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# 6.6: Trapezoids

Difficulty Level: At Grade Created by: CK-12
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What if you were told that the polygon ABCD is an isoceles trapezoid and that one of its base angles measures 38\begin{align*}38^\circ\end{align*} ? What can you conclude about its other angles? After completing this Concept, you'll be able to find the value of a trapezoid's unknown angles and sides given your knowledge of the properties of trapezoids.

### Guidance

A trapezoid is a quadrilateral with exactly one pair of parallel sides. Examples look like:

An isosceles trapezoid is a trapezoid where the non-parallel sides are congruent. The third trapezoid above is an example of an isosceles trapezoid. Think of it as an isosceles triangle with the top cut off. Isosceles trapezoids also have parts that are labeled much like an isosceles triangle. Both parallel sides are called bases.

Recall that in an isosceles triangle, the two base angles are congruent. This property holds true for isosceles trapezoids.

Theorem: The base angles of an isosceles trapezoid are congruent.

The converse is also true: If a trapezoid has congruent base angles, then it is an isosceles trapezoid. Next, we will investigate the diagonals of an isosceles trapezoid. Recall, that the diagonals of a rectangle are congruent AND they bisect each other. The diagonals of an isosceles trapezoid are also congruent, but they do NOT bisect each other.

Isosceles Trapezoid Diagonals Theorem: The diagonals of an isosceles trapezoid are congruent.

The midsegment (of a trapezoid) is a line segment that connects the midpoints of the non-parallel sides. There is only one midsegment in a trapezoid. It will be parallel to the bases because it is located halfway between them. Similar to the midsegment in a triangle, where it is half the length of the side it is parallel to, the midsegment of a trapezoid also has a link to the bases.

##### Investigation: Midsegment Property

Tools Needed: graph paper, pencil, ruler

1. Draw a trapezoid on your graph paper with vertices A(1,5), B(2,5), C(6,1)\begin{align*}A(-1, 5), \ B( 2, 5), \ C(6, 1)\end{align*} and D(3,1)\begin{align*}D(-3, 1)\end{align*}. Notice this is NOT an isosceles trapezoid.
2. Find the midpoint of the non-parallel sides either by using slopes or the midpoint formula. Label them E\begin{align*}E\end{align*} and F\begin{align*}F\end{align*}. Connect the midpoints to create the midsegment.
3. Find the lengths of AB, EF\begin{align*}AB, \ EF\end{align*}, and CD\begin{align*}CD\end{align*}. Can you write a formula to find the midsegment?

Midsegment Theorem: The length of the midsegment of a trapezoid is the average of the lengths of the bases, or EF=AB+CD2\begin{align*}EF=\frac{AB+CD}{2}\end{align*}.

#### Example A

Look at trapezoid TRAP\begin{align*}TRAP\end{align*} below. What is mA\begin{align*}m \angle A\end{align*}?

TRAP\begin{align*}TRAP\end{align*} is an isosceles trapezoid. So, mR=115\begin{align*}m \angle R = 115^\circ\end{align*}. To find mA\begin{align*}m \angle A\end{align*}, set up an equation.

115+115+mA+mP230+2mA2mAmA=360=360mA=mP=130=65\begin{align*}115^\circ + 115^\circ + m \angle A + m \angle P & = 360^\circ\\ 230^\circ + 2m \angle A & = 360^\circ \rightarrow m \angle A = m \angle P\\ 2m \angle A & = 130^\circ\\ m \angle A & = 65^\circ\end{align*}

Notice that mR+mA=115+65=180\begin{align*}m \angle R + m \angle A = 115^\circ + 65^\circ = 180^\circ\end{align*}. These angles will always be supplementary because of the Consecutive Interior Angles Theorem. Therefore, the two angles along the same leg (or non-parallel side) are always going to be supplementary. Only in isosceles trapezoids will opposite angles also be supplementary.

#### Example B

Write a two-column proof.

Given: Trapezoid ZOID\begin{align*}ZOID\end{align*} and parallelogram ZOIM\begin{align*}ZOIM\end{align*}

DI\begin{align*}\angle D \cong \angle I\end{align*}

Prove: ZD¯¯¯¯¯¯¯¯OI¯¯¯¯¯¯\begin{align*}\overline{ZD} \cong \overline{OI}\end{align*}

Statement Reason
1. Trapezoid ZOID\begin{align*}ZOID\end{align*} and parallelogram ZOIM, DI\begin{align*}ZOIM, \ \angle D \cong \angle I\end{align*} Given
2. ZM¯¯¯¯¯¯¯¯¯OI¯¯¯¯¯¯\begin{align*}\overline{ZM} \cong \overline{OI}\end{align*} Opposite Sides Theorem
3. IZMD\begin{align*}\angle I \cong \angle ZMD\end{align*} Corresponding Angles Postulate
4. DZMD\begin{align*}\angle D \cong \angle ZMD\end{align*} Transitive PoC
5. ZM¯¯¯¯¯¯¯¯¯ZD¯¯¯¯¯¯¯¯\begin{align*}\overline{ZM} \cong \overline{ZD}\end{align*} Base Angles Converse
6. ZD¯¯¯¯¯¯¯¯OI¯¯¯¯¯¯\begin{align*}\overline{ZD} \cong \overline{OI}\end{align*} Transitive PoC

#### Example C

Find x\begin{align*}x\end{align*}. All figures are trapezoids with the midsegment.

