polygons. why a hexagon? polygon comes from greek. poly- means "many" -gon means...

Polygons

Why a hexagon?

http://en.wikipedia.org/wiki/File:Regular_Octagon_Inscribed_in_a_Circle.gif

Polygon comes from Greek. poly- means "many"

-gon means "angle"

Polygons 1. 2-dimensional shapes 2. made of straight lines3. Shape is "closed" (all the lines connect up).

Polygon – classified by sides and angles

Polygon (straight sides)

Not a Polygon (has a curve)

Not a Polygon (open, not closed)

Convex Concave

If there are any internal angles greater than 180° then it is concave.

Regular Irregular

Simple Polygon(this one's a Pentagon)

Complex Polygon(also a Pentagon)

complex polygon intersects itself

Regular - all angles are equal and all sides are equal

ConcaveConvex

Not regular polygons

Regular polygons

Complex Polygon (a "star polygon", in

this case, a pentagram)

Concave Octagon

Irregular Hexagon

http://www.mathsisfun.com/geometry/pentagram.html

Common polygons

Sides Namen N-gon3 Triangle4 Quadrilateral5 Pentagon6 Hexagon7 Heptagon8 Octagon

10 Decagon12 Dodecagon

Sides Name9 Nonagon, Enneagon

11 Undecagon, Hendecagon13 Tridecagon, Triskaidecagon

14 Tetradecagon, Tetrakaidecagon

15 Pentadecagon, Pentakaidecagon

16 Hexadecagon, Hexakaidecagon

17 Heptadecagon, Heptakaidecagon

18 Octadecagon, Octakaidecagon

19 Enneadecagon, Enneakaidecagon

20 Icosagon

30 Triacontagon

40 Tetracontagon

50 Pentacontagon

60 Hexacontagon

70 Heptacontagon

80 Octacontagon

90 Enneacontagon

100 Hectogon, Hecatontagon

1,000 Chiliagon

10,000 Myriagon

Sides Name9 Nonagon, Enneagon

11 Undecagon, Hendecagon13 Tridecagon, Triskaidecagon

14 Tetradecagon, Tetrakaidecagon

15 Pentadecagon, Pentakaidecagon

16 Hexadecagon, Hexakaidecagon

17 Heptadecagon, Heptakaidecagon

18 Octadecagon, Octakaidecagon

19 Enneadecagon, Enneakaidecagon

20 Icosagon

30 Triacontagon

40 Tetracontagon

50 Pentacontagon

60 Hexacontagon

70 Heptacontagon

80 Octacontagon

90 Enneacontagon

100 Hectogon, Hecatontagon

1,000 Chiliagon

10,000 Myriagon

Each segment that forms a polygon is a side of the polygon.

The common endpoint of two sides is a vertex of the polygon.

A segment that connects any two nonconsecutive vertices is a diagonal.

The sum of the angles of a quadrilateral is 360 degrees

Quadrilateral

A four-sided polygon with two pairs of parallel sides. The sum of the angles of a parallelogram is 360 degrees

Parallelogram

A four-sided polygon having all right angles. The sum of the angles of a rectangle is 360 degrees.

Rectangle

A four-sided polygon having equal-length sides meeting at right angles. The sum of the angles of a square is 360 degrees

Square

Rhombus

A four-sided polygon having all four sides of equal length. The sum of the angles of a rhombus is 360 degrees.

TrapezoidA four-sided polygon having exactly one pair of parallel sides. The two sides that are parallel are called the bases of the trapezoid. The sum of the angles of a trapezoid is 360 degrees.

Triangle

A three-sided polygon. The sum of the angles of a triangle is 180 degrees.

Equilateral Triangle or Equiangular TriangleA triangle having all three sides of equal length. The angles of an equilateral triangle all measure 60 degrees.

Isosceles TriangleA triangle having two sides of equal length.

Scalene TriangleA triangle having three sides of different lengths.

Acute Triangle

A triangle having three acute angles.

Obtuse Triangle

A triangle having an obtuse angle. One of the angles of the triangle measures more than 90 degrees.

