Post on 17-Dec-2015
Polygons
Why a hexagon?
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
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
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°
(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.
Example 2: Properties of Parallelograms
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.
Example 3: Properties of Parallelograms
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.
diags. bisect each other.
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.
Substitute the given values.
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.
diags. bisect each other.
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
diags. bisect each other.
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.
Divide both sides by 10.
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.
Divide both sides by 7.
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
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.
Step 2 Find PR and QS to determine is PQRS is a rectangle.
Example 3A Continued
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.
Example 3A Continued
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.
Use the diagonals to determine whether a parallelogram with the given vertices is a rectangle, rhombus, or square. Give all the names that apply.
Step 2 Find WY and XZ to determine is WXYZ is a rectangle.
Thus WXYZ is not a square.
Example 3B Continued
Since , WXYZ is not a rectangle.
Step 3 Determine if WXYZ is a rhombus.
Example 3B Continued
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)
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
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.
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
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.
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°
Example 2A Continued
Substitute mCDF for mCBF.
Substitute 52 for mCBF.
Subtract 104 from both sides.
mBCD + mCBF + mCDF = 180°
mBCD + 52° + 52° = 180°
mBCD = 76°
mBCD + mCBF + mCDF = 180°
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.
Example 2B Continued
Substitute.
Simplify.
mABC + mBCD + mABC + mDAB = 360°
mABC + 76° + mABC + 54° = 360°
2mABC = 230°
mABC = 115° Solve.
Kite one pair opp. s
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.
Subtract 78 from both sides.
Divide by 2.
Example 2b
In kite PQRS, mPQR = 78°, and mTRS = 59°. Find mQPS.
Kite one pair opp. s
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.
Subtract 100 from both sides.
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
Example 3B Continued
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.
Example 3B Continued
∆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.
Divide both sides by 3.
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.
Substitute the given values.
Solve.EF = 10.75
Example 7
Find EH.
Trap. Midsegment Thm.
Substitute the given values.
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