Download Polygons - Denise Kapler

Polygons
Why a hexagon?
Polygon comes from Greek.
poly- means "many"
-gon means "angle"
Polygons
1. 2-dimensional shapes
2. made of straight lines
3. Shape is "closed" (all the lines connect up).
Polygon – classified by sides and angles
Polygon
(straight sides)
Convex
Not a Polygon
(has a curve)
Not a Polygon
(open, not closed)
Concave
If there are any internal angles
greater than 180° then it is concave.
Regular
Irregular
Regular - all angles
are equal and all
sides are equal
Simple Polygon
(this one's a Pentagon)
Complex Polygon
(also a Pentagon)
complex polygon intersects itself
Convex
Concave
Regular polygons
Not regular polygons
Concave Octagon
Complex Polygon
(a "star polygon", in
this case, a pentagram)
Irregular
Hexagon
Common polygons
4
5
6
7
8
10
12
Sides
Quadrilateral
Pentagon
Hexagon
Heptagon
Octagon
Decagon
Dodecagon
Name
9
11
Nonagon, Enneagon
Undecagon, Hendecagon
13
Tridecagon, Triskaidecagon
14
Tetradecagon, Tetrakaidecagon
15
Pentadecagon, Pentakaidecagon
16
Hexadecagon, Hexakaidecagon
17
Heptadecagon, Heptakaidecagon
18
Octadecagon, Octakaidecagon
19
Enneadecagon, Enneakaidecagon
20
Icosagon
Sides
Name
9
11
Nonagon, Enneagon
Undecagon, Hendecagon
13
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.
Quadrilateral
The sum of the angles of a quadrilateral is 360 degrees
Parallelogram
A four-sided polygon with two pairs of parallel sides.
The sum of the angles of a parallelogram is 360 degrees
Rectangle
A four-sided polygon having all right angles.
The sum of the angles of a rectangle is 360 degrees.
Square
A four-sided polygon having equal-length sides
meeting at right angles.
The sum of the angles of a square is 360 degrees
Rhombus
A four-sided polygon having all four sides of equal length.
The sum of the angles of a rhombus is 360 degrees.
Trapezoid
A 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 Triangle
A triangle having all three sides of equal length.
The angles of an equilateral triangle all measure 60 degrees.
Isosceles Triangle
A triangle having two sides of equal length.
Scalene Triangle
A 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.
Right Triangle
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.
Pythagorean Theorem
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.
http://www.mathplayground.com/matching_shapes.html
Quadrilateral
Identifying Polygons
If it is a polygon, name it by the number of sides.
polygon, hexagon
Identifying Polygons
If it is a polygon, name it by the number of sides.
polygon, heptagon
Identifying Polygons
If it is a polygon, name it by the number of sides.
not a polygon
Identifying Polygons
If it is a polygon, name it by the number of its sides.
not a polygon
Identifying Polygons
If it is a polygon, name it by the number of its sides.
polygon, nonagon
Identifying Polygons
If it is a polygon, name it by the number of its sides.
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
Classifying Polygons
Regular or irregular? Concave or convex?
irregular, concave
Classifying Polygons
Regular or irregular? Concave or convex?
regular, convex
Classifying Polygons
Regular or irregular? Concave or convex?
regular, convex
Classifying Polygons
Regular or irregular? Concave or 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°
Polygon  Sum Thm.
(7 – 2)180°
A heptagon has 7 sides, so substitute 7 for
n.
900°
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.
(n – 2)180°
Polygon  Sum Thm.
Substitute 16 for n and simplify.
(16 – 2)180° = 2520°
Step 2 Find the measure of one interior angle.
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.
Polygon  Sum Thm.
(5 – 2)180° = 540°
mA + mB + mC + mD + mE = 540°
35c + 18c + 32c + 32c + 18c = 540
135c = 540
c=4
Combine like terms.
Divide both sides by 135.
Polygon  Sum Thm
Substitute.
Example 3C Continued
mA = 35(4°) = 140°
mB = mE = 18(4°) = 72°
mC = mD = 32(4°) = 128°
Example 4
Find the sum of the interior angle measures of a convex 15-gon.
(n – 2)180°
Polygon  Sum Thm.
(15 – 2)180°
A 15-gon has 15 sides, so substitute 15 for
n.
2340°
Simplify.
Example 5
Find the measure of each interior angle of a regular decagon.
Step 1 Find the sum of the interior angle measures.
(n – 2)180°
Polygon  Sum Thm.
Substitute 10 for n and simplify.
(10 – 2)180° = 1440°
Step 2 Find the measure of one interior angle.
