Chapter 6
Quadrilaterals
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Section 6-1
Angles of Polygons
Essential Questions
How do you find and use the sum of the measures of the interior
angles of a polygon?
How do you find and use the sum of the measures of the exterior
angles of a polygon?
Vocabulary
1. Diagonal:
Vocabulary
1. Diagonal: A segment in a polygon that connects a vertex with
another vertex that is nonconsecutive
Theorems
6.1 - Polygon Interior Angles Sum:
6.2 - Polygon Exterior Angles Sum:
Theorems
6.1 - Polygon Interior Angles Sum: The sum of the interior angle
measures of a convex polygon with n sides is found with the formula
S =(n−2)180
6.2 - Polygon Exterior Angles Sum:
Theorems
6.1 - Polygon Interior Angles Sum: The sum of the interior angle
measures of a convex polygon with n sides is found with the formula
S =(n−2)180
6.2 - Polygon Exterior Angles Sum: The sum of the exterior angle
measures of a convex polygon, one at each vertex, is 360°
Polygons and Sides
Polygons and Sides
Types of Polygons
Polygons and Sides
Types of Polygons
# sides 3 4 5 6 7
Polygons and Sides
Types of Polygons
# sides 3 4 5 6 7
Name Triangle Quadrilateral Pentagon Hexagon Heptagon
Polygons and Sides
Types of Polygons
# sides 3 4 5 6 7
Name Triangle Quadrilateral Pentagon Hexagon Heptagon
Types of Polygons
Polygons and Sides
Types of Polygons
# sides 3 4 5 6 7
Name Triangle Quadrilateral Pentagon Hexagon Heptagon
Types of Polygons
# sides 8 9 10 12 n
Polygons and Sides
Types of Polygons
# sides 3 4 5 6 7
Name Triangle Quadrilateral Pentagon Hexagon Heptagon
Types of Polygons
# sides 8 9 10 12 n
Name Octagon Nonagon Decagon Dodecagon n-gon
Example 1
Find the sum of the measures of the interior angles of the following.
a. Nonagon b. 17-gon
Example 1
Find the sum of the measures of the interior angles of the following.
a. Nonagon b. 17-gon
S =(n−2)180
Example 1
Find the sum of the measures of the interior angles of the following.
a. Nonagon b. 17-gon
S =(n−2)180
S =(9−2)180
Example 1
Find the sum of the measures of the interior angles of the following.
a. Nonagon b. 17-gon
S =(n−2)180
S =(9−2)180
=(7)180
Example 1
Find the sum of the measures of the interior angles of the following.
a. Nonagon b. 17-gon
S =(n−2)180
S =(9−2)180
=(7)180
=1260°
Example 1
Find the sum of the measures of the interior angles of the following.
a. Nonagon b. 17-gon
S =(n−2)180
S =(9−2)180
=(7)180
=1260°
S =(n−2)180
Example 1
Find the sum of the measures of the interior angles of the following.
a. Nonagon b. 17-gon
S =(n−2)180
S =(9−2)180
=(7)180
=1260°
S =(n−2)180
S =(17−2)180
Example 1
Find the sum of the measures of the interior angles of the following.
a. Nonagon b. 17-gon
S =(n−2)180
S =(9−2)180
=(7)180
=1260°
S =(n−2)180
S =(17−2)180
=(15)180
Example 1
Find the sum of the measures of the interior angles of the following.
a. Nonagon b. 17-gon
S =(n−2)180
S =(9−2)180
=(7)180
=1260°
S =(n−2)180
S =(17−2)180
=(15)180
=2700°
Example 2
Find the measure of each interior angle of parallelogram RSTU.
Example 2
Find the measure of each interior angle of parallelogram RSTU.
S =(n−2)180
Example 2
Find the measure of each interior angle of parallelogram RSTU.
S =(n−2)180
S =(4−2)180
Example 2
Find the measure of each interior angle of parallelogram RSTU.
S =(n−2)180
S =(4−2)180
=(2)180
Example 2
Find the measure of each interior angle of parallelogram RSTU.