Answers:

1. a) x\begin{align*}x\end{align*} is the average of 12 and 26. 12+262=382=19\begin{align*}\frac{12+26}{2}=\frac{38}{2}=19\end{align*}

b) 24 is the average of x\begin{align*}x\end{align*} and 35.

x+352x+35x=24=48=13\begin{align*}\frac{x+35}{2}&=24\\ x+35&=48\\ x&=13\end{align*}

c) 20 is the average of \begin{align*}5x-15\end{align*} and \begin{align*}2x-8\end{align*}.

\begin{align*}\frac{5x-15+2x-8}{2}&=20\\ 7x-23& =40\\ 7x&=63\\ x&=9\end{align*}

Watch this video for help with the Examples above.

#### Concept Problem Revisited

Given an isosceles trapezoid with \begin{align*} m \angle B = 38^\circ \end{align*}, find the missing angles.

In an isosceles trapezoid, base angles are congruent.

\begin{align*} \angle B \cong \angle C\end{align*} and \begin{align*}\angle A \cong \angle D\end{align*}

\begin{align*}m \angle B = m \angle C = 38^\circ\\ 38^\circ + 38^\circ + m \angle A + m \angle D = 360^\circ\\ m \angle A = m \angle D = 142^\circ\end{align*}

### Vocabulary

A trapezoid is a quadrilateral with exactly one pair of parallel sides. An isosceles trapezoid is a trapezoid where the non-parallel sides and base angles are congruent. The midsegment (of a trapezoid) is a line segment that connects the midpoints of the non-parallel sides.

### Guided Practice

\begin{align*}TRAP\end{align*} an isosceles trapezoid.

Find:

1. \begin{align*}m \angle TPA\end{align*}
2. \begin{align*}m \angle PTR\end{align*}
3. \begin{align*}m \angle PZA\end{align*}
4. \begin{align*}m \angle ZRA\end{align*}

Answers:

1. \begin{align*}\angle TPZ \cong \angle RAZ\end{align*} so \begin{align*}m\angle TPA=20^\circ + 35^\circ = 55^\circ\end{align*}.

2. \begin{align*}\angle TPA\end{align*} is supplementary with \begin{align*}\angle PTR\end{align*}, so \begin{align*}m\angle PTR=125^\circ\end{align*}.

3. By the Triangle Sum Theorem, \begin{align*}35^\circ + 35^\circ + m\angle PZA=180^\circ\end{align*}, so \begin{align*}m\angle PZA=110^\circ\end{align*}.

4. Since \begin{align*}m\angle PZA = 110^\circ\end{align*}, \begin{align*}m\angle RZA=70^\circ\end{align*} because they form a linear pair. By the Triangle Sum Theorem, \begin{align*}m\angle ZRA=90^\circ\end{align*}.

### Practice

1. Can the parallel sides of a trapezoid be congruent? Why or why not?

For questions 2-7, find the length of the midsegment or missing side.

Find the value of the missing variable(s).

Find the lengths of the diagonals of the trapezoids below to determine if it is isosceles.

1. \begin{align*}A(-3, 2), B(1, 3), C(3, -1), D(-4, -2)\end{align*}
2. \begin{align*}A(-3, 3), B(2, -2), C(-6, -6), D(-7, 1)\end{align*}
3. \begin{align*}A(1, 3), B(4, 0), C(2, -4), D(-3, 1)\end{align*}
4. \begin{align*}A(1, 3), B(3, 1), C(2, -4), D(-3, 1)\end{align*}
1. Write a two-column proof of the Isosceles Trapezoid Diagonals Theorem using congruent triangles.

Given: \begin{align*}TRAP\end{align*} is an isosceles trapezoid with \begin{align*}\overline{TR} \ || \ \overline{AP}\end{align*}.

Prove: \begin{align*}\overline{TA} \cong \overline{RP}\end{align*}

1. How are the opposite angles in an isosceles trapezoid related?.
2. List all the properties of a trapezoid.

### Notes/Highlights Having trouble? Report an issue.

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### Vocabulary Language: English

Diagonal

A diagonal is a line segment in a polygon that connects nonconsecutive vertices

midsegment

A midsegment connects the midpoints of two sides of a triangle or the non-parallel sides of a trapezoid.

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Date Created:
Jul 17, 2012
Last Modified:
Aug 02, 2016
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