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# 6.5: Trapezoids and Kites

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Define trapezoids, isosceles trapezoids, and kites.
• Define the midsegments of trapezoids.
• Plot trapezoids, isosceles trapezoids, and kites in the xy\begin{align*}x-y\end{align*} plane.

## Review Queue

1. Draw a quadrilateral with one set of parallel lines.
2. Draw a quadrilateral with one set of parallel lines and two right angles.
3. Draw a quadrilateral with one set of parallel lines and two congruent sides.
4. Draw a quadrilateral with one set of parallel lines and three congruent sides.

Know What? A kite, seen at the right, is made by placing two pieces of wood perpendicular to each other and one piece of wood is bisected by the other. The typical dimensions are included in the picture. If you have two pieces of wood, 36 inches and 54 inches, determine the values of x\begin{align*}x\end{align*} and 2x\begin{align*}2x\end{align*}.

## Trapezoids

Trapezoid: A quadrilateral with exactly one pair of parallel sides.

Isosceles Trapezoid: A trapezoid where the non-parallel sides are congruent.

## Isosceles Trapezoids

Previously, we introduced the Base Angles Theorem with isosceles triangles, which says, the two base angles are congruent. This property holds true for isosceles trapezoids. The two angles along the same base in an isosceles trapezoid are congruent.

Theorem 6-17: The base angles of an isosceles trapezoid are congruent.

If ABCD\begin{align*}ABCD\end{align*} is an isosceles trapezoid, then AB\begin{align*}\angle A \cong \angle B\end{align*} and CD\begin{align*}\angle C \cong \angle D\end{align*}.

Example 1: Look at trapezoid TRAP\begin{align*}TRAP\end{align*} below. What is mA\begin{align*}m \angle A\end{align*}?

Solution: TRAP\begin{align*}TRAP\end{align*} is an isosceles trapezoid. mR=115\begin{align*}m \angle R = 115^\circ\end{align*} also.

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 \qquad \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 from Chapter 3.

Theorem 6-17 Converse: If a trapezoid has congruent base angles, then it is an isosceles trapezoid.

Example 2: Is ZOID\begin{align*}ZOID\end{align*} an isosceles trapezoid? How do you know?

Solution: 4035\begin{align*}40^\circ \neq 35^\circ\end{align*}, ZOID\begin{align*}ZOID\end{align*} is not an isosceles trapezoid.

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

Example 3: Show TA=RP\begin{align*}TA = RP\end{align*}.

Solution: Use the distance formula to show TA=RP\begin{align*}TA = RP\end{align*}.

TA=(28)2+(40)2=(6)2+42=36+16=52RP=(60)2+(40)2    =62+42    =36+16=52\begin{align*}TA & = \sqrt{(2-8)^2 + (4-0)^2} && RP = \sqrt{(6-0)^2 + (4-0)^2}\\ & = \sqrt{(-6)^2 + 4^2} && \ \ \ \ = \sqrt{6^2 + 4^2}\\ & = \sqrt{36 + 16} = \sqrt{52} && \ \ \ \ = \sqrt{36 + 16} = \sqrt{52}\end{align*}

## Midsegment of a Trapezoid

Midsegment (of a trapezoid): 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.

Investigation 6-5: Midsegment Property

Tools Needed: graph paper, pencil, ruler

1. Plot \begin{align*}A(-1, 5), B( 2, 5), C(6, 1)\end{align*} and \begin{align*}D(-3, 1)\end{align*} and connect them. This is NOT an isosceles trapezoid.

2. Find the midpoint of the non-parallel sides by using the midpoint formula. Label them \begin{align*}E\end{align*} and \begin{align*}F\end{align*}. Connect the midpoints to create the midsegment.

3. Find the lengths of \begin{align*}AB\end{align*}, \begin{align*}EF\end{align*}, and \begin{align*}CD\end{align*}. What do you notice?

Midsegment Theorem: The length of the midsegment of a trapezoid is the average of the lengths of the bases.