S =(n−2)180
S =(4−2)180
=(2)180
=360°
Example 2
Find the measure of each interior angle of parallelogram RSTU.
S =(n−2)180
S =(4−2)180
=(2)180
=360°
11x +4+5x +11x +4+5x =360
Example 2
Find the measure of each interior angle of parallelogram RSTU.
S =(n−2)180
S =(4−2)180
=(2)180
=360°
11x +4+5x +11x +4+5x =360
32x +8=360
Example 2
Find the measure of each interior angle of parallelogram RSTU.
S =(n−2)180
S =(4−2)180
=(2)180
=360°
11x +4+5x +11x +4+5x =360
32x +8=360
−8 −8
Example 2
Find the measure of each interior angle of parallelogram RSTU.
S =(n−2)180
S =(4−2)180
=(2)180
=360°
11x +4+5x +11x +4+5x =360
32x +8=360
−8 −8
32x =352
Example 2
Find the measure of each interior angle of parallelogram RSTU.
S =(n−2)180
S =(4−2)180
=(2)180
=360°
11x +4+5x +11x +4+5x =360
32x +8=360
−8 −8
32x =352
32 32
Example 2
Find the measure of each interior angle of parallelogram RSTU.
S =(n−2)180
S =(4−2)180
=(2)180
=360°
11x +4+5x +11x +4+5x =360
32x +8=360
−8 −8
32x =352
32 32
x =11
Example 2
Find the measure of each interior angle of parallelogram RSTU.
S =(n−2)180
S =(4−2)180
=(2)180
=360°
11x +4+5x +11x +4+5x =360
32x +8=360
−8 −8
32x =352
32 32
x =11
m∠R = m∠T =5(11)
Example 2
Find the measure of each interior angle of parallelogram RSTU.
S =(n−2)180
S =(4−2)180
=(2)180
=360°
11x +4+5x +11x +4+5x =360
32x +8=360
−8 −8
32x =352
32 32
x =11
m∠R = m∠T =5(11) =55°
Example 2
Find the measure of each interior angle of parallelogram RSTU.
S =(n−2)180
S =(4−2)180
=(2)180
=360°
11x +4+5x +11x +4+5x =360
32x +8=360
−8 −8
32x =352
32 32
x =11
m∠R = m∠T =5(11) =55°
m∠S = m∠U =11(11)+4
Example 2
Find the measure of each interior angle of parallelogram RSTU.
S =(n−2)180
S =(4−2)180
=(2)180
=360°
11x +4+5x +11x +4+5x =360
32x +8=360
−8 −8
32x =352
32 32
x =11
m∠R = m∠T =5(11) =55°
m∠S = m∠U =11(11)+4
=121+4
Example 2
Find the measure of each interior angle of parallelogram RSTU.
S =(n−2)180
S =(4−2)180
=(2)180
=360°
11x +4+5x +11x +4+5x =360
32x +8=360
−8 −8
32x =352
32 32
x =11
m∠R = m∠T =5(11) =55°
m∠S = m∠U =11(11)+4
=121+4 =125°
Example 3
Park City Mall is designed so that eight walkways meet in a central area in
the shape of a regular octagon. Find the measure of one of the interior
angles of the octagon.
http://www.parkcitycenter.com/directory
Example 3
Park City Mall is designed so that eight walkways meet in a central area in
the shape of a regular octagon. Find the measure of one of the interior
angles of the octagon.
S =(n−2)180
http://www.parkcitycenter.com/directory
Example 3
Park City Mall is designed so that eight walkways meet in a central area in
the shape of a regular octagon. Find the measure of one of the interior
angles of the octagon.
S =(n−2)180
S =(8−2)180
http://www.parkcitycenter.com/directory
Example 3
Park City Mall is designed so that eight walkways meet in a central area in
the shape of a regular octagon. Find the measure of one of the interior
angles of the octagon.
S =(n−2)180
S =(8−2)180
=(6)180
http://www.parkcitycenter.com/directory
Example 3
Park City Mall is designed so that eight walkways meet in a central area in
the shape of a regular octagon. Find the measure of one of the interior
angles of the octagon.