A triangle having a right angle. One of the angles of the triangle measures 90 degrees. The side opposite the right angle is called the hypotenuse. The two sides that form the right angle are called the legs.

Right Triangle

A right triangle has the special property that the sum of the squares of the lengths of the legs equals the square of the length of the hypotenuse.

Pythagorean Theorem

http://www.mathplayground.com/matching_shapes.html

Quadrilateral

Identifying Polygons

If it is a polygon, name it by the number of sides.

polygon, hexagon



polygon, heptagon



not a polygon


If it is a polygon, name it by the number of its sides.

not a polygon



polygon, nonagon



not a polygon

A polygon is concave if any part of a diagonal contains points in the exterior of the polygon. If no diagonal contains points in the exterior, then the polygon is convex. A regular polygon is always convex.

Classifying Polygons

Regular or irregular? Concave or convex?

irregular, convex



irregular, concave



regular, convex



irregular, concave

To find the sum of the interior angle measures of a convex polygon, draw all possible diagonals from one vertex of the polygon. This creates a set of triangles. The sum of the angle measures of all the triangles equals the sum of the angle measures of the polygon.

By the Triangle Sum Theorem, the sum of the interior angle measures of a triangle is 180°.

Example 3A: Finding Interior Angle Measures and Sums in Polygons

Find the sum of the interior angle measures of a convex heptagon.

(n – 2)180°

(7 – 2)180°

900°

Polygon Sum Thm.

A heptagon has 7 sides, so substitute 7 for n.

Simplify.

Example 3B: Finding Interior Angle Measures and Sums in Polygons

Find the measure of each interior angle of a regular 16-gon.

Step 1 Find the sum of the interior angle measures.

Step 2 Find the measure of one interior angle.

(n – 2)180°

(16 – 2)180° = 2520°

Polygon Sum Thm.

Substitute 16 for n and simplify.

The int. s are , so divide by 16.

Example 3C: Finding Interior Angle Measures and Sums in Polygons

Find the measure of each interior angle of pentagon ABCDE.

(5 – 2)180° = 540° Polygon Sum Thm.

mA + mB + mC + mD + mE = 540°Polygon Sum Thm.

35c + 18c + 32c + 32c + 18c = 540 Substitute.

135c = 540 Combine like terms.

c = 4 Divide both sides by 135.

Example 3C Continued

mA = 35(4°) = 140°

mB = mE = 18(4°) = 72°

mC = mD = 32(4°) = 128°

Example 4

Find the sum of the interior angle measures of a convex 15-gon.

(n – 2)180°

(15 – 2)180°

2340°

Polygon Sum Thm.

A 15-gon has 15 sides, so substitute 15 for n.

Simplify.

Find the measure of each interior angle of a regular decagon.

Step 1 Find the sum of the interior angle measures.

Step 2 Find the measure of one interior angle.

Example 5

(n – 2)180°

(10 – 2)180° = 1440°

Polygon Sum Thm.

Substitute 10 for n and simplify.

The int. s are , so divide by 10.

In the polygons below, an exterior angle has been measured at each vertex. Notice that in each case, the sum of the exterior angle measures is 360°.

Example 6: Finding Interior Angle Measures and Sums in Polygons

Find the measure of each exterior angle of a regular 20-gon.

A 20-gon has 20 sides and 20 vertices.

sum of ext. s = 360°.

A regular 20-gon has 20 ext. s, so divide the sum by 20.

The measure of each exterior angle of a regular 20-gon is 18°.

Polygon Sum Thm.

measure of one ext. =

Example 7: Finding Interior Angle Measures and Sums in Polygons

Find the value of b in polygon FGHJKL.

15b° + 18b° + 33b° + 16b° + 10b° + 28b° = 360°

Polygon Ext. Sum Thm.

120b = 360 Combine like terms.

b = 3 Divide both sides by 120.

Find the measure of each exterior angle of a regular dodecagon.

Example 8

A dodecagon has 12 sides and 12 vertices.

sum of ext. s = 360°.

A regular dodecagon has 12 ext. s, so divide the sum by 12.