If \begin{align*}\overline{EF}\end{align*} is the midsegment, then \begin{align*}EF = \frac{AB + CD}{2}\end{align*}.

Example 4: Algebra Connection Find \begin{align*}x\end{align*}. All figures are trapezoids with the midsegment.

a)

b)

c)

Solution:

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

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

\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*}

## Kites

The last quadrilateral to study is a kite. Like you might think, it looks like a kite that flies in the air.

Kite: A quadrilateral with two sets of adjacent congruent sides.

From the definition, a kite could be concave. If a kite is concave, it is called a dart.

The angles between the congruent sides are called vertex angles. The other angles are called non-vertex angles. If we draw the diagonal through the vertex angles, we would have two congruent triangles.

Given: \begin{align*}KITE\end{align*} with \begin{align*}\overline{KE} \cong \overline{TE}\end{align*} and \begin{align*}\overline{KI} \cong \overline{TI}\end{align*}

Prove: \begin{align*}\angle K \cong \angle T\end{align*}

Statement Reason
1. \begin{align*}\overline{KE} \cong \overline{TE}\end{align*} and \begin{align*}\overline{KI} \cong \overline{TI}\end{align*} Given
2. \begin{align*}\overline{EI} \cong \overline{EI}\end{align*} Reflexive PoC
3. \begin{align*}\triangle EKI \cong \triangle ETI\end{align*} SSS
4. \begin{align*}\angle K \cong \angle T\end{align*} CPCTC

Theorem 6-21: The non-vertex angles of a kite are congruent.

If \begin{align*}KITE\end{align*} is a kite, then \begin{align*}\angle K \cong \angle T\end{align*}.

Theorem 6-22: The diagonal through the vertex angles is the angle bisector for both angles.

If \begin{align*}KITE\end{align*} is a kite, then \begin{align*}\angle KEI \cong \angle IET\end{align*} and \begin{align*}\angle KIE \cong \angle EIT\end{align*}.

The proof of Theorem 6-22 is very similar to the proof above for Theorem 6-21.

Kite Diagonals Theorem: The diagonals of a kite are perpendicular.

\begin{align*}\triangle KET\end{align*} and \begin{align*}\triangle KIT\end{align*} triangles are isosceles triangles, so \begin{align*}\overline{EI}\end{align*} is the perpendicular bisector of \begin{align*}\overline{KT}\end{align*} (Isosceles Triangle Theorem, Chapter 4).

Example 5: Find the missing measures in the kites below.

a)

b)

Solution:

a) The two angles left are the non-vertex angles, which are congruent.

\begin{align*}130^\circ + 60^\circ + x + x & = 360^\circ\\ 2x & = 170^\circ\\ x & = 85^\circ \qquad \text{Both angles are}\ 85^\circ.\end{align*}

b) The other non-vertex angle is also \begin{align*}94^\circ\end{align*}. To find the fourth angle, subtract the other three angles from \begin{align*}360^\circ\end{align*}.

\begin{align*}90^\circ + 94^\circ + 94^\circ + x & = 360^\circ\\ x & = 82^\circ\end{align*}

Be careful with the definition of a kite. The congruent pairs are distinct, which means that a rhombus and square cannot be a kite.

Example 6: Use the Pythagorean Theorem to find the length of the sides of the kite.

Solution: Recall that the Pythagorean Theorem is \begin{align*}a^2 + b^2 = c^2\end{align*}, where \begin{align*}c\end{align*} is the hypotenuse. In this kite, the sides are the hypotenuses.

\begin{align*}6^2 + 5^2 & = h^2 && 12^2 + 5^2 = j^2\\ 36 + 25 & = h^2 && 144 + 25 = j^2\\ 61 & = h^2 && \qquad \ 169 = j^2\\ \sqrt{61} & = h && \qquad \quad 13 = j\end{align*}

## Kites and Trapezoids in the Coordinate Plane

Example 7: Determine what type of quadrilateral \begin{align*}RSTV\end{align*} is.

