eTextbook (Basic Geometry and Trigonometry)

7/22/2019 eTextbook (Basic Geometry and Trigonometry)

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Geometry is a branch o mathematics that is concerned with the properties, measurements, and relationshipsand the study o the sizes, shapes, and positions o 2-dimensional plane figures and 3-dimensional solid objectsGeometry(translates to Earth Measurement rom Greek) is linked to many other topics in mathematics and is useddaily or ound everywhere by almost everyone in the fields o art, architecture, engineering, land surveys, astronomysculptures, space, nature, sports, machines, etc.

Geometry has a lot o practical day-to-day uses in the workplace and at home. For example, you use geometry to determinethe quantity o paint needed to paint walls, the amount o carpet neededor your home, the length o ence needed or thegarden, etc. In this chapter, you will learn the most basic orm o geometry, called Euclidean geometry involving pointslines, angles, lengths, areas, volumes the Pythagorean Teorem, basic trigonometric ratios, and their applications.

Learning Outcomes Recognize and use various notations to represent points, lines, line

segments, rays, and angles.

Classiy angles and determine the angle relationships between parallel lineand transversals.

Classiy triangles, quadrilaterals, and polygons based on properties o theisides and angles.

Apply properties o similar and congruent triangles in solving problemsinvolving triangles.

Compute the perimeter and area o plane figures, such as triangles,quadrilaterals, and circles.

Compute the volume and surace area o common, three-dimensionalsolid objects.

Use the Pythagorean Teorem to determinethe length o the unknown sido a right triangle.

Determine the basic trigonometric ratios o angles o right triangles.

Evaluate the exact trigonometric ratios o special angles.

Solve right triangles using the Pythagorean Teorem and trigonometric ratio

Chapter Outline11.1 Lines and Angles

11.2 Classification and Properties o Plane Figures

11.3 Similar and Congruent riangles

11.4 Perimeter and Areas o Plane Geometric Figures

11.5 Volume and Surace Area o Common Solid Objects

11.6 Pythagorean Teorem

11.7 Primary rigonometric Ratios

Chapter 11 Basic Geometry and Trigonometry


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11.1 Lines and Angles

Introduction

Geometry is a branch o Mathematics dealing with the study o relative positions,properties, and relations o Geometric objects (such as points, lines, angles,suraces, solids, and calculations involving lengths, angles, perimeters, areas,and volumes o such objects). Geometry can be traced as ar back as the earlyhistorical era, to the ancient Egyptians and Babylonians. However, geometrywas revolutionized by the ancient Greeks, including Pythagoras, Plato, andmost notably, Euclid, who invented Euclidean Geometry, which is the ocus othis chapter.

Euclidean Geometry begins with the notion o a point. Recall rom Chapter 8 thata pointin the Cartesian plane represents a locationin the plane, determined byits x-coordinate, representing its horizontal position with respect to the origin,and itsy-coordinate, representing its vertical position with respect to the origin.It has no dimensions; that is, it has no length, width, or height.

We label a point in the Cartesian plane using a dot, a letter (most ofen P), andordered coordinates in brackets. Labelling the point P(3, 5) in the Cartesianplane is illustrated in Exhibit 11.1-a.

Ofen, when working with Euclidean Geometry, we care only about the relativeposition o the point to other points, and not its specific position in the Cartesianplane. As such, we ofen omit the coordinates and label the point using a dotand a letter (.P).

II I

III IV

origin

P(3, 5)

5

5 5 5

10

10

10

10

X-axis

Y-axis

(0, 0)

Exhibit 11.1-a Labelling a point on the Cartesian Plane

Copyright 2013 Vretta Inc. 11.1 Lines and


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Lines, Line Segments, and Rays

A line is an object that has only one dimension, length. A line is created byjoining two points,includesall the points that all directly between them, andextends indefinitely in opposite directions. Tereore, a line is straight, has no

gaps, and extends infinitely in both directions. It is denoted with the names othe two points over-lined with a double-arrowhead. It has no end-points.

Line A B Line AB(or BA)

A line segment is the portion o a line bound between two points. A line segmentis created by joining two points and includesall the points that all directlybetween them. It is denoted with the names o the two points, over-lined witha straight line. It has two end-points.

Line Segment A B Line Segment B(or BA)

A ray is the portion o a line bound in one direction by a point. A ray is createdby joining two points, includes all the points that all directly between them,and extends indefinitely in one direction only. It is denoted with the names othe two points, over-lined with a single arrowhead. It has only one end-point.

Ray

A B Ray AB

A B

Ray BA

Note: When labeling a ray, the order of the letters matters. For example, rayABoriginates at point A and extends indefinitely in the direction of point B, whileray BAoriginates at point B and extends indefinitely in the direction of point A.

IdentifyingLines, Line Segments, and Rays

Identiy and label the ollowing geometric objects:

(i) (ii) (iii) (iv)

P

QM N

C

D

XY

Solution

(i) Line PQ or QP (ii) Line NM (iii) Line segment CDorDC (iv) Ray YX

Angle Measures in Degrees

An angle is ormed when two rays intersect at their endpoints. Te point ointersection is called the vertex o the angle and the two rays are called the sides

o the angle. Te angle is identified by the symbol , ollowed by the letters o

the three points o the two rays, with the vertex in the middle.

For example, rays BA and BC orm the angle ABC or CBA. When the

Example 11.1-a

apter 11 | Basic Geometry and Trigonometry Copyright 2013 Vretta Inc.


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context is clear, we may simply reer to this angle as B.

CSide

Side Intersecting ray sVertex

B

A ABC(or CBA) =

Or simply B = When naming an angle,the vertex is alwayswritten in the middle.