The measure of each exterior angle of a regular dodecagon is 30°.

Polygon Sum Thm.

measure of one ext.

Example 9

Find the value of r in polygon JKLM.

4r° + 7r° + 5r° + 8r° = 360° Polygon Ext. Sum Thm.

24r = 360 Combine like terms.

r = 15 Divide both sides by 24.

Example 10: Art Application

Ann is making paper stars for party decorations. What is the measure of 1?

1 is an exterior angle of a regular pentagon. By the Polygon Exterior Angle Sum Theorem, the sum of the exterior angles measures is 360°.

A regular pentagon has 5 ext. , so divide the sum by 5.

Find the value of each variable.

1. x 2. y 3. z2 184

Prove and apply properties of parallelograms.

Use properties of parallelograms to solve problems.

Objectives

A quadrilateral with two pairs of parallel sides is a parallelogram. To write the name of a parallelogram, you use the symbol .

If a quadrilateral is a parallelogram• its opposite sides are congruent• its opposite angles are congruent• its consecutive angles are supplementary

If a quadrilateral is a parallelogram• diagonals bisect each other• each diagonal splits the parallelogram

into two congruent triangles.

http://www.mathwarehouse.com/geometry/quadrilaterals/parallelograms/

http://www.ies.co.jp/math/products/geo1/applets/para/para.html

Example 1: Properties of Parallelograms

Def. of segs.

Substitute 74 for DE.

In CDEF, DE = 74 mm, DG = 31 mm, and mFCD = 42°. Find CF.

CF = DE

CF = 74 mm

opp. sides

Substitute 42 for mFCD.


Subtract 42 from both sides.

mEFC + mFCD = 180°

mEFC + 42 = 180

mEFC = 138°

In CDEF, DE = 74 mm, DG = 31 mm, and mFCD = 42°. Find mEFC.

cons. s supp.


Substitute 31 for DG.

Simplify.

In CDEF, DE = 74 mm, DG = 31 mm, and mFCD = 42°. Find DF.

DF = 2DG

DF = 2(31)

DF = 62

diags. bisect each other.

Example 4

In KLMN, LM = 28 in., LN = 26 in., and mLKN = 74°. Find KN.

Def. of segs.

Substitute 28 for DE.

LM = KN

LM = 28 in.

opp. sides

Example 5

Def. of angles.

In KLMN, LM = 28 in., LN = 26 in., and mLKN = 74°. Find mNML.

Def. of s.

Substitute 74° for mLKN.

NML LKN

mNML = mLKN

mNML = 74°

opp. s

Example 6

In KLMN, LM = 28 in., LN = 26 in., and mLKN = 74°. Find LO.

Substitute 26 for LN.

Simplify.

LN = 2LO

26 = 2LO

LO = 13 in.


Example 7: Using Properties of Parallelograms to Find Measures

WXYZ is a parallelogram. Find YZ.

Def. of segs.

Substitute the given values.

Subtract 6a from both sides and add 4 to both sides.

Divide both sides by 2.

YZ = XW

8a – 4 = 6a + 10

2a = 14

a = 7

YZ = 8a – 4 = 8(7) – 4 = 52

opp. s

Example 8: Using Properties of Parallelograms to Find Measures

WXYZ is a parallelogram. Find mZ .

Divide by 27.

Add 9 to both sides.

Combine like terms.


mZ + mW = 180°

(9b + 2) + (18b – 11) = 180

27b – 9 = 180

27b = 189

b = 7

mZ = (9b + 2)° = [9(7) + 2]° = 65°

cons. s supp.

Example 9

EFGH is a parallelogram.Find JG.

Substitute.

Simplify.

EJ = JG

3w = w + 8

2w = 8

w = 4 Divide both sides by 2.

JG = w + 8 = 4 + 8 = 12

Def. of segs.


Example 10

EFGH is a parallelogram.Find FH.

Substitute.

Simplify.

FJ = JH

4z – 9 = 2z

2z = 9

z = 4.5 Divide both sides by 2.

Def. of segs.