S =(n−2)180
S =(8−2)180
=(6)180
=1080° http://www.parkcitycenter.com/directory
Example 3
Park City Mall is designed so that eight walkways meet in a central area in
the shape of a regular octagon. Find the measure of one of the interior
angles of the octagon.
S =(n−2)180
S =(8−2)180
=(6)180
=1080°
1080°
8 http://www.parkcitycenter.com/directory
Example 3
Park City Mall is designed so that eight walkways meet in a central area in
the shape of a regular octagon. Find the measure of one of the interior
angles of the octagon.
S =(n−2)180
S =(8−2)180
=(6)180
=1080°
1080°
8
=135° http://www.parkcitycenter.com/directory
Example 4
The measure of an interior angle of a regular polygon is 150°. Find the
number of sides in the polygon.
Example 4
The measure of an interior angle of a regular polygon is 150°. Find the
number of sides in the polygon.
150n=(n−2)180
Example 4
The measure of an interior angle of a regular polygon is 150°. Find the
number of sides in the polygon.
150n=(n−2)180
150n=180n−360
Example 4
The measure of an interior angle of a regular polygon is 150°. Find the
number of sides in the polygon.
150n=(n−2)180
150n=180n−360
−180n −180n
Example 4
The measure of an interior angle of a regular polygon is 150°. Find the
number of sides in the polygon.
150n=(n−2)180
150n=180n−360
−180n −180n
−30n= −360
Example 4
The measure of an interior angle of a regular polygon is 150°. Find the
number of sides in the polygon.
150n=(n−2)180
150n=180n−360
−180n −180n
−30n= −360
−30 −30
Example 4
The measure of an interior angle of a regular polygon is 150°. Find the
number of sides in the polygon.
150n=(n−2)180
150n=180n−360
−180n −180n
−30n= −360
−30 −30
n=12
Example 4
The measure of an interior angle of a regular polygon is 150°. Find the
number of sides in the polygon.
150n=(n−2)180
150n=180n−360
−180n −180n
−30n= −360
−30 −30
n=12
There are 12 sides to the polygon
Example 5
Find the value of x in the diagram.
m∠1=5x +5, m∠2=5x, m∠3= 4x −6,
m∠4=5x −5, m∠5= 4x +3, m∠6=6x −12,
m∠7=2x +3
Example 5
Find the value of x in the diagram.
m∠1=5x +5, m∠2=5x, m∠3= 4x −6,
m∠4=5x −5, m∠5= 4x +3, m∠6=6x −12,
m∠7=2x +3
5x +5+5x +4x −6+5x −5+4x +3+6x −12+2x +3=360
Example 5
Find the value of x in the diagram.
m∠1=5x +5, m∠2=5x, m∠3= 4x −6,
m∠4=5x −5, m∠5= 4x +3, m∠6=6x −12,
m∠7=2x +3
5x +5+5x +4x −6+5x −5+4x +3+6x −12+2x +3=360
31x −12=360
Example 5
Find the value of x in the diagram.
m∠1=5x +5, m∠2=5x, m∠3= 4x −6,
m∠4=5x −5, m∠5= 4x +3, m∠6=6x −12,
m∠7=2x +3
5x +5+5x +4x −6+5x −5+4x +3+6x −12+2x +3=360
31x −12=360
31x =372
Example 5
Find the value of x in the diagram.
m∠1=5x +5, m∠2=5x, m∠3= 4x −6,
m∠4=5x −5, m∠5= 4x +3, m∠6=6x −12,
m∠7=2x +3
5x +5+5x +4x −6+5x −5+4x +3+6x −12+2x +3=360
31x −12=360
31x =372
x =12

Geometry Section 6-1

  • 1.
  • 2.
  • 3.
    Essential Questions How doyou find and use the sum of the measures of the interior angles of a polygon? How do you find and use the sum of the measures of the exterior angles of a polygon?
  • 4.
  • 5.
    Vocabulary 1. Diagonal: Asegment in a polygon that connects a vertex with another vertex that is nonconsecutive
  • 6.
