lesson 6-5 properties of kites

Properties of Kites 339

Lesson

6-5Properties of Kites

Lesson 6-5

Vocabulary

ends of a kite

symmetry diagonal

BIG IDEA Kites include rhombuses, and rhombuses include squares, so all properties of kites are properties of these other fi gures as well.

Lockheed Martin’s F-117A Nighthawk was the fi rst airplane design whose shape and form contributed to its stealth. The angles between surfaces actually scatter and defl ect radar signals making it hard to detect in the air. A view from above reveals many quadrilaterals from the hierarchy. One of its most obvious features is its symmetry. This lesson focuses on kites such as FAST, which is outlined on the body of the airplane. The symmetry of kites can be used to deduce some of the characteristics of this quadrilateral.

Constructing a Kite

Because kites must have two distinct pairs of congruent sides, one way to construct a kite precisely is with circles, as in the Activity below.

Activity

MATERIALS DGS

Step 1 Construct �A, point B on �A, and point C not on �A.

Step 2 Construct �C with radius CB.

(continued on next page)

Mental Math

F

A

S

T

� ⎯ � VU is a symmetry line in

the fi gure below.

V

Z

YU

1.7

32˚

Find:

a. m∠Z.

b. m∠ZUV.

c. ZY.

SMP_SEGEO_C06L05_339-344.indd 339SMP_SEGEO_C06L05_339-344.indd 339 3/26/08 12:03:04 PM3/26/08 12:03:04 PM

340 Polygons and Symmetry

Chapter 6

Step 3 Let D be the other point of intersection of �C and �A.

Step 4 Construct polygon ABCD.

Step 5 After Step 4, your construction should look like the one at the right. Drag point C. For most of the positions of C, ABCD is a kite. Why?

Step 6 For what positions of C is ABCD a nonconvex kite?

Step 7 For what positions of C is ABCD not a kite?

Step 8 Make ABCD a convex kite. Construct diagonals __

AC and ___

BD . Name the point where these diagonals intersect point E. Experiment with your sketch to come up with a relationship between the diagonals.

The Activity shows that one way to construct a kite starts with intersecting circles. A kite can also be constructed by taking the union of two isosceles triangles and removing the common side. Points A and C are then ends of the kite. The ends of a kite are the common endpoints of its congruent sides.

Refl ection Symmetry of a Kite

Because kites can be described using isosceles triangles, many of the theorems from Lesson 6-2 apply to kites.

Kite Symmetry Theorem

The line containing the ends of a kite is a symmetry line for the kite.

Given ABCD is a kite with ends B and D.

Prove � �� BD is a symmetry line for ABCD.

Drawing

D

A

C

B

Proof It is given that ABCD is a kite with ends B and D. Construct the auxiliary segment

__ AC .

__ AC has exactly one perpendicular bisector;

construct it and call it m.

D

A

C

Bm

D

A

C

Bm



Lesson 6-5

Because m is the perpendicular bisector of __

AC , rm

(A) = C and rm

(C) = A by the defi nition of refl ection. B and D are the ends of the kite, so AB = BC and AD = DC. So �ABC and �ADC are isosceles triangles. By the Isosceles Triangle Symmetry Theorem, the perpendicular bisector of the base also bisects the vertex angle, which means it contains the vertex.

Thus, the line m contains B and D, so we can name it � �� BD . By the defi nition of refl ection, r

B �

��

D(B) = B and r

B �

��

D(D) = D. By the Figure

Transformation Theorem, rB �

��

D(ABCD) = CBAD, which means that

� �� BD is a symmetry line of ABCD.

The diagonal of a kite that is on the symmetry line and contains the ends of the kite is the symmetry diagonal of the kite. The above proof has also demonstrated the following theorem.

Kite Diagonal Theorem

The symmetry line of a kite is the perpendicular bisector of the other diagonal and bisects the two angles at the ends of the kite.

Example 1

Given kite KITE with ends K and T. If EM = 8, m∠EKT = 40, and

m∠ITK = 23.5, fi nd as many other lengths and angle measures as you can.

Solution −−

KT is a symmetry diagonal for the kite, so

r −−

KT (∠EKT) = ∠IKT and m∠IKT = ? . Also,

r −−

KT (∠ITK) = ∠ETK, so m∠ETK = ? .

By the Angle Addition Assumption, m∠ETI = 47 and

m∠EKI = ? . Because KT is the perpendicular bisector of EI,

MI = ? and EI = ? .

The four angles with vertex M each have measure 90. So, from

the Triangle-Sum Theorem, m∠TIE = ? and m∠KEI = ? .

Then, because base angles of isosceles triangles are congruent,

m∠ ? = ? and m∠ ? = ? . By the Angle Addition

Assumption, m∠TEK = m∠TIK = 116.5.

GUIDED

E K

M

T

I

40˚8

23.5˚



Chapter 6

Symmetry of Rhombuses

The Kite Diagonal Theorem applies to rhombuses and squares, as well as to kites, because they are below kites in the quadrilateral hierarchy. When a kite is a rhombus, any of the opposite vertices are a pair of ends. Thus, a rhombus has two symmetry diagonals. Consequently, the diagonals of a rhombus are perpendicular, and they bisect each other.

Rhombus Diagonal Theorem

Each diagonal of a rhombus is a symmetry line of the rhombus and the perpendicular bisector of the other diagonal.