FH = (4z – 9) + (2z) = 4(4.5) – 9 + 2(4.5) = 18


When you are drawing a figure in the coordinate plane,

the name ABCD gives the order of the vertices.

Example 11

In PNWL, NW = 12, PM = 9, and mWLP = 144°. Find each measure.

1. PW 2. mPNW

18 144°

Example 12

QRST is a parallelogram. Find each measure.

2. TQ 3. mT

28 71°

Example 13

Three vertices of ABCD are A (2, –6), B (–1, 2), and C(5, 3).

Find the coordinates of vertex D.

(8, –5)

PracticeJustify each statement.

1.

2.

Evaluate each expression for x = 12 and y = 8.5.

3. 2x + 7

4. 16x – 9

5. (8y + 5)°

Reflex Prop. of

Conv. of Alt. Int. s Thm.

31

183

73°

PracticeSolve for x.

1. 16x – 3 = 12x + 13

2. 2x – 4 = 90

ABCD is a parallelogram. Find each measure.

3. CD 4. mC

4

47

14 104°

Prove and apply properties of rectangles, rhombuses, and squares.

Use properties of rectangles, rhombuses, and squares to solve problems.

Objectives

A second type of special quadrilateral is a rectangle. A rectangle is a quadrilateral with four right angles.

Since a rectangle is a parallelogram by Theorem 6-4-1, a rectangle “inherits” all the properties of parallelograms that you learned in Lesson 6-2.

Example 1: Craft Application

A woodworker constructs a rectangular picture frame so that JK = 50 cm and JL = 86 cm. Find HM.

Rect. diags.

Def. of segs.

Substitute and simplify.

KM = JL = 86

diags. bisect each other

Example 1a

Carpentry The rectangular gate has diagonal braces. Find HJ.

Def. of segs.

Rect. diags.

HJ = GK = 48

Example 1b

Carpentry The rectangular gate has diagonal braces. Find HK.

Def. of segs.

Rect. diags.

JL = LG

JG = 2JL = 2(30.8) = 61.6 Substitute and simplify.

Rect. diagonals bisect each other

A rhombus is another special quadrilateral. A rhombus is a quadrilateral with four congruent sides.

Like a rectangle, a rhombus is a parallelogram. So you can apply the properties of parallelograms to rhombuses.

Example 2A: Using Properties of Rhombuses to Find Measures

TVWX is a rhombus. Find TV.

Def. of rhombus

Substitute given values.

Subtract 3b from both sides and add 9 to both sides.


WV = XT

13b – 9 = 3b + 4

10b = 13

b = 1.3

Example 2A Continued

Def. of rhombus

Substitute 3b + 4 for XT.

Substitute 1.3 for b and simplify.

TV = XT

TV = 3b + 4

TV = 3(1.3) + 4 = 7.9

Rhombus diag.

Example 2B: Using Properties of Rhombuses to Find Measures

TVWX is a rhombus. Find mVTZ.

Substitute 14a + 20 for mVTZ.

Subtract 20 from both sides and divide both sides by 14.

mVZT = 90°

14a + 20 = 90°

a = 5

Example 2B Continued

Rhombus each diag. bisects opp. s

Substitute 5a – 5 for mVTZ.

Substitute 5 for a and simplify.

mVTZ = mZTX

mVTZ = (5a – 5)°

mVTZ = [5(5) – 5)]° = 20°

Example 2a

CDFG is a rhombus. Find CD.

Def. of rhombus

Substitute

Simplify

Substitute

Def. of rhombus

Substitute

CG = GF

5a = 3a + 17

a = 8.5

GF = 3a + 17 = 42.5

CD = GF

CD = 42.5

Example 2b

CDFG is a rhombus. Find the measure.

mGCH if mGCD = (b + 3)°and mCDF = (6b – 40)°

mGCD + mCDF = 180°

b + 3 + 6b – 40 = 180°

7b = 217°

b = 31°

Def. of rhombus

Substitute.

Simplify.


Example 2b Continued

mGCH + mHCD = mGCD

2mGCH = mGCDRhombus each diag. bisects opp. s

2mGCH = (b + 3)

2mGCH = (31 + 3)

mGCH = 17°

Substitute.