    Theorems 6.1 - PolygonInterior Angles Sum: 6.2 - Polygon Exterior Angles Sum:
  • 7.
    Theorems 6.1 - PolygonInterior Angles Sum: The sum of the interior angle measures of a convex polygon with n sides is found with the formula S =(n−2)180 6.2 - Polygon Exterior Angles Sum:
  • 8.
    Theorems 6.1 - PolygonInterior Angles Sum: The sum of the interior angle measures of a convex polygon with n sides is found with the formula S =(n−2)180 6.2 - Polygon Exterior Angles Sum: The sum of the exterior angle measures of a convex polygon, one at each vertex, is 360°
  • 9.
  • 10.
  • 11.
    Polygons and Sides Typesof Polygons # sides 3 4 5 6 7
  • 12.
    Polygons and Sides Typesof Polygons # sides 3 4 5 6 7 Name Triangle Quadrilateral Pentagon Hexagon Heptagon
  • 13.
    Polygons and Sides Typesof Polygons # sides 3 4 5 6 7 Name Triangle Quadrilateral Pentagon Hexagon Heptagon Types of Polygons
  • 14.
    Polygons and Sides Typesof Polygons # sides 3 4 5 6 7 Name Triangle Quadrilateral Pentagon Hexagon Heptagon Types of Polygons # sides 8 9 10 12 n
  • 15.
    Polygons and Sides Typesof Polygons # sides 3 4 5 6 7 Name Triangle Quadrilateral Pentagon Hexagon Heptagon Types of Polygons # sides 8 9 10 12 n Name Octagon Nonagon Decagon Dodecagon n-gon
  • 16.
    Example 1 Find thesum of the measures of the interior angles of the following. a. Nonagon b. 17-gon
  • 17.
    Example 1 Find thesum of the measures of the interior angles of the following. a. Nonagon b. 17-gon S =(n−2)180
  • 18.
    Example 1 Find thesum of the measures of the interior angles of the following. a. Nonagon b. 17-gon S =(n−2)180 S =(9−2)180
  • 19.
    Example 1 Find thesum of the measures of the interior angles of the following. a. Nonagon b. 17-gon S =(n−2)180 S =(9−2)180 =(7)180
  • 20.
    Example 1 Find thesum of the measures of the interior angles of the following. a. Nonagon b. 17-gon S =(n−2)180 S =(9−2)180 =(7)180 =1260°
  • 21.
    Example 1 Find thesum of the measures of the interior angles of the following. a. Nonagon b. 17-gon S =(n−2)180 S =(9−2)180 =(7)180 =1260° S =(n−2)180
  • 22.
    Example 1 Find thesum of the measures of the interior angles of the following. a. Nonagon b. 17-gon S =(n−2)180 S =(9−2)180 =(7)180 =1260° S =(n−2)180 S =(17−2)180
  • 23.
    Example 1 Find thesum of the measures of the interior angles of the following. a. Nonagon b. 17-gon S =(n−2)180 S =(9−2)180 =(7)180 =1260° S =(n−2)180 S =(17−2)180 =(15)180
  • 24.
    Example 1 Find thesum of the measures of the interior angles of the following. a. Nonagon b. 17-gon S =(n−2)180 S =(9−2)180 =(7)180 =1260° S =(n−2)180 S =(17−2)180 =(15)180 =2700°
  • 25.
    Example 2 Find themeasure of each interior angle of parallelogram RSTU.
  • 26.
    Example 2 Find themeasure of each interior angle of parallelogram RSTU. S =(n−2)180
  • 27.
    Example 2 Find themeasure of each interior angle of parallelogram RSTU. S =(n−2)180 S =(4−2)180
  • 28.
    Example 2 Find themeasure of each interior angle of parallelogram RSTU. S =(n−2)180 S =(4−2)180 =(2)180
  • 29.
    Example 2 Find themeasure of each interior angle of parallelogram RSTU. S =(n−2)180 S =(4−2)180 =(2)180 =360°
  • 30.
