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6. Quadrilaterals & Other Polygons
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United States of AmericaFL
Grade 912

6.03 Properties of special parallelograms

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Practice
Lesson

Concept summary

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:

Square

A quadrilateral with four right angles and four congruent sides.

A quadrilateral with four congruent sides and four right angles
Rhombus

A quadrilateral containing four congruent sides.

A quadrilateral with four congruent sides

These are two examples of rhombi:

The following theorems relate to the special parallelograms above:

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.

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

Worked examples

Example 1

Patricia draws a quadrilateral and covers it up. She tells Glen that the quadrilateral consists of right angles only. From this information, determine the most specific classification that Glen can determing for the figure.

Solution

There are two parallelograms that have four right angles: rectangles and squares. We are only given information about the angles of the shape and nothing about the side lengths. Using this information and the definition of special parallelograms, we can determine that the described figure is definitely a rectangle, but not necessarily a square.

So the most specific classification we can give is to say that Patricia has drawn a rectangle.

Reflection

If we were told that the side lengths were also equal length, then the figure would have been a square. Patricia could have drawn a square in this case, as a square is rectangle, but we do not know anything about the side lengths.

Example 2

Given:

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

Complete the following:

a

Solve for x.

Approach

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.

Solution

\displaystyle \left(5x+8\right) + \left(5x+8\right)\displaystyle =\displaystyle \left(12x+2\right)
\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
\displaystyle x\displaystyle =\displaystyle 7Symmetric property of equality
b

Solve for m\angle ABC.

Approach

\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.

Solution

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

Reflection

We can use other theorems for special parallelograms to solve for missing side lengths, angles, and diagonals.

Example 3

Consider the quadrilateral ABCD.

Quadrilateral A B C D with diagonals B D and A C intersecting at point E and perpendicular with each other. The diagonals also bisect each other so that segment B E, E D, C E and A E are all marked congruent.
a

Classify the quadrilateral.

Approach

We can identify which properties are noted on the diagram and use the list of properties for each type of special parallelogram to classify it.

Solution

We can see that the diagonals are perpendicular to one another and bisect one another. More than just bisecting one another though, BE=ED=AE=EC, so \overline{BD} \cong \overline{AC} using the segment addition postulate. This means that the diagonals are congruent and perpendicular, so ABCD is a rhombus.

Reflection

This diagram is not drawn to scale, so we need to use the actual given information and not try to determine by what it looks like.

b

Find BC.

Approach

ABCD is a rhombus, so we can use the other properties of rhombus.

Solution

Another property of rhombus is that all sides are congruent. This means that \overline{AB} \cong \overline{BC}, and since AB=6 \text{ ft}, BC=6 \text{ ft}.

Outcomes

MA.912.GR.1.4

Prove relationships and theorems about parallelograms. Solve mathematical and real-world problems involving postulates, relationships and theorems of parallelograms.

MA.912.LT.4.8

Construct proofs, including proofs by contradiction.

MA.912.LT.4.10

Judge the validity of arguments and give counterexamples to disprove statements.

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