Substitute.

Simplify and divide both sides by 2.

A square is a quadrilateral with four right angles and four congruent sides. In the exercises, you will show that a square is a parallelogram, a rectangle, and a rhombus. So a square has the properties of all three.

Example 3: Verifying Properties of Squares

Show that the diagonals of square EFGH are congruent perpendicular bisectors of each other.

Example 3 Continued

Step 1 Show that EG and FH are congruent.

Since EG = FH,

Example 3 Continued

Step 2 Show that EG and FH are perpendicular.

Since ,

The diagonals are congruent perpendicular bisectors of each other.

Example 3 Continued

Step 3 Show that EG and FH are bisect each other.

Since EG and FH have the same midpoint, they bisect each other.

Example 4

The vertices of square STVW are S(–5, –4), T(0, 2), V(6, –3) , and W(1, –9) . Show that the diagonals of square STVW are congruent perpendicular bisectors of each other.

111

slope of SV =

slope of TW = –11

SV ^ TW

SV = TW = 122 so, SV @ TW .

Step 1 Show that SV and TW are congruent.

Example 4 Continued

Since SV = TW,

Step 2 Show that SV and TW are perpendicular.

Example 4 Continued

Since

The diagonals are congruent perpendicular bisectors of each other.

Step 3 Show that SV and TW bisect each other.

Since SV and TW have the same midpoint, they bisect each other.

Example 4 Continued

Example 5: Using Properties of Special Parallelograms in Proofs

Prove: AEFD is a parallelogram.

Given: ABCD is a rhombus.

E is the midpoint of , and F is the midpoint of .

Example 5 Continued

||

Example 6

Given: PQTS is a rhombus with diagonal

Prove:

Example 6 Continued

Statements Reasons

1. PQTS is a rhombus. 1. Given.

2. Rhombus → eachdiag. bisects opp. s

3. QPR SPR 3. Def. of bisector.

4. Def. of rhombus.

5. Reflex. Prop. of

6. SAS

7. CPCTC

2.

4.

5.

7.

6.

Example 7

A slab of concrete is poured with diagonal spacers. In rectangle CNRT, CN = 35 ft, and NT = 58 ft. Find each length.

1. TR 2. CE

35 ft 29 ft

Example 8

PQRS is a rhombus. Find each measure.

3. QP 4. mQRP

42 51°

Example 9

The vertices of square ABCD are A(1, 3), B(3, 2), C(4, 4), and D(2, 5). Show that its diagonals are congruent perpendicular bisectors of each other.

Example 10

ABE CDF

Given: ABCD is a rhombus. Prove:

Practice

1. Find AB for A (–3, 5) and B (1, 2).

2. Find the slope of JK for J(–4, 4) and K(3, –3).

ABCD is a parallelogram. Justify each statement.

3. ABC CDA

4. AEB CED

5

–1

Vert. s Thm.

opp. s

Prove that a given quadrilateral is a rectangle, rhombus, or square.

Objective

When you are given a parallelogram with certainproperties, you can use the theorems below to determine whether the parallelogram is a rectangle.

Example 1: Carpentry Application

A manufacture builds a mold for a desktop so that , , and mABC = 90°. Why must ABCD be a rectangle?

Both pairs of opposites sides of ABCD are congruent, so ABCD is a . Since mABC = 90°, one angle ABCD is a right angle. ABCD is a rectangle by Theorem 6-5-1.

Example 1a

A carpenter’s square can be used to test that an angle is a right angle. How could the contractor use a carpenter’s square to check that the frame is a rectangle?

Both pairs of opp. sides of WXYZ are , so WXYZ is a parallelogram. The contractor can use the carpenter’s square to see if one of WXYZ is a right . If one angle is a right , then by Theorem 6-5-1 the frame is a rectangle.

Below are some conditions you can use to determine whether a parallelogram is a rhombus.

To prove that a given quadrilateral is a square, it is sufficient to show that the figure is both a rectangle and a rhombus. You will explain why this is true in Exercise 43.

