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6. Quadrilaterals & Other Polygons
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Geometry

6.03 Trapezoids

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Lesson

Concept summary

A trapezoid is a quadrilateral with at least one pair of parallel sides.

A trapezoid labelled with its parts. Each of two parallel sides islabelled Base, and each of the two non parallel sides is labelled Leg. Top base angles are labelled 1 and 2 consecutively. Bottom base angles are labelled 3 and 4 respectively.

The parallel sides are called bases and the nonparallel sides are called legs.

Consecutive angles in a trapezoid whose common side is a base are base angles. A trapezoid has two pairs of base angles.

In this case, \angle 1 and \angle 2 are base angles to the top base and \angle 3 and \angle 4 are base angles to the bottom base.

A midsegment of a trapezoid is a line segment that bisects both legs.

Trapezoid midsegment theorem

The midsegment of a trapezoid is parallel to each base and the length of the midsegment is equal to the sum of the length of the bases divided by two.

\overline{PS} \parallel \overline{MN}, \overline{MN}\parallel \overline{QR}

\text{and}

\text{Length of midsegment}=\dfrac{b_1+b_2}{2}

\text{or}

\text{Length of midsegment}=\dfrac{1}{2}\left(b_1+b_2\right)

Trapezoid P Q R S is drawn with top base Q R labelled b sub 2 and bottom base P S labelled b sub 1. M is the midpoint of Q P and N is the midpoint of R S. Midsegment M N is also drawn.

An isosceles trapezoid is a trapezoid with congruent legs.

The following theorems are related to isosceles trapezoids:

Isosceles trapezoid diagonals theorem

If a quadrilateral is an isosceles trapezoid, then its diagonals are congruent.

An isosceles trapezoid with its 2 diagonals marked congruent.
Isosceles trapezoid base angles theorem

If a quadrilateral is an isosceles trapezoid, then each pair of base angles is congruent

An isosceles trapezoid. Each pair of base angles is marked congruent.

There are several types of quadrilaterals. Below is a flowchart showing the different types of quadrilaterals. One shape points to another that has one extra condition.

A flowchart showing the different types of quadrilaterals. Starting with an irregular quadrilateral an arrow points to trapezoid in which two sides are marked parallel. An arrow points to an isosceles trapezoid where a pair of opposite sides are marked parallel and non parallel sides are marked congruent. Arrow points to a parallelogram where 2 pairs of opposite sides are marked parallel with each other. On the left of parallelogram, an arrow points to a rectangle in which all angles are right angles. On the right of the parallelogram, arrow points to a rhombus with all four sides marked congruent and 2 pairs of opposite sides are marked parallel with each other. Lastly, arrows point to a square with all four angles marked as right angles.

Worked examples

Example 1

ABCD is a trapezoid. Solve for x.

Trapezoid A B C D with segment A B and segment D C marked as parallel. A B measures 3 x + 2 and D C measures 2 x minus 2. A midsegment is drawn with a length of 15 which divides A D into two congruent segments as well as side B C.

Approach

We want to use the trapezoid midsegment theorem to relate the three given side lengths and then solve for x.

\text{Length of midsegment}=\dfrac{b_1+b_2}{2}

The base lengths are 3x+2 and 2x-2. The midsegment length is 15.

Solution

\displaystyle \text{Length of midsegment}\displaystyle =\displaystyle \dfrac{b_1+b_2}{2}
\displaystyle 15\displaystyle =\displaystyle \dfrac{\left(3x+2\right)+\left(2x-2\right)}{2}Substitution
\displaystyle 15\displaystyle =\displaystyle \dfrac{5x}{2}Combining like terms in the numerator on RHS
\displaystyle 30\displaystyle =\displaystyle 5xMultiply both sides of equation by 2
\displaystyle 6\displaystyle =\displaystyle xDivide both sides of equation by 5
\displaystyle x\displaystyle =\displaystyle 6Symmetric property of equality

Example 2

Given: Trapezoid PQRS

Trapezoid P Q R S is drawn in which side P S and Side Q R are marked congruent. Angle P Q R measures 63 degrees while angle P S R measures open parenthesis 12 x + 9 close parenthesis.

Solve for x.

Approach

From the diagram, we know that trapezoid PQRS is an isosceles trapezoid since the legs are congruent.

Each pair of base angles are congruent in isosceles triangles. In this case, \angle PQR \cong \angle QPS and \angle PSR \cong \angle SRQ.

All quadrilaterals' interior angles sum to 360 \degree.

Using the above information, we have m\angle PQR + m\angle QPS + m\angle PSR + m\angle SRQ=360

We can rewrite this equation using substitution to be:

m\angle PQR + m\angle PQR + m\angle PSR + m\angle PSR=360

Combining like terms, we get 2\left(m\angle PQR\right) + 2\left(m\angle PSR\right)=360

Next, we want to substitute given values into the above equation and solve for x.

Solution

\displaystyle 2\left(m\angle PQR\right) + 2\left(m\angle PSR\right)\displaystyle =\displaystyle 360
\displaystyle 2\left(m\angle PQR + m\angle PSR\right)\displaystyle =\displaystyle 360Factor out the common factor of 2
\displaystyle 2\left(12x+9+63\right)\displaystyle =\displaystyle 360Substitution
\displaystyle \left(12x+9+63\right)\displaystyle =\displaystyle 180Divide both sides of equation by 2
\displaystyle 12x+72\displaystyle =\displaystyle 180Combine like terms
\displaystyle 12x\displaystyle =\displaystyle 108Subtract 72 from both sides of equation
\displaystyle x\displaystyle =\displaystyle 9Divide both sides of the equation by 12

Reflection

Bases of a trapezoid are parallel so \angle QPS and \angle PSR are supplementary using the consecutive interior angles theorem. We could have also used theorems of parallel lines to solve for x.

Outcomes

G.MP1

Make sense of problems and persevere in solving them.

G.MP3

Construct viable arguments and critique the reasoning of others.

G.MP4

Model with mathematics.

G.MP5

Use appropriate tools strategically.

G.MP6

Attend to precision.

G.MP7

Look for and make use of structure.

G.MP8

Look for and express regularity in repeated reasoning.

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