R and Q S drawn. The diagonals bisect angles P, Q, R, and S.

Squares have the same properties as both a rectangle and rhombus.

A square with its diagonals drawn. The diagonals are perpendicular to each other. The four segments formed from each of the vertex to the point of intersection of the diagonals are all congruent.

Note that these theorems are for parallelograms, so if we are only told that a polygon is a quadrilateral, then they may not meet the conditions stated.

Examples

Example 1

List all classifications of quadrilaterals that apply to the figures. Explain your reasoning.

a
A quadrilateral with 4 right interior angles.
Worked Solution
Create a strategy

Recall what we know about parallelograms:

  • A quadrilateral is a parallelogram if and only if its opposite sides are congruent.

  • A quadrilateral is a parallelogram if and only if its opposite angles are congruent.

  • In a parallelogram, consecutive angles will be supplementary.

  • A quadrilateral is a parallelogram if and only if its diagonals bisect each other.

Recall the special types of parallelograms:

  • A rectangle is a quadrilateral with four right angles.

  • A square is a quadtrilateral with four right angles and four congruent sides.

  • A rhombus is a quadtrilateral containing four congruent sides.

Apply the idea

The quadrilateral is a parallelogram and rectangle. A parallelogram's opposite angles are congruent, and a rectangle is a quadrilateral with four right angles by definition.

b
A quadrilateral with 4 congruent sides.
Worked Solution
Create a strategy

We are given information about the side lengths of the figure. Use this information to classify the figure.

Apply the idea

The figure is a parallelogram and rhombus. A parallelogram's opposite sides are congruent, and a quadrilateral with four congruent sides is a rhombus by definition.

c
A quadrilateral with 4 right interior angles and 4 congruent sides.
Worked Solution
Apply the idea

The figure can be classified as a parallelogram, square, rhombus, and rectangle because it has four right angles and consecutive congruent sides.

Reflect and check

Most specifically, since the angle measures are all right angles and the sides are all congruent, the figure is a square.

d
A quadrilateral with 2 pairs of adjacent congruent sides.
Worked Solution
Create a strategy

This figure cannot be classified as a parallelogram because its opposite sides are not congruent.

Apply the idea

The figure is a quadrilateral because it has four sides.

Example 2

Consider the diagram that illustrates the rhombus diagonals theorem: If a parallelogram is a rhombus, then its diagonals are perpendicular bisectors of one another.

A diagram showing 4 steps: step 1 shows rhombus A B C D; step 2 shows rhombus A B C D with diagonals A C and B D drawn, A C and B D intersect at point E, A E and E C are congruent as well as B E and E D; step 3 shows the same rhombus A B C D from step 2 with these additional information, angles D A C and A C D are congruent as well as angles A E D and D E C and angles A D E and E D C; step 4 shows the same rhombus A B C D from step 2 with these additional information, angles D A C and A C D are congruent as well as angles A D E and E D C, angles A E D and D E C are right angles.
a

Add reasoning to each step of the diagram.

Worked Solution
Create a strategy

Start with the given statement, and use the initial step to identify what is given. From there, the diagram provides more information to use when attempting to prove that the diagonals are perpendicular bisectors of one another.

Apply the idea
A figure titled Step 1 showing rhombus A B C D.

We are given a parallelogram that is a rhombus, where all sides are congruent and opposite sides are parallel to one another.

A figure titled Step 2 showing rhombus A B C D with diagonals A C and B D drawn. A C and B D intersect at point E. A E and E C are congruent as well as B E and E D.

The diagonals of the rhombus bisect one another since the diagonals of a parallelogram bisect one another.

A figure titled Step 3 showing rhombus A B C D with diagonals A C and B D drawn. A C and B D intersect at point E. A E and E C are congruent as well as B E and E D. Angles D A C and A C D are congruent as well as angles A E D and D E C and angles A D E and E D C.

The triangles formed by the diagonals are congruent by SSS congruence, so we know that their corresponding angles are also congruent by CPCTC.

A figure titled Step 4 showing rhombus A B C D with diagonals A C and B D drawn. A C and B D intersect at point E. A E and E C are congruent as well as B E and E D. Angles D A C and A C D are congruent as well as angles A D E and E D C. Angles A E D and D E C are right angles.