You can also prove that a given quadrilateral is arectangle, rhombus, or square by using the definitions of the special quadrilaterals.

In order to apply Theorems 6-5-1 through 6-5-5, the quadrilateral must be a parallelogram.

Example 2A: Applying Conditions for Special Parallelograms

Determine if the conclusion is valid. If not, tell what additional information is needed to make it valid.

Given:

Conclusion: EFGH is a rhombus.

The conclusion is not valid. By Theorem 6-5-3, if one pair of consecutive sides of a parallelogram are congruent, then the parallelogram is a rhombus. By Theorem 6-5-4, if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. To apply either theorem, you must first know that ABCD is a parallelogram.

Example 2b


Given: ABC is a right angle.

Conclusion: ABCD is a rectangle.

The conclusion is not valid. By Theorem 6-5-1, if one angle of a parallelogram is a right angle, then the parallelogram is a rectangle. To apply this theorem, you need to know that ABCD is a parallelogram .

Example 3A: Identifying Special Parallelograms in the Coordinate Plane

Use the diagonals to determine whether a parallelogram with the given vertices is a rectangle, rhombus, or square. Give all the names that apply.

P(–1, 4), Q(2, 6), R(4, 3), S(1, 1)


Step 1 Graph PQRS.

Step 2 Find PR and QS to determine is PQRS is a rectangle.


Since , the diagonals are congruent. PQRS is a rectangle.

Step 3 Determine if PQRS is a rhombus.

Step 4 Determine if PQRS is a square.

Since PQRS is a rectangle and a rhombus, it has four right angles and four congruent sides. So PQRS is a square by definition.


Since , PQRS is a rhombus.

Example 3B: Identifying Special Parallelograms in the Coordinate Plane

W(0, 1), X(4, 2), Y(3, –2), Z(–1, –3)

Step 1 Graph WXYZ.


Step 2 Find WY and XZ to determine is WXYZ is a rectangle.

Thus WXYZ is not a square.


Since , WXYZ is not a rectangle.

Step 3 Determine if WXYZ is a rhombus.


Since (–1)(1) = –1, , PQRS is a rhombus.

Example4


K(–5, –1), L(–2, 4), M(3, 1), N(0, –4)

Step 1 Graph KLMN.

Example 4 Continued

Example 4 Continued

Step 2 Find KM and LN to determine is KLMN is a rectangle.

Since , KMLN is a rectangle.

Step 3 Determine if KLMN is a rhombus.

Since the product of the slopes is –1, the two lines are perpendicular. KLMN is a rhombus.

Example 4 Continued

Step 4 Determine if PQRS is a square.

Since PQRS is a rectangle and a rhombus, it has four right angles and four congruent sides. So PQRS is a square by definition.

Example 4 Continued

Example 5


P(–4, 6) , Q(2, 5) , R(3, –1) , S(–3, 0)

Example 5 Continued

Step 1 Graph PQRS.

Step 2 Find PR and QS to determine is PQRS is a rectangle.

Example 5 Continued

Since , PQRS is not a rectangle. Thus PQRS is not a square.

Step 3 Determine if KLMN is a rhombus.

Example 5 Continued

Since (–1)(1) = –1, are perpendicular and congruent. KLMN is a rhombus.

Example 6

Given that AB = BC = CD = DA, what additional information is needed to conclude that ABCD is a square?

Example 7


Given: PQRS and PQNM are parallelograms.

Conclusion: MNRS is a rhombus.

valid

Example 8

Use the diagonals to determine whether a parallelogram with vertices A(2, 7), B(7, 9), C(5, 4), and D(0, 2) is a rectangle, rhombus, or square. Give all the names that apply.

AC ≠ BD, so ABCD is not a rect. or a square. The slope of AC = –1, and the slope of BD= 1, so AC BD. ABCD is a rhombus.

PracticeSolve for x.

1. x2 + 38 = 3x2 – 12

2. 137 + x = 180

3.

4. Find FE.

5 or –5

43

156

Use properties of kites to solve problems.

