Geometry Section 6-1 1112

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Angles of Polygon

Transcript of Geometry Section 6-1 1112

Chapter 6Quadrilaterals

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Tuesday, April 10, 2012

Section 6-1Angles of Polygons

Tuesday, April 10, 2012

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?

Tuesday, April 10, 2012

Vocabulary

1. Diagonal:

Tuesday, April 10, 2012

Vocabulary

1. Diagonal: A segment in a polygon that connects a vertex with another vertex that is nonconsecutive

Tuesday, April 10, 2012

Theorems

6.1 - Polygon Interior Angles Sum:

6.2 - Polygon Exterior Angles Sum:

Tuesday, April 10, 2012

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:

Tuesday, April 10, 2012

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°

Tuesday, April 10, 2012

Polygons and Sides

Tuesday, April 10, 2012

Polygons and Sides

Types of PolygonsTypes of PolygonsTypes of PolygonsTypes of PolygonsTypes of PolygonsTypes of Polygons

# sides 3 4 5 6 7

Name Triangle Quadrilateral Pentagon Hexagon Heptagon

Tuesday, April 10, 2012

Polygons and Sides

Types of PolygonsTypes of PolygonsTypes of PolygonsTypes of PolygonsTypes of PolygonsTypes of Polygons

# sides 3 4 5 6 7

Name Triangle Quadrilateral Pentagon Hexagon Heptagon

Tuesday, April 10, 2012

Polygons and Sides

Types of PolygonsTypes of PolygonsTypes of PolygonsTypes of PolygonsTypes of PolygonsTypes of Polygons

# sides 3 4 5 6 7

Name Triangle Quadrilateral Pentagon Hexagon Heptagon

Tuesday, April 10, 2012

Polygons and Sides

Types of PolygonsTypes of PolygonsTypes of PolygonsTypes of PolygonsTypes of PolygonsTypes of Polygons

# sides 3 4 5 6 7

Name Triangle Quadrilateral Pentagon Hexagon Heptagon

Types of PolygonsTypes of PolygonsTypes of PolygonsTypes of PolygonsTypes of PolygonsTypes of Polygons

# sides 8 9 10 12 n

Name Octagon Nonagon Decagon Dodecagon n-gon

Tuesday, April 10, 2012

Polygons and Sides

Types of PolygonsTypes of PolygonsTypes of PolygonsTypes of PolygonsTypes of PolygonsTypes of Polygons

# sides 3 4 5 6 7

Name Triangle Quadrilateral Pentagon Hexagon Heptagon

Types of PolygonsTypes of PolygonsTypes of PolygonsTypes of PolygonsTypes of PolygonsTypes of Polygons

# sides 8 9 10 12 n

Name Octagon Nonagon Decagon Dodecagon n-gon

Tuesday, April 10, 2012

Polygons and Sides

Types of PolygonsTypes of PolygonsTypes of PolygonsTypes of PolygonsTypes of PolygonsTypes of Polygons

# sides 3 4 5 6 7

Name Triangle Quadrilateral Pentagon Hexagon Heptagon

Types of PolygonsTypes of PolygonsTypes of PolygonsTypes of PolygonsTypes of PolygonsTypes of Polygons

# sides 8 9 10 12 n

Name Octagon Nonagon Decagon Dodecagon n-gon

Tuesday, April 10, 2012

Example 1Find the sum of the measures of the interior angles of the following.

a. Nonagon b. 17-gon

Tuesday, April 10, 2012

Example 1Find the sum of the measures of the interior angles of the following.

a. Nonagon b. 17-gon

S = (n−2)180

Tuesday, April 10, 2012

Example 1Find 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

Tuesday, April 10, 2012

Example 1Find 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

Tuesday, April 10, 2012

Example 1Find 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°

Tuesday, April 10, 2012

Example 1Find 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

Tuesday, April 10, 2012

Example 1Find 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

Tuesday, April 10, 2012

Example 1Find 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

Tuesday, April 10, 2012

Example 1Find 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°

Tuesday, April 10, 2012

Example 2Find the measure of each interior angle of parallelogram RSTU.

Tuesday, April 10, 2012

Example 2Find the measure of each interior angle of parallelogram RSTU.

S = (n−2)180

Tuesday, April 10, 2012

Example 2Find the measure of each interior angle of parallelogram RSTU.

S = (n−2)180

S = (4−2)180

Tuesday, April 10, 2012

Example 2Find the measure of each interior angle of parallelogram RSTU.

S = (n−2)180

S = (4−2)180

= (2)180

Tuesday, April 10, 2012

Example 2Find the measure of each interior angle of parallelogram RSTU.

S = (n−2)180

S = (4−2)180

= (2)180 =360°

Tuesday, April 10, 2012

Example 2Find 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

Tuesday, April 10, 2012

Example 2Find 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

Tuesday, April 10, 2012

Example 2Find 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

Tuesday, April 10, 2012

Example 2Find 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

Tuesday, April 10, 2012

Example 2Find 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

Tuesday, April 10, 2012

Example 2Find 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 =11Tuesday, April 10, 2012

Example 2Find 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)

Tuesday, April 10, 2012

Example 2Find 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°

Tuesday, April 10, 2012

Example 2Find 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

Tuesday, April 10, 2012

Example 2Find 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

Tuesday, April 10, 2012

Example 2Find 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°

Tuesday, April 10, 2012

Example 3Park 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

Tuesday, April 10, 2012

Example 3Park 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

S = (n−2)180

Tuesday, April 10, 2012

Example 3Park 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

S = (n−2)180

S = (8−2)180

Tuesday, April 10, 2012

Example 3Park 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

S = (n−2)180

S = (8−2)180

= (6)180

Tuesday, April 10, 2012

Example 3Park 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

S = (n−2)180

S = (8−2)180

= (6)180

=1080°

Tuesday, April 10, 2012

Example 3Park 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

S = (n−2)180

S = (8−2)180

= (6)180

=1080° 1080°

8

Tuesday, April 10, 2012

Example 3Park 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

S = (n−2)180

S = (8−2)180

= (6)180

=1080° 1080°

8 =135°

Tuesday, April 10, 2012

Example 4The measure of an interior angle of a regular polygon is 150°. Find the

number of sides in the polygon.

Tuesday, April 10, 2012

Example 4The measure of an interior angle of a regular polygon is 150°. Find the

number of sides in the polygon.

150n = (n−2)180

Tuesday, April 10, 2012

Example 4The 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

Tuesday, April 10, 2012

Example 4The 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

Tuesday, April 10, 2012

Example 4The 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

Tuesday, April 10, 2012

Example 4The 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

Tuesday, April 10, 2012

Example 4The 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

Tuesday, April 10, 2012

Example 4The 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

Tuesday, April 10, 2012

Example 5Find 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

Tuesday, April 10, 2012

Example 5Find 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

Tuesday, April 10, 2012

Example 5Find 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

Tuesday, April 10, 2012

Example 5Find 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

Tuesday, April 10, 2012

Example 5Find 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

Tuesday, April 10, 2012

Check Your Understanding

Review #1-11 on p. 393

Tuesday, April 10, 2012

Problem Set

Tuesday, April 10, 2012

Problem Set

p. 394 #13-37 odd, 49, 59

"They can because they think they can." - VirgilTuesday, April 10, 2012