Te size o the angle is measured in degrees (denoted with the symbol ),where one revolution o a circle is 360. 1 is a 1/360 slice o one revolution oa circle. Imagine a circle centered at point B, divided into 360 equal sectorsthrough B the degree measure o angle ABC is the number o sectors thatcan fit in the wedge ormed between rays BA and BC . Exhibit 11.1-b shows acircle divided into 36 sectors, where each sector represents 10

90 80 7060

50130

120110

100

40140

30150

20160

10

0/360

350

340

330

320

310300

290280270260250

240

230

220

210

200

190

180

170

B A

C

Exhibit 11.1-b Circle divided into 36 sectors, where

each sector represents 10

60120

A protractor is used to measure anddraw angles in degrees.

wo rays rom the center o a circleextending in opposite directions createa line which divide the circle into twoequal halves, thus, the angle betweentwo opposite rays has an angle measure

equal to1

2

a revolution or360

1802

=

180

wo perpendicular lines through the centero a circle cut the circle into our equalquadrants, thus the angle between twoperpendicular rays has an angle measure

equal to 14

a revolution or 360 904

= .

90



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Classification of Angles

Angles are classified according to their size in degrees.

Right Angles

( 90 = )

Any angle measure o 90 is

called a right angle, and isdenoted with a small square atthe vertex.

Straight angles

( 180 = )

Any angle measure o 180 iscalled a straight angle, and isdenoted with a semi-circle wherethe rays meet.

Acute angles

(0 90 < < )

Any angle less than a right angle(i.e., with a degree measureless than 90) is called an acute

angle.

Obtuse angles

(90 180 < < )

Any angle greater than a rightangle but less than a straightangle (i.e., with a degree measurebetween 90 and 180) is calledan obtuse angle.

Reflex angles

(180 360 < < )

Any angle greater than a straightangle (i.e., with a degree measurebetween 180 and 360 is called areflex angle. In any pair o rays,

there is one angle that is at most180 and one that is at least 180.

Classiying Angles

Identiy the ollowing angles as acute, right, obtuse, straight, or reflex:

(i) (ii) (iii) (iv) (v)

Solution (i) Obtuse (ii) Acute (iii) Straight (iv) Right (v) Reflex

Supplementary and Complementary Angles

Angle pairs whose measures sum to right angle (90) or straight angle (180)are given special names:

Example 11.1-b



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Supplementaryangles

( 180 + = )

wo angles are calledsupplementary anglesi their

sum is 180.Each angle is called a supplemento the other.

Complementaryangles

( 90 + = )

wo angles are calledcomplementary anglesi their sum is 90.

Each angle is called acomplement o the other.

Note: Since the sum of complementary angles is 90, each angle must be acute (i.e.,less than 90). As a result, only acute angles have complements.

Complementary and Supplementary Angles

Determine the supplement and complement (i possible) o the ollowing angles:

(i) 30 (ii) 45 (iii) 72 (iv) 90 (v) 126

Solution(i) Supplement o 30 = 180 30 = 150

Complement o 30= 90 30 = 60

(ii) Supplement o 45= 180 45 = 135Complement o 45= 90 45 = 45Note: A 45-angle is sel-complementary.

(iii) Supplement o 72= 180 72 = 108

Complement o 72= 90 72 = 18(iv) Supplement o 90= 180 90 = 90

Note: A 90 (right)-angle is sel-supplementary.Since 90 is not acute, it does not have a complementary angle.

(v) Supplement o 126= 180 126 = 54Since 126 is not acute, it does not have a complementary angle.

Opposite and Adjacent Angles

When two lines intersect at a point P, they create our angles. Every pair oconsecutive angles, called adjacent angles, are supplementary, since each line

orms a straight angle (180) at point P and the other line cuts it into two angles,which thereore sum to 180. As a result, the angle opposite to each other,calledopposite angles,are always equal (congruent).

b a

Pc

d

Adjacent Angle Opposite Angle

a+ b= 180 b+ c= 180

a = 180 b c= 180 b Tereore, a= c

b+ c = 180 c+ d= 180

b= 180 c d= 180 c Tereore, b = d

Example 11.1-c



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Note: When two lines intersect, the adjacent angles are supplementary (sum to180) and opposite angles are congruent (equal).

Opposite and Adjacent Angles

Determine the measures o the three unknown angles in the ollowing diagram:

a

b c

=52

(i) Since angle a is adjacent to angle 52 = , it is supplementary to angle a.

Tereore, 180 52 128= = a .(ii) Since angle bis opposite to angle

52 = , it is congruent to angle b.

Tereore, 52b = .(iii) Since angle cis adjacent to angle 52 = , it is supplementary to angle c.

Tereore, 180 52 128= = c .Parallel Lines and Transversal Angles

When a line (called the transversal) intersects two distinct parallel lines, theangles it orms with each o the two parallel lines are congruent.

Parallel lines never

intersect even whenextended. Tey areidentified by arrows.

Angles a, b, c, anddare congruent toangles e,f,g, andh, respectively.

b ac d

f eg

Transversal

Two distinct

parallel lines h

[a =e, b =f, c =g, d =h]

Tis means that there are special relationships with special names betweenthe angles ormed by the transversal and each o the parallel lines, as classifiedbelow:

Correspondingangles

( = )

Te angles ormed on the same

corner o the intersectionbetween the transversal and eacho the parallel lines are calledcorresponding angles, and theyare congruent.

Example 11.1-d



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Co-Interiorangles

( 180 + = )

Te angles ormed on the sameside o the transversal and on theinterior o the parallel lines arecalled co-interior angles, andthey are supplementary.

Alternate angles

( = )

Te angles ormed on oppositesides o the transversal and onthe interior o the parallel linesare called alternate angles, andthey are congruent.