Use properties of trapezoids to solve problems.

Objectives

A kite is a quadrilateral with exactly two pairs of congruent consecutive sides.

Example 1: Problem-Solving Application

Lucy is framing a kite with wooden dowels. She uses two dowels that measure 18 cm, one dowel that measures 30 cm, and two dowels that measure 27 cm. To complete the kite, she needs a dowel to place along .

She has a dowel that is 36 cm long.

About how much wood will she have left after cutting the last dowel?

Example 1 Continued

The answer will be the amount of wood Lucy has left after cutting the dowel.

The diagonals of a kite are perpendicular, so the four triangles are right triangles. Let N represent the intersection of the diagonals. Use the Pythagorean Theorem and the properties of kites to find , and . Add these lengths to find the length of .

Solve

N bisects JM.

Pythagorean Thm.

Pythagorean Thm.

Example 1 Continued

Lucy needs to cut the dowel to be 32.4 cm long. The amount of wood that will remain after the cut is,

36 – 32.4 3.6 cm

Lucy will have 3.6 cm of wood left over after the cut.

Example 1 Continued

Kite cons. sides

Example 2A: Using Properties of Kites

In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mBCD.

∆BCD is isos. 2 sides isos. ∆

isos. ∆ base s

Def. of s

Polygon Sum Thm.

CBF CDF

mCBF = mCDF

mBCD + mCBF + mCDF = 180°


Substitute mCDF for mCBF.

Substitute 52 for mCBF.



mBCD + 52° + 52° = 180°

mBCD = 76°


Kite one pair opp. s

Example 2B: Using Properties of Kites

Def. of s

Polygon Sum Thm.

In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mABC.

ADC ABC

mADC = mABC

mABC + mBCD + mADC + mDAB = 360°

mABC + mBCD + mABC + mDAB = 360°

Substitute mABC for mADC.


Substitute.

Simplify.

mABC + mBCD + mABC + mDAB = 360°

mABC + 76° + mABC + 54° = 360°

2mABC = 230°

mABC = 115° Solve.


Example 2C: Using Properties of Kites

Def. of s

Add. Post.

Substitute.

Solve.

In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mFDA.

CDA ABC

mCDA = mABC

mCDF + mFDA = mABC

52° + mFDA = 115°

mFDA = 63°

Example 2a

In kite PQRS, mPQR = 78°, and mTRS = 59°. Find mQRT.

Kite cons. sides

∆PQR is isos. 2 sides isos. ∆

isos. ∆ base s

Def. of s

RPQ PRQ

mQPT = mQRT

Example 2a Continued

Polygon Sum Thm.

Substitute 78 for mPQR.

mPQR + mQRP + mQPR = 180°

78° + mQRT + mQPT = 180°

Substitute. 78° + mQRT + mQRT = 180°

78° + 2mQRT = 180°

2mQRT = 102°

mQRT = 51°

Substitute.


Divide by 2.

Example 2b

In kite PQRS, mPQR = 78°, and mTRS = 59°. Find mQPS.


Add. Post.

Substitute.

Substitute.

QPS QRS

mQPS = mQRT + mTRS

mQPS = mQRT + 59°

mQPS = 51° + 59°

mQPS = 110°

Example 2c

Polygon Sum Thm.

Def. of s

Substitute.

Substitute.

Simplify.

In kite PQRS, mPQR = 78°, and mTRS = 59°. Find each mPSR.

mSPT + mTRS + mRSP = 180°

mSPT = mTRS

mTRS + mTRS + mRSP = 180°

59° + 59° + mRSP = 180°

mRSP = 62°

A trapezoid is a quadrilateral with exactly one pair of parallel sides. Each of the parallel sides is called a base. The nonparallel sides are called legs. Base angles of a trapezoid are two consecutive angles whose common side is a base.

If the legs of a trapezoid are congruent, the trapezoid is an isosceles trapezoid. The following theorems state the properties of an isosceles trapezoid.

Theorem 6-6-5 is a biconditional statement.So it is true both “forward” and “backward.”