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°.
measure of one ext.  =
Polygon  Sum Thm.
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°.
Example 7: Finding Interior Angle Measures and Sums in Polygons
Find the value of b in polygon FGHJKL.
Polygon Ext.  Sum Thm.
15b° + 18b° + 33b° + 16b° + 10b° + 28b° = 360°
120b = 360
b=3
Combine like terms.
Divide both sides by 120.
Example 8
Find the measure of each exterior angle of a regular dodecagon.
A dodecagon has 12 sides and 12 vertices.
sum of ext. s = 360°.
measure of one ext.
Polygon  Sum Thm.
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°.
Example 9
Find the value of r in polygon JKLM.
4r° + 7r° + 5r° + 8r° = 360°
24r = 360
r = 15
Combine like terms.
Divide both sides by 24.
Polygon Ext.  Sum Thm.
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
2
3. z
4
18
Objectives
Prove and apply properties of parallelograms.
Use properties of parallelograms to solve
problems.
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.ies.co.jp/math/products/geo1/applets/para/para.html
http://www.mathwarehouse.com/geometry/quadrilaterals/parallelograms/
Example 1: Properties of Parallelograms
In
CDEF, DE = 74 mm,
DG = 31 mm, and mFCD = 42°. Find CF.
 opp. sides 
CF = DE
Def. of  segs.
CF = 74 mm
Substitute 74 for DE.
Example 2: Properties of Parallelograms
In
CDEF, DE = 74 mm,
DG = 31 mm, and mFCD = 42°. Find mEFC.
mEFC + mFCD = 180°
mEFC + 42 = 180
mEFC = 138°
 cons. s supp.
Substitute 42 for mFCD.
Subtract 42 from both sides.
Example 3: Properties of Parallelograms
In
CDEF, DE = 74 mm,
DG = 31 mm, and mFCD = 42°. Find DF.
DF = 2DG
 diags. bisect each other.
DF = 2(31)
Substitute 31 for DG.
DF = 62
Simplify.
Example 4
In
KLMN, LM = 28 in.,
LN = 26 in., and mLKN = 74°. Find KN.
 opp. sides 
LM = KN
Def. of  segs.
LM = 28 in.
Substitute 28 for DE.
Example 5
In
KLMN, LM = 28 in.,
LN = 26 in., and mLKN = 74°. Find mNML.
NML  LKN
 opp. s 
mNML = mLKN
Def. of  s.
mNML = 74°
Substitute 74° for mLKN.
Def. of  angles.
Example 6
In
KLMN, LM = 28 in.,
LN = 26 in., and mLKN = 74°. Find LO.
LN = 2LO
 diags. bisect each other.
26 = 2LO
Substitute 26 for LN.
LO = 13 in.
Simplify.
Example 7: Using Properties of Parallelograms to Find Measures
WXYZ is a parallelogram. Find YZ.
 opp. s 
Def. of  segs.
YZ = XW
8a – 4 = 6a + 10
2a = 14
a=7
Substitute the given values.
Subtract 6a from both sides and add 4 to both sides.
Divide both sides by 2.
YZ = 8a – 4 = 8(7) – 4 = 52
Example 8: Using Properties of Parallelograms to Find Measures
WXYZ is a parallelogram. Find mZ .
 cons. s supp.
mZ + mW = 180°
(9b + 2) + (18b – 11) = 180
Substitute the given values.
27b – 9 = 180
27b = 189
b=7
mZ = (9b + 2)° = [9(7) + 2]° = 65°
Combine like terms.
Add 9 to both sides.
Divide by 27.
Example 9
EFGH is a parallelogram.
Find JG.
 diags. bisect each other.
EJ = JG
3w = w + 8
2w = 8
w=4
JG = w + 8 = 4 + 8 = 12
Def. of  segs.
Substitute.
Simplify.
Divide both sides by 2.
Example 10
EFGH is a parallelogram.
Find FH.
 diags. bisect each other.
FJ = JH
4z – 9 = 2z
2z = 9
z = 4.5
Def. of  segs.
Substitute.
Simplify.
Divide both sides by 2.
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
mWLP = 144°. Find each measure.
1. PW
18
2. mPNW
144°
Example 12
QRST is a parallelogram. Find each measure.
2. TQ
28
3. mT
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)
Practice
Justify each statement.
1.
Reflex Prop. of 
2.
Conv. of Alt. Int. s Thm.
Evaluate each expression for x = 12 and y = 8.5.
3. 2x + 7
31
4. 16x – 9
183
5. (8y + 5)°
73°
Practice
Solve for x.