For example, consider the angles ormed by two distinct parallel lines and atransversal.

b a c d

f eg h

Alternate angles have a patternthat look like the letter Z:

Co-Interior angles have a patternthat look like the letter C:

Corresponding angles have a pattern

that look like the letter F:

Opposite angles have a pattern

that looks like the letter X:

Opposite Angles are Equal Corresponding Angles are Equal

a = c b = d e = g f = h

a = e b = f c = g d = h

Co-Interior Angles are Supplementary Alternate Angles are Equal

d + e=180 c + f = 180

d = f c = e

Identifying Relationships Between Angles

State the relationship to angle o each o the five unknown angles a, b, c, d,and eidentified in the ollowing diagram, and also state whether the angle iscongruent or supplementary to :

e

a

b

dc

Example 11.1-e



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Solution (i) Angle ais opposite angle , hence it is congruent to .

(ii) Angle bis adjacent to angle , hence it is supplementary to .

(iii) Angle cis co-interior to angle , hence it is supplementary to .

(iv) Angle dis alternate to angle , hence it is congruent to .(v) Angle eis corresponding to angle , hence it is congruent to .

Calculating the Measure of Transversal Angles

Calculate the angle measure o the five unknown angles identified in Example11.1-e, given that angle = 105:

Solution (i) Since angle a is congruent to , a= 105.

(ii) Since angle bis supplementary to , b= 75.

(iii) Since angle cis supplementary to , c= 75.

(iv) Since angle dis congruent to , d= 105.

(v) Since angle eis congruent to , e= 105.

An Application of Transversal Angles Intersections of Roads

Alder Road, Birch Street, and Cedar Avenue are all straight roads that run indifferent directions their intersections orm a triangle. I Alder Road intersectsBirch Street at an angle o 72 and Cedar Avenue at an internal angle o 47, bothas measured rom within the triangle. Using the angle relationship learned in thissection, find the angle o intersection between Birch Street and Cedar Avenue.

Solution Step 1: Draw a diagram representing theintersection o roads and mark theknown angles. Name the triangleas XYZ and let be the angle ointersection between Birch Streetand Cedar Avenue.

Alder Road

Birch Street Cedar Avenue

72 47

X

Y Z

Step 2: o make use o the anglerelationships that we learned inthis section, draw an imaginaryroad, parallel to Alder Road,thatruns through X, the intersectiono Birch Street and Cedar Avenue.

Alder Road

Birch Street Cedar Avenue

72 47

X

Y Z

a b

Step 3: Calculate the alternate transversal angles that are ormed and use those tocalculate the angle o intersection between Birch Street and Cedar Avenue. a= 72 (Alternate Angle) b= 47 (Alternate Angle)

Example 11.1-f

Example 11.1-g

0



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Step 4: Te three angles a, , and bat the vertex X o the triangle XYZ mustbe equal to 180 (angles in a straight line).

180a b+ + = 180 a b =

= 180 72 47 = 61

Tereore, the angle o intersection between Birch Street and Cedar Avenue is 61.

Te above example demonstrates that the three internal angles o a triangle mustadd up to 180. We will examine this urther now as we begin to analyze planefigures in the next section.

11 .1 Exercises Answers to odd-numbered problems are available online



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11.1 Exercises Answers to odd-numbered problems are available online

1. Draw and label the following geometric objects:

a. Line EF b. Line segment GH c. Ray JK

2. Draw and label the following geometric objects:

a. Line ST b. Line segment UV c. Ray XW

3. Identify and name the following geometric objects:

Grada.

A B

Nu b. L

M

c.

Y

Z

4. Identify and name the following geometric objects:

Grada.

C

D

N m b.

O

N

c.Q R

For the figures shown in problems 5 to 8, answer the following questions:

a. Name the angle using the three-letter naming convention (e.g. ABC). b. Classify the angle as acute, right, or obtuse.

c. Determine the approximate angle measure using a protractor. d. Calculate the supplement and complement (if applicable) of the angle.

5. Grada.

P

D

C

N b.

Q

A B

6. Grada.

MO

N

b.

N

M

L

7. Grada.X

Y

Z

Numb.

QR

P



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2

8. Grada.

U

TS

Numb.

X

V

W

9. Determine the measure of the angle that is complementary to: a. 32.5 b. 18

10. Determine the measure of the angle that is complementary to: a. 83.1 b. 5

11. Determine the measure of the angle that is supplementary to: a. 123.4 b. 89

12. Determine the measure of the angle that is supplementary to: a. 7.8 b. 92

For the figures shown in Problems 13 to 16, determine which pairs of angles are congruent:

13. Grada. A C

O

B D

Numb.

q

ts

r

14. Grada.

O

PQ

M N

a Numb.

u

vw

x

yz

15.

ab

cd

e

f g

16.h

m n o

p

i j

k



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For the figures shown in problems 17 and 18, determine the value of the unknown angles, given theangle measure of :

17. Grada.

= 132

a

b

c d

Numb.

a

b c

= 120

= 70

18.Grad

a.a

b d

c

= 26

Num

b.

a

b

c

80

145

19. Grada.

a

bc

d

= 59 = 58

Numb.

a b

c

80 20

50

20. Grada.

a

bd

c

= 72 = 96

Numb.

a

b c30 25

120



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21. A small island is situated at the south of Lois Lake, separated from the mainland by two tributariesof Lois Lake: Crag Creek to the West and Slip Stream to the East. A straight highway called RiverRoad connects the island to the mainland in either direction. Te River Road bridge over CragCreek forms an angle of 77 with the creek, and the bridge over Slip Stream forms an angle of 71with the stream, both on the islands side. Assuming that both Crag Creek and Slip Stream are

fairly straight, determine the angle that they form with each other when they branch off Lois Lake?