Since the triangles are congruent, we know that \angle AED \cong \angle CED and they are linear pairs along \overline{AC}. Since they are congruent and linear pairs, they must each be 90 \degree. This means that \overline{AC} \perp \overline{BD}, so the diagonals are perpendicular bisectors of one another.

b

Write a formal proof of the theorem.

Worked Solution
Create a strategy

Use the diagram from part (a), definitions, and theorems to write a formal proof.

Apply the idea

Consider the rhombus ABCD:

Rhombus A B C D with diagonals A C and B D drawn. The diagonals intersect at point E.
To prove: \overline{AC} is a perpendicular bisector of \overline{BD}
StatementsReasons
1.ABCD is a rhombusGiven
2.\overline{AC} is a bisector of \overline{BD}If a quadrilateral is a parallelogram, then its diagonals bisect each other.
3.\overline{AE} \cong \overline{EC} and \overline{BE} \cong \overline{ED}Definition of bisecting
4.\overline{AB} \cong \overline{BC} \cong \overline{CD} \cong \overline{DA}Sides of rhombus are congruent
5.\triangle AED \cong \triangle AEB \cong \triangle CEB \cong \triangle CEDSide-side-side congruency theorem
6.\angle AED \cong \angle AEB \cong \angle CEB \cong \angle CEDCorresponding parts of congruent triangles theorem
7.m\angle AED = m\angle DECDefinition of congruence
8.m\angle AED + m\angle DEC=180 \degree\angle AED and \angle DEC are a linear pair
9.2 \cdot m\angle AED =180 \degreeSubstitution
10.m\angle AED =90 \degreeDivide by 2
11.m\angle AED = m\angle AEB = m\angle CEB = m\angle CED=90\degreeDefinition of congruence
12.\overline{AC} \perp \overline{BD}Definition of perpendicular
13.\overline{AC} and \overline{BD} are perpendicular bisectors of one anotherPerpendicular and bisect one another

Example 3

Prove that in the given rectangle ABCD, \overline{AC}\cong \overline{BD}:

Rectangle A B C D with diagonals A C and B D drawn.
Worked Solution
Create a strategy

It may be useful to mark up a diagram with information that we are given and that we already know using the definition of a rectangle, then use that to help build the proof.

Rectangle A B C D with diagonals A C and B D drawn. A B and D C are congruent as well as A D and B C. Angles A, B, C and D are right angles.
Apply the idea
A flow chart proof with 6 levels and the reason below each step. The third level has 2 steps. On the first level, the step is labeled A B C D, with reason Given. On the second level, the step is A B C D is parallelogram and angle A D C and angle B C D are right angles, with reason Definition of a rectangle. On the third level, the left step is labeled A D is congruent to B C, with reason If a quadrilateral is parallelogram, then its opposite sides are congruent, and the right step is labeled angle A D C is congruent to angle B C D, with reason All right angles are congruent. The left step on the thirs level proceeds to the fourth level while the right step goes directly to the fifth level. On the fourth level, the step is labeled D C is congruent to D C, with reason Reflexive property of congruence. On the fifth level, the step is labeled triangle A C D is congruent to triangle B D C, with reason Side angle side congruency theorem. On the sixth level, the step is labeled A C is congruent to B D, with reason C P C T C.

Example 4

Given:

  • ABCD is a rhombus
  • m\angle ACD=(5x+8)\degree
  • m\angle BCD=(12x+2)\degree

Complete the following:

Rhombus A B C D with diagonal A C drawn.
a

Solve for x.

Worked Solution
Create a strategy

Since parallelogram ABCD is a rhombus, the diagonals in a rhombus bisect a pair of opposite angles.

In other words, m\angle ACD = m\angle ACB and m\angle ACD+ m\angle ACB = m\angle BCD

We want to use this information to write an equation relating the two given angles.

\left(5x+8\right) + \left(5x+8\right)=\left(12x+2\right)

We want to use the above equation to solve for x.

Apply the idea
\displaystyle \left(5x+8\right) + \left(5x+8\right)\displaystyle =\displaystyle \left(12x+2\right)m\angle ACD+ m\angle ACB = m\angle BCD
\displaystyle 10x+16 \displaystyle =\displaystyle 12x+2Combine like terms
\displaystyle 10x+14\displaystyle =\displaystyle 12xSubtract 2 from both sides of equation
\displaystyle 14\displaystyle =\displaystyle 2xSubtract 10x from both sides of equation
\displaystyle 7\displaystyle =\displaystyle xDivide both sides of equation by 2
b

Solve for m\angle ABC.