1. 16x – 3 = 12x + 13
2. 2x – 4 = 90
4
47
ABCD is a parallelogram. Find each measure.
3. CD
4. mC
14
104°
Objectives
Prove and apply properties of rectangles,
rhombuses, and squares.
Use properties of rectangles, rhombuses, and
squares to solve problems.
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. 
KM = JL = 86
Def. of  segs.
 diags. bisect each other
Substitute and simplify.
Example 1a
Carpentry The rectangular gate has diagonal braces.
Find HJ.
Rect.  diags. 
HJ = GK = 48
Def. of  segs.
Example 1b
Carpentry The rectangular gate has diagonal braces.
Find HK.
Rect.  diags. 
Rect.  diagonals bisect each other
JL = LG
JG = 2JL = 2(30.8) = 61.6
Def. of  segs.
Substitute and simplify.
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.
WV = XT
13b – 9 = 3b + 4
10b = 13
b = 1.3
Def. of rhombus
Substitute given values.
Subtract 3b from both sides and add 9 to both sides.
Divide both sides by 10.
Example 2A Continued
TV = XT
Def. of rhombus
TV = 3b + 4
Substitute 3b + 4 for XT.
TV = 3(1.3) + 4 = 7.9
Substitute 1.3 for b and simplify.
Example 2B: Using Properties of Rhombuses to Find Measures
TVWX is a rhombus. Find mVTZ.
mVZT = 90°
14a + 20 = 90°
a=5
Rhombus  diag. 
Substitute 14a + 20 for mVTZ.
Subtract 20 from both sides and divide both sides by
14.
Example 2B Continued
mVTZ = mZTX
Rhombus  each diag. bisects opp. s
mVTZ = (5a – 5)°
Substitute 5a – 5 for mVTZ.
mVTZ = [5(5) – 5)]°
= 20°
Substitute 5 for a and simplify.
Example 2a
CDFG is a rhombus. Find CD.
CG = GF
5a = 3a + 17
a = 8.5
GF = 3a + 17 = 42.5
Def. of rhombus
Substitute
Simplify
Substitute
CD = GF
Def. of rhombus
CD = 42.5
Substitute
Example 2b
CDFG is a rhombus.
Find the measure.
mGCH if mGCD = (b + 3)°
and mCDF = (6b – 40)°
Def. of rhombus
mGCD + mCDF = 180°
b + 3 + 6b – 40 = 180°
Substitute.
7b = 217°
b = 31°
Simplify.
Divide both sides by 7.
Example 2b Continued
mGCH + mHCD = mGCD
2mGCH = mGCD
Rhombus  each diag. bisects opp.
s
2mGCH = (b + 3)
Substitute.
2mGCH = (31 + 3)
Substitute.
mGCH = 17°
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
,
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.
The diagonals are congruent perpendicular bisectors of 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.
SV = TW =
122 so, SV  TW .
1
slope of SV =
11
slope of TW = –11
SV  TW
Example 4 Continued
Step 1 Show that SV and TW are congruent.
Since SV = TW,
Example 4 Continued
Step 2 Show that SV and TW are perpendicular.
Since
Example 4 Continued
Step 3 Show that SV and TW bisect each other.
Since SV and TW have the same midpoint, they bisect each other.
The diagonals are congruent perpendicular bisectors of each other.
Example 5: Using Properties of Special Parallelograms in Proofs
Given: ABCD is a rhombus.
E is the midpoint of
F is the midpoint of
, and
.
Prove: AEFD is a parallelogram.
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.
2. Rhombus → each
diag. bisects opp. s
3. QPR  SPR
3. Def. of  bisector.
4.
4. Def. of rhombus.
5.
5. Reflex. Prop. of 
6.
6. SAS
7.
7. CPCTC
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
42
4. mQRP
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
Given: ABCD is a rhombus.
Prove:
ABE  CDF

Practice
1. Find AB for A (–3, 5) and B (1, 2).
5
2. Find the slope of JK for J(–4, 4) and K(3, –3).
–1
ABCD is a parallelogram. Justify each statement.
3. ABC  CDA
4. AEB  CED
 opp. s 
Vert. s Thm.
Objective
Prove that a given quadrilateral is a rectangle,
rhombus, or square.
When you are given a parallelogram with certain
properties, 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 mABC = 90°. Why must
ABCD be a rectangle?
Both pairs of opposites sides of ABCD are congruent, so ABCD is a . Since mABC =
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.
In order to apply Theorems 6-5-1 through 6-5-5,
the quadrilateral must be a parallelogram.