22. Te South-West corner of the intersection of Main and Queen forms an angle of 104. Furtherdown Main Street, the South-West corner of the intersection of Main and King forms an angle of63. Determine the acute angle formed by the intersection of Queen and King, assumingthat all three roads are perfectly straight.

For the figures shown in Problems 23 and 24, use transversal angles and the fact that the sum of the threeinternal angles of a triangle always equals 180 to:

23. Grada. Calculate the value of

136

92

Numb. Calculate the value of a, b, and c

a

b

c

110

50

24. Grada. Calculate the value of

63

121

Numb. Calculate the value of a, b, and c

a

b

c

80

120

4



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11.2 Classification and Properties of Plane Figures

IntroductionTe study o Geometry that deals with the objects or figures that are flat(2-dimensions) and that can be drawn in the Cartesian plane is known as PlaneGeometry. In Plane Geometry, we study the properties and relations o planefigures such as triangles, quadrilaterals, polygons, and circles. A plane figure iscontinuous and closed, meaning that it can be drawn without lifing the pencilrom the page and that the start-point is the same as the end-point o the object.

A ew examples o plane figures are shown below:

Exhibit 11.2-a Example of Plane Figures

POLYGONS

A polygon is a plane figure that is created by joining a finite number o linesegments together at their vertices; i.e., a polygon is a plane figure that is boundby three or more straight edges, known as sides. Te first 6 shapes in Exhibit11.2-a are polygons. Te circle (i.e., the 7th shape) in Exhibit 11.2-a is not apolygon, as it is not ormed by joining a finite number o line segments together.However, the circle is a special shape and you will learn o its properties in thenext section.

A simple polygonis a polygon which does not intersect itsel. Te first 5 shapesin Exhibit 11.1-c are simple polygons. A polygon that is not simple (i.e., itintersects itsel) is called a complex polygon. Te hourglass shape (i.e., the 6thshape) in Exhibit 11.2-a is an example o a complex polygon.

A convex polygonis a simple polygon whose internal angles are all less than180. Te first 4 shapes in Exhibit 11.2-a are convex polygons. Every simplepolygon that is not convex is called a concave polygon. Te star shape (i.e. the5th shape) in Exhibit 11.2-a is an example o a concave polygon.

A regular convex polygon is a convex polygon whose sides are all the samelength and whose internal angles have the same measure. Te first 3 shapes inExhibit 11.2-aare regular convex polygons.

Polygons are named according to the number o sidesthat they have. Te firsteight regular convex polygons are shown below:

Copyright 2013 Vretta Inc. 11.2 Classification and Properties of Plane


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6

Exhibit 11.2-b Names of first eight regular convex polygons

An internal angle o a simple polygon is an angle at a vertex where two linesegments meet, as measured rom the inside o the simple polygon.

I lines are drawn rom one vertex o ann-sided polygon to a vertex across romit, there will be (n 2) triangles that can be drawn within the polygon. Forexample:

Tereore, the sum o the internal angles o any (n-sided polygon) = (n 2)180.

Tis is known as theInternal Angles Teorem (IA) Part 1.

Tereore, every internal angle in a regular n-sided convex polygon =

( 2) 180n

n

.

Tis is known as the Internal Angles Teorem (IA) Part 2.

Internal Angles of Regular Convex Polygons

Using the Internal Angles Teorem, calculate the measure o each internal angleo the first eight regular convex polygons, listed in Exhibit 11.2-b

Solution Using( 2) 180n

n

=

Name of Polygon Number ofsides (n)

Each internal angle

(i) riangle 3 (3 2) 180 180

603 3

= = =

(ii) Square 4(4 2) 180 360

904 4

= = =

(iii) Pentagon 5 (5 2) 180 540

1085 5

= = =

Triangle (3 sides) Square (4 sides) Pentagon (5 sides) Hexagon (6 sides)

Heptagon (7 sides) Octagon (8 sides)

Decagon (10 sides)Nonagon (9 sides)

n = 5

(3 Triangles)

n = 7

(5 Triangles)

Example 11.2-a



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Solutioncontinued

(iv) Hexagon 6(6 2) 180 720

1206 6

= = =

(v) Heptagon 7 (7 2) 180 900

128.67 7

= =

(vi) Octagon 8 (8 2) 180 1080

1358 8

= = =

(vii) Nonagon 9 (9 2) 180 1260

1409 9

= = =

(viii) Decagon 10 (10 2) 180 1440

14410 10

= = =

Verifying a Special Case of the Internal Angles Theorem

A trapezoid is any our-sided convex polygon with one pair o opposite sidesthat are parallel to each other (see diagram below). Use the properties o parallellines and transversal angles to prove that the Internal Angles Teorem (Part 1)holds true or all trapezoids.

A B

DC c

a b

d

Solution Te IA-1 states that the sum o the internal angles o any 4-sided convex polygonis (4 2) 180 = 360.

Since the line segmentABis parallel to the line segment CD, angles aand careco-interior, transversal angles.

i.e., a + c = 180 (1)Similarly, b and d are also co-interior angles, hence supplementary.

i.e., b + d = 180 (2)Adding (1) and (2): a + b + c + d = 360.Tereore, the sum o all our angles in the trapezoid is 360, which equals the

result o the IA-1 ormula. Hence, the ormula is valid or all trapezoids.

An external angle o a simple polygon is the external angle obtained by extendingone o the sides at a vertex where two line segments meet, and measuring theangle ormed outside the simple polygon.

For example,

an n-sided Polygon will haven external angles.

Example 11.2-b



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Tere are nvertices and the sum o the internal angle and the external angle ateach vertex is supplementary, 180.

ie., the sum o all interior and exterior angles o an n-sided Polygon = n 180.However, the sum o all the interior angles = (n 2) 180.