Solution: Find the lengths of all the sides.

\begin{align*}RS & = \sqrt{(-5 - 2)^2 + (7 - 6)^2} && ST = \sqrt{(2 - 5)^2 + (6-(-3))^2}\\ & = \sqrt{(-7)^2 + 1^2} && \quad \ = \sqrt{(-3)^2 + 9^2}\\ & = \sqrt{50} && \quad \ = \sqrt{90}\end{align*}

\begin{align*}RV & = \sqrt{(-5-(-4))^2 + (7-0)^2} && VT = \sqrt{(-4-5)^2 + (0-(-3))^2}\\ & = \sqrt{(-1)^2 + 7^2} && \quad \ \ = \sqrt{(-9)^2 + 3^2}\\ & = \sqrt{50} && \quad \ \ = \sqrt{90}\end{align*}

From this we see that the adjacent sides are congruent. Therefore, \begin{align*}RSTV\end{align*} is a kite.

Example 8: Determine what type of quadrilateral \begin{align*}ABCD\end{align*} is. \begin{align*}A(-3, 3), B(1, 5), C(4, -1), D(1, -5)\end{align*}.

Solution: First, graph \begin{align*}ABCD\end{align*}. This will make it easier to figure out what type of quadrilateral it is. From the graph, we can tell this is not a parallelogram. Find the slopes of \begin{align*}\overline{BC}\end{align*} and \begin{align*}\overline{AD}\end{align*} to see if they are parallel.

Slope of \begin{align*}\overline{BC} = \frac{5-(-1)}{1-4} = \frac{6}{-3} = -2\end{align*}

Slope of \begin{align*}\overline{AD} = \frac{3-(-5)}{-3-1} = \frac{8}{-4} = -2\end{align*}

\begin{align*}\overline{BC} \| \overline{AD}\end{align*}, so \begin{align*}ABCD\end{align*} is a trapezoid. To determine if it is an isosceles trapezoid, find \begin{align*}AB\end{align*} and \begin{align*}CD\end{align*}.

\begin{align*}AB & = \sqrt{(-3-1)^2 + (3 - 5)^2} && ST = \sqrt{(4 - 1)^2 + (-1-(-5))^2}\\ & = \sqrt{(-4)^2 + (-2)^2} && \quad \ = \sqrt{3^2 + 4^2}\\ & = \sqrt{20} = 2 \sqrt{5} && \quad \ = \sqrt{25} = 5\end{align*}

\begin{align*}AB \neq CD\end{align*}, therefore this is only a trapezoid.

Example 9: Determine what type of quadrilateral \begin{align*}EFGH\end{align*} is. \begin{align*}E(5, -1), F(11, -3), G(5, -5), H(-1, -3)\end{align*}

Solution: To contrast with Example 8, we will not graph this example. Let’s find the length of all four sides.

\begin{align*}EF & = \sqrt{(5 - 11)^2 + (-1-(-3))^2} && \ \sqrt{FG} = \sqrt{(11-5)^2 + (-3-(-5))^2}\\ & = \sqrt{(-6)^2 + 2^2} = \sqrt{40} && \qquad \ \ = \sqrt{6^2 + 2^2} = \sqrt{40}\\ GH & = \sqrt{(5-(-1))^2 + (-5-(-3))^2} && \quad HE = \sqrt{(-1-5)^2 +(-3-(-1))^2}\\ & = \sqrt{6^2 + (-2)^2} = \sqrt{40} && \qquad \ \ = \sqrt{(-6)^2 + (-2)^2} = \sqrt{40}\end{align*}

All four sides are equal. This quadrilateral is either a rhombus or a square. Let’s find the length of the diagonals.

\begin{align*}EG & = \sqrt{(5 - 5)^2 + (-1-(-5))^2} && \ FH = \sqrt{(11-(-1))^2 + (-3-(-3))^2}\\ & = \sqrt{0^2 + 4^2} && \qquad = \sqrt{12^2 + 0^2}\\ & = \sqrt{16} = 4 && \qquad = \sqrt{144} = 12\end{align*}

The diagonals are not congruent, so \begin{align*}EFGH\end{align*} is a rhombus.