    Example 2 Find themeasure of each interior angle of parallelogram RSTU. S =(n−2)180 S =(4−2)180 =(2)180 =360° 11x +4+5x +11x +4+5x =360
  • 31.
    Example 2 Find themeasure of each interior angle of parallelogram RSTU. S =(n−2)180 S =(4−2)180 =(2)180 =360° 11x +4+5x +11x +4+5x =360 32x +8=360
  • 32.
    Example 2 Find themeasure of each interior angle of parallelogram RSTU. S =(n−2)180 S =(4−2)180 =(2)180 =360° 11x +4+5x +11x +4+5x =360 32x +8=360 −8 −8
  • 33.
    Example 2 Find themeasure of each interior angle of parallelogram RSTU. S =(n−2)180 S =(4−2)180 =(2)180 =360° 11x +4+5x +11x +4+5x =360 32x +8=360 −8 −8 32x =352
  • 34.
    Example 2 Find themeasure of each interior angle of parallelogram RSTU. S =(n−2)180 S =(4−2)180 =(2)180 =360° 11x +4+5x +11x +4+5x =360 32x +8=360 −8 −8 32x =352 32 32
  • 35.
    Example 2 Find themeasure of each interior angle of parallelogram RSTU. S =(n−2)180 S =(4−2)180 =(2)180 =360° 11x +4+5x +11x +4+5x =360 32x +8=360 −8 −8 32x =352 32 32 x =11
  • 36.
    Example 2 Find themeasure of each interior angle of parallelogram RSTU. S =(n−2)180 S =(4−2)180 =(2)180 =360° 11x +4+5x +11x +4+5x =360 32x +8=360 −8 −8 32x =352 32 32 x =11 m∠R = m∠T =5(11)
  • 37.
    Example 2 Find themeasure of each interior angle of parallelogram RSTU. S =(n−2)180 S =(4−2)180 =(2)180 =360° 11x +4+5x +11x +4+5x =360 32x +8=360 −8 −8 32x =352 32 32 x =11 m∠R = m∠T =5(11) =55°
  • 38.
    Example 2 Find themeasure of each interior angle of parallelogram RSTU. S =(n−2)180 S =(4−2)180 =(2)180 =360° 11x +4+5x +11x +4+5x =360 32x +8=360 −8 −8 32x =352 32 32 x =11 m∠R = m∠T =5(11) =55° m∠S = m∠U =11(11)+4
  • 39.
    Example 2 Find themeasure of each interior angle of parallelogram RSTU. S =(n−2)180 S =(4−2)180 =(2)180 =360° 11x +4+5x +11x +4+5x =360 32x +8=360 −8 −8 32x =352 32 32 x =11 m∠R = m∠T =5(11) =55° m∠S = m∠U =11(11)+4 =121+4
  • 40.
    Example 2 Find themeasure of each interior angle of parallelogram RSTU. S =(n−2)180 S =(4−2)180 =(2)180 =360° 11x +4+5x +11x +4+5x =360 32x +8=360 −8 −8 32x =352 32 32 x =11 m∠R = m∠T =5(11) =55° m∠S = m∠U =11(11)+4 =121+4 =125°
  • 41.
    Example 3 Park CityMall is designed so that eight walkways meet in a central area in the shape of a regular octagon. Find the measure of one of the interior angles of the octagon. http://www.parkcitycenter.com/directory
  • 42.
    Example 3 Park CityMall is designed so that eight walkways meet in a central area in the shape of a regular octagon. Find the measure of one of the interior angles of the octagon. S =(n−2)180 http://www.parkcitycenter.com/directory
  • 43.
    Example 3 Park CityMall is designed so that eight walkways meet in a central area in the shape of a regular octagon. Find the measure of one of the interior angles of the octagon. S =(n−2)180 S =(8−2)180 http://www.parkcitycenter.com/directory
  • 44.
    Example 3 Park CityMall is designed so that eight walkways meet in a central area in the shape of a regular octagon. Find the measure of one of the interior angles of the octagon. S =(n−2)180 S =(8−2)180 =(6)180 http://www.parkcitycenter.com/directory
  • 45.