Tereore, the sum o all the exteriorangles o an nsided Polygon = n 180 (n 2)180= n 180 n 180 + 2 180= 360

Tereore, the sum o the external angles ormed by extending the sides o anyn-sided, simple polygon = 360.

Tis is known as the External Angles Teorem (EA) Part 1.

Tereore, every external angle in a regular n-sided convex polygon =360

n

.

Tis is known as the External Angles Teorem (EA) Part 2.

External Angles of Regular Convex Polygons

Using the 2ndpart o the External Angles Teorem, calculate the measure o eachexternal angle o the first eight regular convex polygons, listed in Exhibit 11.2-b.

Solution Using360

n

=

Name of PolygonNumber of

sides (n)Each internal angle

(i) riangle 3 360

1203

= =

(ii) Square 4 360 904

= =

(iii) Pentagon 5 360

725

= =

(iv) Hexagon 6 360

60

6

= =

(v) Heptagon 7 360

51.47

=

(vi) Octagon 8 360

458

= =

(vii) Nonagon 9 360

409

= =

(viii) Decagon 10 360

3610

= =

Note: Te internal angle and the external angle at every vertex of a convex polygonare supplementary, as each pair of internal and external angles together form astraight line.

Example 11.2-c

8



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An Application of the External Angles Theorem Navigation

A plane takes off, heading due west. Shortly afer take-off, it turns 60 to the north(clockwise). Later on, it turns another 75 in the same (clockwise) direction. Aew minutes later, it makes another turn o 80 in the same direction. Finally, it

makes one last turn in the same direction and heads back to its take-off point,flying in to the airstrip bearing due south. Find the bearing change (change inangle) o the final turn.

Solution Based on the given solution, draw a picture o the situation:

60

75

80 ?

Using EA 1, the sum oall exterior angles = 360.

Let the final external angle be .

i.e., 60 + 75 + 80 + + 90 = 360

= 360 305 = 55.

Tereore, the bearing changeo the final turn is 55.

Classification and Properties of Triangles

We will now examine one type o convex polygon triangles. A triangle (literallymeaning three-angles) is any polygon with 3 sides and 3 internal angles. Wewill now look at the different sub-categories and classifications o triangles, andthe various properties o the figures.

Using the IA-1, the sum o the three internal angles

o a triangle equals (3 2) 180 = 1 180 = 180.Tereore, since the sum o the internal angles equals180, each internal angle must be less than 180, whichmeans every triangle is a convex polygon.

Tere are two ways to classiy triangles by anglemeasure and by side length.

Classification of Triangles by Angle Measures

Acute riangle

< 90

< 90 < 90

A triangle with all three angles lessthan 90 (acute angle) is called an acutetriangle.

Right riangle = 90

A triangle with one angle at 90 (rightangle) is called a right triangle.

Since the sum o the three angles is180 and one angle is 90, this meansthat the other two angles must addupto 90 thereore, they are acute andcomplementary.

Example 11.2-d

A

B C

A + B + C = 180



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Obtuseriangle

> 90

A triangle with an angle greater than90 (obtuse angle) is called an obtusetriangle.

Since the sum o the three angles is 180and one angle is greater than 90, this

means that the other two angles mustadd upto less than 90 thereore, theyare acute.

Equilateralriangle

= 60

A triangle that has sides o equallengthsis called an equilateral triangle.

Since an equilateral triangle is a regularpolygon o 3 sides, by the IA-2, eachangle 60. Tereore, every equilateraltriangle is also an acute triangle.

Isoscelesriangle

< 90

A triangle that has 2 sides o equallengths is called an isosceles triangle.

Te angles opposite the equal sides oan isosceles triangle will have equalmeasure.

An isosceles triangle may be acute,right, or obtuse (but the equal angleswill be acute).

Scaleneriangle

A triangle withsides o different lengthsis called a scalene triangle.

A scalene triangle may be acute, right,or obtuse

Exhibit 11.2-c Classification of Triangles

AcuteTriangles

RightTriangles

Based onAngle

Measures

ObtuseTriangles

EquilateralTriangles

IsoscelesTriangles

Based onSide

Lengths

ClassifyingTriangles

ScaleneTriangles

0



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Classifying Triangles

Classiy the ollowing triangles by angle measure and by side length:

(i)

5

5

(ii)

6

66

(iii) 18

4

(iv)

80

4

4

(v) 106

37

(vi) 26

64

Solution(i) Right and Isosceles triangle.

(ii) Acute and Equilateral triangle.(iii) Obtuse and Scalene triangle.

(iv) Acute and Isosceles triangle.

(v) Obtuse and Isosceles triangle.

(vi) Right and Scalene triangle.

Calculating Unknown Angles in a Triangle

Use the IA-1 to calculate the measure o the unknown angle in each o theollowing triangles. Ten classiy the triangle, both by side length and by angle:

(i) XYZ, YXZ = 30,

XYZ = 120

(ii) ABC, BAC = 35, ACB = 55(iii) RS, RS = 60, SR = 60

Solution (i) XZY = 180 (30+ 120) = 30Since Y > 90, XYZ is Obtuse, and since YXZ = XZY ,XYZ is Isosceles.

(ii) ABC = 180 (35+ 55) = 90Since ABC = 90, ABC is Right, and since no angles are equal,ABC is Scalene.

(iii) SR = 180 (60+ 60) = 60Since all angles are less than 90, RS is Acute, and since all angles areequal, RS is Equilateral.