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 a
rectangle, rhombus, or square by using the definitions of the special quadrilaterals.
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
Determine if the conclusion is valid. If not, tell what additional information is
needed to make it valid.
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)
Example 3A Continued
Step 1 Graph
PQRS.
Example 3A Continued
Step 2 Find PR and QS to determine is PQRS is a rectangle.
Since
, the diagonals are congruent. PQRS is a rectangle.
Example 3A Continued
Step 3 Determine if PQRS is a rhombus.
Since
, 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.
Example 3B: 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.
W(0, 1), X(4, 2), Y(3, –2), Z(–1, –3)
Step 1 Graph
WXYZ.
Example 3B Continued
Step 2 Find WY and XZ to determine is WXYZ is a rectangle.
Since
, WXYZ is not a rectangle.
Thus WXYZ is not a square.
Example 3B Continued
Step 3 Determine if WXYZ is a rhombus.
Since (–1)(1) = –1,
, PQRS is a rhombus.
Example4
Use the diagonals to determine whether a parallelogram with the given vertices is a
rectangle, rhombus, or square. Give all the names that apply.
K(–5, –1), L(–2, 4), M(3, 1), N(0, –4)
Example 4 Continued
Step 1 Graph
KLMN.
Example 4 Continued
Step 2 Find KM and LN to determine is KLMN is a rectangle.
Since
, KMLN is a rectangle.
Example 4 Continued
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 5
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(–4, 6) , Q(2, 5) , R(3, –1) , S(–3, 0)
Example 5 Continued
Step 1 Graph
PQRS.
Example 5 Continued
Step 2 Find PR and QS to determine is PQRS is a rectangle.
Since
, PQRS is not a rectangle. Thus PQRS is not a square.
Example 5 Continued
Step 3 Determine if KLMN is a rhombus.
Since (–1)(1) = –1,
rhombus.
are perpendicular and congruent. KLMN is a
Example 6
Given that AB = BC = CD = DA, what additional information is needed to conclude that
ABCD is a square?
Example 7
Determine if the conclusion is valid. If not, tell what additional information is needed
to make it valid.
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.
Practice
Solve for x.
1. x2 + 38 = 3x2 – 12
5 or –5
2. 137 + x = 180
43
3.
156
4. Find FE.
Objectives
Use properties of kites to solve problems.
Use properties of trapezoids to solve problems.
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
.
Example 1 Continued
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 2A: Using Properties of Kites
In kite ABCD, mDAB = 54°, and
mBCD.
mCDF = 52°. Find
Kite  cons. sides 
∆BCD is isos.
2  sides isos. ∆
CBF  CDF
isos. ∆ base s 
mCBF = mCDF
mBCD + mCBF + mCDF = 180°
Def. of   s
Polygon  Sum Thm.
Example 2A Continued
mBCD + mCBF + mCDF = 180°
mBCD + mCBF + mCDF = 180°
mBCD + 52° + 52° = 180°
mBCD = 76°
Substitute mCDF for mCBF.
Substitute 52 for mCBF.
Subtract 104 from both
sides.
Example 2B: Using Properties of Kites
In kite ABCD, mDAB = 54°, and
mABC.
ADC  ABC
mADC = mABC
mCDF = 52°. Find
Kite  one pair opp. s 
Def. of  s
Polygon  Sum Thm.
mABC + mBCD + mADC + mDAB = 360°
Substitute mABC for mADC.
mABC + mBCD + mABC + mDAB = 360°
Example 2B Continued
mABC + mBCD + mABC + mDAB = 360°
mABC + 76° + mABC + 54° = 360°
Substitute.
2mABC = 230°
mABC = 115°
Simplify.
Solve.
Example 2C: Using Properties of Kites
In kite ABCD, mDAB = 54°, and
mFDA.
CDA  ABC
mCDA = mABC
mCDF + mFDA = mABC
52° + mFDA = 115°
mFDA = 63°
mCDF = 52°. Find
Kite  one pair opp. s 
Def. of  s
 Add. Post.
Substitute.
Solve.
Example 2a
In kite PQRS, mPQR = 78°, and mTRS = 59°. Find
mQRT.
Kite  cons. sides 
∆PQR is isos.
RPQ  PRQ
mQPT = mQRT
2  sides  isos. ∆
isos. ∆  base s 
Def. of  s
Example 2a Continued
Polygon  Sum Thm.
mPQR + mQRP + mQPR = 180°
78° + mQRT + mQPT = 180°
78° + mQRT + mQRT = 180°
78° + 2mQRT = 180°
2mQRT = 102°
mQRT = 51°
Substitute 78 for mPQR.