Know What? Revisited If the diagonals (pieces of wood) are 36 inches and 54 inches, \begin{align*}x\end{align*} is half of 36, or 18 inches. Then, \begin{align*}2x\end{align*} is 36.

## Review Questions

• Questions 1 and 2 are similar to Examples 1, 2, 5 and 6.
• Questions 3 and 4 use the definitions of trapezoids and kites.
• Questions 5-10 are similar to Example 4.
• Questions 11-16 are similar to Examples 5 and 6.
• Questions 17-22 are similar to Examples 4-6.
• Questions 23 and 24 are similar to Example 3.
• Questions 25-28 are similar to Examples 7-9.
• Questions 29 and 30 are similar to the proof of Theorem 6-21.
1. \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 ZRA\end{align*}
4. \begin{align*}m \angle PZA\end{align*}

2. \begin{align*}KITE\end{align*} is a kite. Find:
1. \begin{align*}m \angle ETS\end{align*}
2. \begin{align*}m \angle KIT\end{align*}
3. \begin{align*}m \angle IST\end{align*}
4. \begin{align*}m \angle SIT\end{align*}
5. \begin{align*}m \angle ETI\end{align*}

3. Writing Can the parallel sides of a trapezoid be congruent? Why or why not?
4. Writing Besides a kite and a rhombus, can you find another quadrilateral with perpendicular diagonals? Explain and draw a picture.

For questions 5-10, find the length of the midsegment or missing side.

For questions 11-16, find the value of the missing variable(s). All figures are kites.

Algebra Connection For questions 17-22, 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*}

For questions 25-28, determine what type of quadrilateral \begin{align*}ABCD\end{align*} is. \begin{align*}ABCD\end{align*} could be any quadrilateral that we have learned in this chapter. If it is none of these, write none.

1. \begin{align*}A(1, -2), B(7, -5), C(4, -8), D(-2, -5)\end{align*}
2. \begin{align*}A(6, 6), B(10, 8), C(12, 4), D(8, 2)\end{align*}
3. \begin{align*}A(-1, 8), B(1, 4), C(-5, -4), D(-5, 6)\end{align*}
4. \begin{align*}A(5, -1), B(9, -4), C(6, -10), D(3, -5)\end{align*}

Fill in the blanks to the proofs below.

1. Given: \begin{align*}\overline{KE} \cong \overline{TE}\end{align*} and \begin{align*}\overline{KI} \cong \overline{TI}\end{align*} Prove: \begin{align*}\overline{EI}\end{align*} is the angle bisector of \begin{align*}\angle KET\end{align*} and \begin{align*}\angle KIT\end{align*}

Statement Reason
1. \begin{align*}\overline{KE} \cong \overline{TE}\end{align*} and \begin{align*}\overline{KI} \cong \overline{TI}\end{align*}
2. \begin{align*}\overline{EI} \cong \overline{EI}\end{align*}
3. \begin{align*}\triangle EKI \cong \triangle ETI\end{align*}
4. CPCTC
5. \begin{align*}\overline{EI}\end{align*} is the angle bisector of \begin{align*}\angle KET\end{align*} and \begin{align*}\angle KIT\end{align*}
1. Given: \begin{align*}\overline{EK} \cong \overline{ET}, \overline{KI} \cong \overline{IT}\end{align*} Prove: \begin{align*}\overline{KT} \bot \overline{EI}\end{align*}

Statement Reason
1. \begin{align*}\overline{KE} \cong \overline{TE}\end{align*} and \begin{align*}\overline{KI} \cong \overline{TI}\end{align*}
2. Definition of isosceles triangles
3. \begin{align*}\overline{EI}\end{align*} is the angle bisector of \begin{align*}\angle KET\end{align*} and \begin{align*} \angle KIT\end{align*}
4. Isosceles Triangle Theorem
5. \begin{align*}\overline{KT} \bot \overline{EI}\end{align*}

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