Constructing Triangles

Given that BAC = 37, a = 5 cm, and b = 8 cm, draw two different trianglesABC, such that

(i) ABC is an obtuse, isosceles triangle

(ii) ABC is an acute, scalene triangle

Example 11.2-e

Example 11.2-f

Example 11.2-g



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Solution(i)

(ii)

An Application of Triangles Distances Between CitiesTe flying distance rom oronto to Sudbury is the same as it is rom oronto toOttawa approximately 345 km. Te angle rom oronto between Sudbury andOttawa is 76. What kind o triangle is created between the three cities?

Solution Since the distance between oronto and Sudbury is equal to the distance betweenoronto and Ottawa, the angles opposite to these sides, as represented in thediagram, are equal.

Let be the equal angles. 76 180 + + = 2 180 76 104 = = 10452

2

= =

Tereore, the angles are 52, 52, and 76.

Tereore, two sides are equal and all three angles are less than 90. Tereore, thetriangle created is an acute, isosceles triangle.

Classification and Properties of Quadrilaterals

We will now examine another class o convex polygons and their properties convex quadrilaterals.

AB

C

a = 5 cm b = 8 cm

37

37

AB

C

a = 5cm b = 8 cm 37

Example 11.2-h

345 km

345 km76

Su ury

Ottawa

Toronto

2



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A quadrilateral(literally meaning our-sided) isany polygon with 4 sides and 4 internal angles. Inthis section, we will examine convex quadrilateralsonly, in which each o the internal angles is less than180.

Tere are two main classes o quadrilaterals: parallelograms, which have specialproperties, and non-parallelograms.

A parallelogram is a quadrilateral with oppositesides that are parallel. As a result, in a parallelogram,the opposite sides are equal, the opposite angles areequal, and the adjacent angles are supplementary(+ = 180).

Classification of Quadrilaterals that are Parallelograms

Within the class o parallelograms, there are several sub-classes:

Rectangle

Opposite sides are parallel.

Opposite sides are oequal lengths.

All angles are equal (90).

Rhombus


All sides are o equal lengths.

Opposite angles are equal.

Square


All sides are o equal lengths.

All angles are equal (90).

Note: A parallelogram that is neither a rectangle, nor a square, nor a rhombus isknown simply as a parallelogram.

Classification of Quadrilaterals that are Non-ParallelogramsWithin the class o non-parallelograms, there are two sub-classes:

rapezoid One set o opposite sides are parallel.

Parallel sides are not o equal lengths.



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4

Kite

wo sets o adjacent sidesare o equal lengths.

One pair o opposite angles is equal.

Diagonals meet at right angles.

Note: A quadrilateral that is a non-parallelogram, which is neither a trapezoid nora kite is known as a general quadrilateral.

Classifying Quadrilaterals

ParallelogramQuadrilaterals

(Two Pairs

Non-Parallelogram-Quadrilaterals

(One or No Pair

One pair of

parallel sides

No pair of

parallel sides

All sides areof equallengths

Opposite sidesare equal and

parallel

of Parallel sides)of Parallel Sides)

All anglesare equal to

90

SQUARES

Oppositeangles are

equal

RHOMBUS

All anglesare equal

to 90

RECTANGLE

Oppositeangles are

equal

PARALLELOGRAMS

Parallel sidesare not of equal

lenghts

TRAPEZOIDS(TRAPEZIUM)

Two sets ofadjacent

sides are equalKITES

One or no set ofadjacent sides

are of equal

lenghtsGENERAL

QUADRILATERALS

Exhibit 11.2-d Types of Quadrilaterals

Classifying Quadrilaterals

Classiy the ollowing quadrilaterals:

(i)

(ii) (iii)

Solution (i) One pair o opposite side is parallel rapezoid

(ii) All sides are o equal lengths and all angles are equal (90) Square

(iii) Opposite sides are parallel Parallelogram

Calculating Unknown Angles in a Quadrilateral

Use the IA-1 and the properties o various quadrilaterals to calculate the

Example 11.2-j



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measure o the unknown angle(s) in each o the ollowing:

(i) WXYZ is a general quadrilateral, XWZ =72, WXY = 106, XYZ = 55.

(ii) ABCD is a parallelogram and ADC = 25.(iii) QRS is a kite, where QR = 80and RS = 50.

Solution Using the IA-1, the sum o all our angles o a quadrilateral equals(4 2) 180 2 180 360 = = .

(i) Since WXYZ is a quadrilateral, the our anglesadd up to 360.Tereore, WZY

Z = 360 (72 106 55 ) 360 233 127 + + = =

(ii) Since ABCD is a parallelogram, theopposite angles are congruent and adjacentangles are supplementary.Tereore, ABC = ADC B = D= 25and BAD = BCD A = C= 180- 25=155

(iii) Since QRS is a kite, one pair o opposite angles are equal,and since RQ Q S RS, QRS R = QS. Let represent the measure o each o thetwo equal angles:80 + +50 + = 3602= 360 - 130 = 230

=230

2

=115

Tereore, QRS = QS R = = 115.Identifying Quadrilaterals Based on Angle Measures

For the ollowing quadrilaterals, use the IA-1 to find the missing angle measureand then classiy each quadrilateral based on their angle measures:

(i) EFGH, given that E = 64, F = 116, and H = 90(ii) MNOP, given that M = 112, N = 58, and O = 112(iii) SUV, given that S = 45, U = 45, and V = 135

72 106

X

Y

Z

W 55

25

A B

CD

80

Q

R

S

T

50



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6

Solution Using the IA-1, the the sum o all our angles o a quadrilateral is 360.

(i) G = 360 (64 116 90 ) 90 + + =

Tereore, G + H = 180

E + F = 180 Since adjacent angles are supplementary,one opposite pair o sides is parallel.Since opposite angles are not congruent, both oppositesides are not parallel. Hence, EFGH is a trapezoid.