Substitute.
Substitute.
Subtract 78 from both sides.
Divide by 2.
Example 2b
In kite PQRS, mPQR = 78°, and mTRS = 59°. Find
mQPS.
QPS  QRS
mQPS = mQRT + mTRS
Kite  one pair opp. s 
 Add. Post.
mQPS = mQRT + 59°
Substitute.
mQPS = 51° + 59°
Substitute.
mQPS = 110°
Example 2c
In kite PQRS, mPQR = 78°, and mTRS = 59°. Find each
mPSR.
mSPT + mTRS + mRSP = 180°
Polygon  Sum Thm.
mSPT = mTRS
Def. of  s
Substitute.
mTRS + mTRS + mRSP = 180°
59° + 59° + mRSP = 180°
mRSP = 62°
Substitute.
Simplify.
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.”
Example 3A: Using Properties of Isosceles Trapezoids
Find mA.
mC + mB = 180°
Same-Side Int. s Thm.
100 + mB = 180
Substitute 100 for mC.
mB = 80°
A  B
mA = mB
mA = 80°
Subtract 100 from both sides.
Isos. trap. s base 
Def. of  s
Substitute 80 for mB
Example 3B: Using Properties of Isosceles Trapezoids
KB = 21.9m and MF = 32.7. Find FB.
Isos.  trap. s base 
KJ = FM
Def. of  segs.
KJ = 32.7
Substitute 32.7 for FM.
KB + BJ = KJ
21.9 + BJ = 32.7
BJ = 10.8
Seg. Add. Post.
Substitute 21.9 for KB and 32.7 for KJ.
Subtract 21.9 from both sides.
Example 3B Continued
Same line.
KFJ  MJF
Isos. trap.  s base 
Isos. trap.  legs 
∆FKJ  ∆JMF
SAS
BKF  BMJ
CPCTC
FBK  JBM
Vert. s 
Example 3B Continued
Isos. trap.  legs 
∆FBK  ∆JBM
AAS
CPCTC
FB = JB
Def. of  segs.
FB = 10.8
Substitute 10.8 for JB.
Check It Out! Example 3a
Find mF.
mF + mE = 180°
E  H
mE = mH
mF + 49° = 180°
mF = 131°
Same-Side Int. s Thm.
Isos. trap. s base 
Def. of  s
Substitute 49 for mE.
Simplify.
Check It Out! Example 3b
JN = 10.6, and NL = 14.8. Find KM.
Isos. trap. s base 
KM = JL
JL = JN + NL
Def. of  segs.
Segment Add Postulate
KM = JN + NL
Substitute.
KM = 10.6 + 14.8 = 25.4
Substitute and simplify.
Example 4A: Applying Conditions for Isosceles Trapezoids
Find the value of a so that PQRS is isosceles.
S  P
mS = mP
2a2
– 54 =
a2
+ 27
a2
= 81
a = 9 or a = –9
Trap. with pair base s   isosc.
trap.
Def. of  s
Substitute 2a2 – 54 for mS and a2 + 27 for mP.
Subtract a2 from both sides and add 54 to both sides.
Find the square root of both sides.
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.
AD = BC
12x – 11 = 9x – 2
3x = 9
x=3
Def. of  segs.
Substitute 12x – 11 for AD and 9x – 2 for BC.
Subtract 9x from both sides and add 11 to both sides.
Divide both sides by 3.
Example 5
Find the value of x so that PQST is isosceles.
Q  S
mQ = mS
2x2 + 19 = 4x2 – 13
32 = 2x2
x = 4 or x = –4
Trap. with pair base s   isosc.
trap.
Def. of  s
Substitute 2x2 + 19 for mQ and 4x2 – 13 for mS.
Subtract 2x2 and add 13 to both sides.
Divide by 2 and simplify.
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.
Substitute the given values.
EF = 10.75
Solve.
Example 7
Find EH.
Trap. Midsegment Thm.
16.5 =
1
(225 + EH)
Substitute the given values.
Simplify.
33 = 25 + EH
Multiply both sides by 2.
13 = EH
Subtract 25 from both sides.
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?
about 191.2 in.
In kite HJKL, mKLP = 72°,
and mHJP = 49.5°. Find each
measure.
2. mLHJ
3. mPKL
81°
18°
Practice
Use the diagram for Items 4 and 5.
4. mWZY = 61°. Find mWXY.
119°
5. XV = 4.6, and WY = 14.2. Find VZ.
9.6
6. Find LP.
18