Isos. trap. s base

Example 3A: Using Properties of Isosceles Trapezoids

Find mA.

Same-Side Int. s Thm.

Substitute 100 for mC.


Def. of s

Substitute 80 for mB

mC + mB = 180°

100 + mB = 180

mB = 80°

A B

mA = mB

mA = 80°

Example 3B: Using Properties of Isosceles Trapezoids

KB = 21.9m and MF = 32.7. Find FB.

Isos. trap. s base

Def. of segs.

Substitute 32.7 for FM.

Seg. Add. Post.

Substitute 21.9 for KB and 32.7 for KJ.

Subtract 21.9 from both sides.

KJ = FM

KJ = 32.7

KB + BJ = KJ

21.9 + BJ = 32.7

BJ = 10.8


Same line.

Isos. trap. s base

Isos. trap. legs

SAS

CPCTC

Vert. s

KFJ MJF

BKF BMJ

FBK JBM

∆FKJ ∆JMF

Isos. trap. legs

AAS

CPCTC

Def. of segs.

Substitute 10.8 for JB.


∆FBK ∆JBM

FB = JB

FB = 10.8

Isos. trap. s base

Same-Side Int. s Thm.

Def. of s

Substitute 49 for mE.

mF + mE = 180°

E H

mE = mH

mF = 131°

mF + 49° = 180°

Simplify.

Check It Out! Example 3a

Find mF.

Check It Out! Example 3b

JN = 10.6, and NL = 14.8. Find KM.

Def. of segs.

Segment Add Postulate

Substitute.

Substitute and simplify.

Isos. trap. s base

KM = JL

JL = JN + NL

KM = JN + NL

KM = 10.6 + 14.8 = 25.4

Example 4A: Applying Conditions for Isosceles Trapezoids

Find the value of a so that PQRS is isosceles.

a = 9 or a = –9

Trap. with pair base s isosc. trap.

Def. of s

Substitute 2a2 – 54 for mS and a2 + 27 for mP.

Subtract a2 from both sides and add 54 to both sides.

Find the square root of both sides.

S P

mS = mP

2a2 – 54 = a2 + 27

a2 = 81

Example 4B: Applying Conditions for Isosceles Trapezoids

AD = 12x – 11, and BC = 9x – 2. Find the value of x so that ABCD is isosceles.

Diags. isosc. trap.

Def. of segs.

Substitute 12x – 11 for AD and 9x – 2 for BC.

Subtract 9x from both sides and add 11 to both sides.


AD = BC

12x – 11 = 9x – 2

3x = 9

x = 3

Example 5

Find the value of x so that PQST is isosceles.

Subtract 2x2 and add 13 to both sides.

x = 4 or x = –4 Divide by 2 and simplify.

Trap. with pair base s isosc. trap.Q S

Def. of s

Substitute 2x2 + 19 for mQ and 4x2 – 13 for mS.

mQ = mS

2x2 + 19 = 4x2 – 13

32 = 2x2

The midsegment of a trapezoid is the segment whose endpoints are the midpoints of the legs. In Lesson 5-1, you studied the Triangle Midsegment Theorem. The Trapezoid Midsegment Theorem is similar to it.

Example 6: Finding Lengths Using Midsegments

Find EF.

Trap. Midsegment Thm.


Solve.EF = 10.75

Example 7

Find EH.

Trap. Midsegment Thm.


Simplify.

Multiply both sides by 2.33 = 25 + EH

Subtract 25 from both sides.13 = EH

116.5 = (25 + EH)2

Practice

1. Erin is making a kite based on the pattern below. About how much binding does Erin need to cover the edges of the kite?

In kite HJKL, mKLP = 72°,and mHJP = 49.5°. Find eachmeasure.

2. mLHJ 3. mPKL

about 191.2 in.

81° 18°

Practice

Use the diagram for Items 4 and 5.

4. mWZY = 61°. Find mWXY.

5. XV = 4.6, and WY = 14.2. Find VZ.

6. Find LP.

119°

9.6

18

polygons. why a hexagon? polygon comes from greek. poly- means "many" -gon means...

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