(ii) P = 360 (112 58 112 ) 78 + + = Te act that there is one pair o congruentopposite angles does not give us enoughinormation to determine the type oquadrilateral. However, we can narrow downthe choices to two: a kite or a general

quadrilateral. ***Draw a kite.(iii) T = 360 (45 45 135 ) 135 + + =

Since both pairs o opposite angles arecongruent, SUV is either a parallelogram or arhombus (we cannot tell which withoutknowing the side lengths). ***Draw a rhombus.

Constructing Quadrilaterals

Jeremy labels a point A on his paper and draws a

straight line 20 cm long to another point B. Fromthere, he uses a compass to measure a 90 angle rom ABand draws a line rom point B to a third point C,that is perpendicular to ABand 15 cm long. How manydifferent types o quadrilaterals can Jeremy create byplotting his ourth point D and then connecting the line segments CDandDA?

Solution Since the lengths o two sides are different, Jeremy cannot create a square or arhombus; since the angle is a right angle, he cannot create a general parallelogram,either. However, he can create 4 other kinds o quadrilaterals:

Option A: Jeremy can create a rectangleby measuringout another right angle rom point C anddrawing a line segment CDparallel to ABand 20 cm long. 20cm

15cmA B

CD20cm

Option B: Jeremy can create a trapezoid bymeasuring out another right anglerom point C and drawing a linesegment CDparallel to ABbut oa length other than 20 cm.

20 cm15cm

A B

CD30cm

64 116

90

E

F

G

H

N

O

P

M

58112

112

45

45

S T

UV

Example 11.2-l

20 cm

15 cm

A B

C



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Solutioncontinued Option C: Jeremy can create a kiteby drawing a

dashed line rom point A to point C,drawing a line segment rom point B toa ourth point D that is perpendicular toACand twice the length rom B to AC.

20cm

20cm15cm

15cm

A B

C

D

Option D: Jeremy can create a general quadrilateralby placing point D in any locationthat is any distance, other than 15 cm,away rom C and not parallel to AB.

20cm

15cmA B

C

D

11 .2 Exercises Answers to odd-numbered problems are available online



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Use the first part o the Internal or External Angle Teorem (as appropriate) to determine the measure

o the unknown angle or the figures shown in Problems 1 to 4:

1. 5-sided shape, internal angle

122

146

97

114

2. 8-sided shape, internal angle

138

152

135

127

105

131

144

3. 9- sided shape, external angle

43

61

26

38

52

84

4. 6- sided shape, internal angle

48

63

19

68

70

Use the second part o the Internal and External Angle Teorems (or regular polygons) to determinethe measure o the internal and external angles or the figures shown in Problems 5 and 6:

5. a. Dodecagon (12 sides) b. Icosagon (20 sides)6. b. Hexadecagon (16 sides) a. Hectogon (100 sides)

For the problems 7 and 8, the internal or external angle measure o a regular polygon is given. Use thesecond part o the External Angle Teorem to determine the number o sides in the regular polygon.

7. a.

External angle measure is 12 b.

Internal angle measure is 175 - Hint: first find thesize o the external angle measure.

8. a. External angle measure is 15 b. Internal angle measure is 175 - Hint: first find thesize o the external angle measure.

9. A sailboat in a race heads West on the opening stretch o the race. At the first checkpoint, the boatmakes a 66 turn to port (lef) and sails towards the second checkpoint, where it then makes a 112turn to port. It then continues toward the third checkpoint, makes a 75 turn to port and heads to theourth and final checkpoint, where it makes a final turn to port until it aces due West again, and headsback toward the starting line to complete the circuit. Determine the degree measure o the final turn.



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8

10. Te owners o a house with a backyard in the shape o an irregular hexagon (6 sides) are putting upa ence around their yard, except or one side o their yard which is tree-lined. Using a city survey,the owners have laid down guidelines and have determined the angles at each o the corners,except where the ence meets the trees. Use the diagram and measurements below to determine theunknown angle.

11. In an acute isosceles triangle, the measure o the unique angle is 2 degrees less than three-fifhs othe measures o each o the other two angles. Find the degree measures o all three internal angles.

12. In a parallelogram, the degree measure o the larger pair o congruent angles is 5 degrees morethan 6 times the degree measure o the smaller pair o congruent angles. Determine the degreemeasures o both pairs o congruent angles.

13. Use the Internal Angle Teorem (Part 1) or triangles to show that any external angle o a triangleis equal to the sum o the opposite two internal angles.

14. In an obtuse scalene triangle, the measure o the larger internal acute angle is 60. Te measure o

the external angle to the obtuse angle is 6 degrees greater than 3 times the measure o the smallestinternal acute angle. Determine the measure o the internal obtuse angle.

Classiy the triangles shown in Problems 15 and 16 by side length and angle measure:

15. Grada. A

B C

88

46

Gradb.D

E F

5

13

12

c.G J

H

30 30

16. Grada.

N

M

P

60 60

Gradb.

R

Q

S

10

10

c.

U

T

V

31

23

125

150

Fence

145

x



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Classiy the quadrilaterals shown in Problems 17 and 18:

17. Grada.A B

C D

b.

E F

G H

15 cm

15 cm

7 cm7 cm

c.L

M

N

K

98

98

18. Grada.

P Q

R S

b.

T U

V W

87

5

5

5 5

c.