    Example 3 Park CityMall is designed so that eight walkways meet in a central area in the shape of a regular octagon. Find the measure of one of the interior angles of the octagon. S =(n−2)180 S =(8−2)180 =(6)180 =1080° http://www.parkcitycenter.com/directory
  • 46.
    Example 3 Park CityMall is designed so that eight walkways meet in a central area in the shape of a regular octagon. Find the measure of one of the interior angles of the octagon. S =(n−2)180 S =(8−2)180 =(6)180 =1080° 1080° 8 http://www.parkcitycenter.com/directory
  • 47.
    Example 3 Park CityMall is designed so that eight walkways meet in a central area in the shape of a regular octagon. Find the measure of one of the interior angles of the octagon. S =(n−2)180 S =(8−2)180 =(6)180 =1080° 1080° 8 =135° http://www.parkcitycenter.com/directory
  • 48.
    Example 4 The measureof an interior angle of a regular polygon is 150°. Find the number of sides in the polygon.
  • 49.
    Example 4 The measureof an interior angle of a regular polygon is 150°. Find the number of sides in the polygon. 150n=(n−2)180
  • 50.
    Example 4 The measureof an interior angle of a regular polygon is 150°. Find the number of sides in the polygon. 150n=(n−2)180 150n=180n−360
  • 51.
    Example 4 The measureof an interior angle of a regular polygon is 150°. Find the number of sides in the polygon. 150n=(n−2)180 150n=180n−360 −180n −180n
  • 52.
    Example 4 The measureof an interior angle of a regular polygon is 150°. Find the number of sides in the polygon. 150n=(n−2)180 150n=180n−360 −180n −180n −30n= −360
  • 53.
    Example 4 The measureof an interior angle of a regular polygon is 150°. Find the number of sides in the polygon. 150n=(n−2)180 150n=180n−360 −180n −180n −30n= −360 −30 −30
  • 54.
    Example 4 The measureof an interior angle of a regular polygon is 150°. Find the number of sides in the polygon. 150n=(n−2)180 150n=180n−360 −180n −180n −30n= −360 −30 −30 n=12
  • 55.
    Example 4 The measureof an interior angle of a regular polygon is 150°. Find the number of sides in the polygon. 150n=(n−2)180 150n=180n−360 −180n −180n −30n= −360 −30 −30 n=12 There are 12 sides to the polygon
  • 56.
    Example 5 Find thevalue of x in the diagram. m∠1=5x +5, m∠2=5x, m∠3= 4x −6, m∠4=5x −5, m∠5= 4x +3, m∠6=6x −12, m∠7=2x +3
  • 57.
    Example 5 Find thevalue of x in the diagram. m∠1=5x +5, m∠2=5x, m∠3= 4x −6, m∠4=5x −5, m∠5= 4x +3, m∠6=6x −12, m∠7=2x +3 5x +5+5x +4x −6+5x −5+4x +3+6x −12+2x +3=360
  • 58.
    Example 5 Find thevalue of x in the diagram. m∠1=5x +5, m∠2=5x, m∠3= 4x −6, m∠4=5x −5, m∠5= 4x +3, m∠6=6x −12, m∠7=2x +3 5x +5+5x +4x −6+5x −5+4x +3+6x −12+2x +3=360 31x −12=360
  • 59.
    Example 5 Find thevalue of x in the diagram. m∠1=5x +5, m∠2=5x, m∠3= 4x −6, m∠4=5x −5, m∠5= 4x +3, m∠6=6x −12, m∠7=2x +3 5x +5+5x +4x −6+5x −5+4x +3+6x −12+2x +3=360 31x −12=360 31x =372
  • 60.
    Example 5 Find thevalue of x in the diagram. m∠1=5x +5, m∠2=5x, m∠3= 4x −6, m∠4=5x −5, m∠5= 4x +3, m∠6=6x −12, m∠7=2x +3 5x +5+5x +4x −6+5x −5+4x +3+6x −12+2x +3=360 31x −12=360 31x =372 x =12