Worked Solution
Create a strategy

\angle BCD and \angle ABC are consecutive angles.

Since ABCD is a rhombus and a rhombus is a parallelogram, we know that consecutive angles are supplementary.

So m\angle BCD + m\angle ABC = 180\degree.

Use the given information and the value for x we solved for in part (a) to find m\angle BCD. Then use that information and the above equation to solve for m\angle ABC.

Apply the idea

Substituting 7 for x, we get m\angle BCD=\left(12 \left(7 \right )+2 \right)\degree.

So m\angle BCD = 86 \degree. Now we want to use this angle measure and the fact that consecutive angles are supplementary to solve for m\angle ABC.

\displaystyle m\angle BCD + m\angle ABC\displaystyle =\displaystyle 180\degreeConsecutive angles in a parallelogram are supplementary
\displaystyle 86\degree + m\angle ABC\displaystyle =\displaystyle 180\degreeSubstitution
\displaystyle m\angle ABC\displaystyle =\displaystyle 94\degreeSubtract 86\degree from both sides of the equation

Example 5

If \overline{AB} \cong \overline{CD}, show that ABCD is not a rhombus.

Quadrilateral A B C D. A B has a length of 2 x plus 5, B C has a length of 3 x minus 2, C D has a length of 4 x minus 17, and D A has a length of 27.
Worked Solution
Create a strategy

We know that a rhombus has four congruent sides by definition, so to show that the figure is not a rhombus, we need to show that the side lengths are not congruent.

If \overline{AB} \cong \overline{CD}, we know that AB = CD by the definition of congruence. Start by solving the equation 2x+5=4x-17 for x, then use the value of x to determine the side lengths of the quadrilateral.

Apply the idea
\displaystyle 2x+5\displaystyle =\displaystyle 4x-17\overline{AB} = \overline{CD}
\displaystyle 2x+22\displaystyle =\displaystyle 4xAdd 17 to both sides of the equation
\displaystyle 22\displaystyle =\displaystyle 2xSubtract 2x from both sides of the equation
\displaystyle x\displaystyle =\displaystyle 11Divide both sides of the equation by 2

Since x=11, we know that BC=3x-2=3 \left(11 \right) -2= 22-2 = 31. Since AD \neq BC, ABCD is not a rhombus because we have shown that at least two of the sides are not congruent.

Example 6

Use geometric constructions and properties of rhombuses to verify that parallelogram WXYZ is a rhombus.

Worked Solution
Create a strategy

If a parallelogram has perpendicular diagonals, it's a rhombus. So, we may construct perpendicular bisectors for the diagonals of the parallelogram to verify that it is a rhombus.

Apply the idea

\overline{X'Z'}\perp \overline{WY} at the midpoint of \overline{WY}, so \overline{X'Z'} is the perpendicular bisector of \overline{WY}.

The diagonals \overline{WY} and \overline{X'Z'} are perpendicular bisectors of each other and pass through the intersection point E'. Since they coincide with the diagonals \overline{WY} and \overline{XZ} and the intersection point E, parallelogram WXYZ is a rhombus.

Idea summary

Use the theorems relating to the special parallelogrmas to solve problems:

  • Rectangle diagonals theorem: A parallelogram is a rectangle if and only if its diagonals are congruent.

  • Rhombus diagonals theorem: A parallelogram is a rhombus if and only if its diagonals are perpendicular.

  • Rhombus opposite angles theorem: A parallelogram is a rhombus if and only if each diagonal bisects a pair of opposite angles.

Outcomes

G.PC.1

The student will prove and justify theorems and properties of quadrilaterals, and verify and use properties of quadrilaterals to solve problems, including the relationships between the sides, angles, and diagonals.

G.PC.1a

Solve problems, using the properties specific to parallelograms, rectangles, rhombi, squares, isosceles trapezoids, and trapezoids.

G.PC.1b

Prove and justify that quadrilaterals have specific properties, using coordinate and algebraic methods, such as the slope formula, the distance formula, and the midpoint formula.

G.PC.1c

Prove and justify theorems and properties of quadrilaterals using deductive reasoning.

G.PC.1d

Use congruent segment, congruent angle, angle bisector, perpendicular line, and/or parallel line constructions to verify properties of quadrilaterals.

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