In rhombus PEDR at the right, ___

PD is the perpendicular bisector of ___ ER and

___ ER is the perpendicular bisector of

___ PD . Because of this and

due to the symmetry of any rhombus, many angles in PEDR have the same measure, and many segments have the same length. For instance, O is the midpoint of both

___ PD and

___ ER . Also, because of the

two symmetry lines, opposite angles of a rhombus are congruent. That is, ∠PED � ∠PRD and ∠EDR � ∠EPR.

Example 2

In rhombus PEDR, if m∠RED = 52, fi nd each angle measure.

a. m∠REP b. m∠DRE c. m∠ERP d. m∠ODE

Solution

a. Each diagonal is an angle bisector of the vertex angles.

Thus, m∠REP = 52.

b. �DER is isosceles with base −−

ER . So, m∠DER = m∠DRE = 52.

c. Because � ⎯⎯ � RE bisects ∠PRD, m∠ERP = 52.

d. −−

PD ⊥ −−

ER , so ∠EOD is a right angle.

By the Triangle-Sum Theorem, m∠ODE = 38.

Completing the Hierarchy of Quadrilaterals

Notice in Example 2 that there are two pairs of congruent alternate interior angles: ∠DER and ∠PRE, and ∠ERD and ∠REP. From the fi rst pair,

____ ED ‖

___ PR , and from the second pair,

___ PE ‖

____ RD . This allows

us to prove the fi nal connection in our hierarchy of quadrilaterals. That is, if a quadrilateral is a rhombus, then it is a parallelogram.

P

E D

O

R



Lesson 6-5

Example 3

Prove that if a quadrilateral is a rhombus, then it is a parallelogram.

Given RHOM is a rhombus.

Prove RHOM is a parallelogram.

Proof Conclusions Justifi cations

1. RHOM is a rhombus. 1. Given2.

___ RO and

___ HM are symmetry

diagonals of RHOM.2. ?

3. ∠1 � ∠2 and ∠4 � ∠3. 3. Symmetric Figures Theorem

4. ∠2 � ∠4 4. ?

5. ∠1 � ∠4 and ∠2 � ∠3. 5. ?

6. ___

RH ‖ ___

OM and ___

MR ‖ ___

OH . 6. ?

7. RHOM is a parallelogram. 7. ?

Questions

COVERING THE IDEAS

1. Which quadrilaterals in the hierarchy of quadrilaterals inherit characteristics from kites?

2. True or False Every kite is also a rhombus.

3. In kite BELO at the right, BE = BO and EL = OL. a. Name the ends of BELO. b. Name its symmetry line. c. Fill in the Blank r � � BL (∠BEL) = ?

4. Explain why, from their defi nitions, every square is a kite.

5. a. In the picture of an F-117A at the right, if m∠B = 20, what other angle measure do you know?

b. What is the symmetry line for the airplane?

6. Draw a nonconvex kite and mark the diagram to show how the Kite Diagonal Theorem applies to nonconvex kites.

APPLYING THE MATHEMATICS

7. Fill in the Blanks Complete the following proof that the opposite angles of a rhombus ABCD are equal in measure.

The refl ection image of ∠ADC over � �� AC is ? , so m∠ADC = ? because refl ections preserve ? . Similarly, the refl ection image of ∠DCB over ? is ? , so m∠DCB = ? .

GUIDED

R H

OM

21

3 4

B

E

O

L

D

C

B

A

DC

BA



Chapter 6

8. Suppose that P and Q are points on the perpendicular bisector of ___

XY . Prove that PYQX is a kite.

9. In the fi gure at the right, � �� AC bisects both ∠BCD and ∠BAD. a. Explain why D must be the refl ection image of B over � �� AC . b. Use Part a to prove that ABCD is a kite.

10. In kite KITE, ___

KT is the symmetry diagonal, T is on the x-axis, K = (–5, 0), and I = (0, 9).

a. What are the coordinates of point E? b. What are the possible coordinates of point T? c. For what coordinates of point T is the kite convex?

REVIEW

11. Name all types of quadrilaterals in the quadrilateral hierarchy that are isosceles trapezoids. (Lesson 6-4)

12. True or False Every square is a trapezoid. (Lesson 6-4)

13. In �O at the right, m∠ABC = 27 and m∠BAD = 63. Prove that m∠ABC = m∠ABD. (Lesson 6-3)

14. Suppose that � �� AD is the symmetry line of ∠BAC, where D lies in the interior of ∠BAC. Let � �� AE be the symmetry line of ∠DAB, where E lies in the interior of ∠DAB. (Lesson 6-1)

a. Draw a picture of this situation. b. How is m∠EAB related to m∠BAC?

15. Is there exactly one circle through any three points? Explain why or why not. (Lesson 5-6)

16. The fi gure at the right shows the hierarchy of quadrilaterals, drawn as a network. Is this network traversable? If not, explain why not. If so, describe a possible path that traverses the network. (Lesson 1-3)

EXPLORATION

17. The name “kite” is given to the quadrilateral studied in this lesson because it resembles a real-life kite. Why do you think that many kites in real life are shaped this way?

A

B

D

C

AD

OC

B

Q

KP

H

S

R

I

T


lesson 6-5 properties of kites

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