W

Y Z

X

120 150

30

In Problems 19 to 22, determine the missing angle(s) or each quadrilateral ABCD:

19. ABCD is a rectangle

20. ABCD is a rhombus, with A = 7721. ABCD is a kite, with AB = BC , AD= DC, A = 105, and D = 5222. ABCD is a trapezoid, with AB parallel to CD, A = 93, and B = 116In each o the ollowing exercises, classiy the quadrilateral ABCD based on the properties given:

23. AB = BC= CD= DAand A = 9024. AB = CD, BC = DA, and A = 10525. AB = BC = CD= 15 cm, DA= 27 cm, and BC is parallel to DA

26. AB = BC = 20 cm, AD= DC= 30 cm, and A = C.In Problems 27 to 30 state the names o all the possible quadrilateral based on the given property:

27. a. 4 equal angles b. 4 equal sides

28. a. 4 right angles b. No equal sides

29. a. 2 pairs o parallel sides b. 2 pairs o equal angles

30. a. 1 pair o paralles sides b. 1 pair o equal angles



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In Problems 1 and 2, for each pair of similar trianglesname the proportional sides and congruent angles.

1. a. ABC~ DEF b. PQR ~ RST

2. a. UVW ~ XYZ b. GHI ~ JKL

In Problems 3 and 4,for each pair of congruent triangles, name the equal side and equal angles.

3. a. ABC,XYZ b. DEF,RST

4. a. UVW,JKL b. XYZ,JKL

For Problems 5 to 10, identify the pairs of triangles that are similar and state the rule used todetermine this.

5. a.4 6

8

Grb.ad 5

7

9

rc. 6

9

12

6. a.

45

70Grb.ad

70

60 rc.

65 70

7. a.10

6Grb.ad

8

8

rc. 3

5

8. a.74

7

Grb.ad

rc.

6060

9. a.

3

9

70

18

Grb.ad 7.5 10

15

rc.92

62

Copyright 2013 Vretta Inc. 11.3 Similar and Congruent Tr


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0

10. a.

10

12

Grb.ad

15 12

rc. 8

10

For Problems 11 to 16, identify the pair of triangles that are congruent and state the rule used todetermine this.

11. a. Grb.ad

rc.

12. a. Grb.ad

rc.

13. a. Grb.ad

rc.

14. a. Grb.ad

rc.

15. a. Grb.ad

rc.

16. a. Grb.ad

rc.

For Problems 17 to 22, determine whether each pair of triangles is congruent, similar, or neither.

17. 10

14165

87

18. 11

59

157

12



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19.

8 1813.5

6 20.

457

5 45

7

5

21. 13 22.

In the following problems, triangles and quadrilaterals are divided into two triangular pieces.Determine if the resulting pieces are congruent, similar, or neither.

23.

30

30

24.

100

100

25. 26. 7.5 cm

7.5 cm

In Problems 27 to 30, solve each pair of similar triangles completely, calculating the unknown sidelengths (rounded to the nearest tenth as necessary) and angle measures.

27.

A

B

C

D E

F

5

9

83

41

6

4

x

y

28.

6

5

9

32

39

H

I

LK

J

x

y

G

29.

N

O

P 60

8

Q

5

4

7

M

x

y

30. S9

T

R

U

15

45

45

45

25

25

25

20

V

w

y

x

z



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2

31. A flagpole casts a 3.4 m shadow. Melanie, who is 160 cm tall, stands beside the flagpole, and hershadow of 64 cm long. Draw a diagram and calculate the height of the flagpole.

32. A tree casts a 4.5m shadow. At the same time, a stick 55 cm casts a shadow 90 cm long. Calculatethe height of the tree in metres and centimeters.

33. In an outdoor play at night, a spotlight isplaced 8 m behind a sheet that is 5.5 m high.As an actress, who is 1.65 m tall, walks fromthe sheet towards the spotlight, it casts hershadow onto the wall. How far away fromthe spotlight is the actress when her shadowis the entire height of the sheet?

5.5 m

1.65 m

x

8 m

34. A man is standing 12 m away from lamppost. If his shadow is 2.2 m long, how tallis he?

10 m

h

2.2 m

12 m

35. A streetlight situated 7.4 m above the streetcasts a shadow on a pedestrian out taking alate-night walk. If the pedestrian is 1.8 mtall, how long is his shadowwhen he is 7 maway from the streetlight?

5.5 m

1.8 m

x

7 m

36. Two ladders of lengths 4 m and 9 m areleaning at the same angle against a wall. Ifthe 4 m ladder reaches 3.2 m up the wall,how much further up the wall does the 9 mladder reach? 3.2 m

y

4 m

9 m

37. Calculate the heightyin the diagram below.

6 m

y

15 m

38. Calculate the distancexin the diagram below.

3 m

x

9 m



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11.3 Similar and Congruent Triangles

IntroductionGeometric shapes, also known as figures, are an important part o the study ogeometry. Recognizing and using congruent and similar shapes make calculationsand design work easier. For example, in most design work, rather than usingdifferent shapes, a ew shapes are copied and used in different positions and/orproduced in different sizes to complete the design.

When a shape is obtained rom another figure by means o enlargement orreduction, its size will be different rom the original one, but it remains the sameshape as the original one.

Similar figures have the same shape and retain the same angles at corresponding

vertices (congruent). Tey may or may not have the same size, but the lengths o thecorresponding sides will have the same ratios between the figures.

Congruent figures have sides with lengths that are in proportion (congruent) andequal angles at corresponding vertices (congruent).

It is important to note that two figures can be similar, but not congruent; however,two figures cannot be congruent and not similar.

Te triangle is one o the basic shapes in geometry. It is the simplest shape within aclassification o polygons. All triangles have three sides and three angles, but theycome in many different shapes and sizes.

In Section 11.2, you learned that the triangles can be classified as acute, obtuse, or

right based on their angles. riangles can also be classified as equilateral, isosceles,or scalene based on the lengths o their sides. In this section, you will learn thecharacteristics o a triangles sides and angles that are used to classiy pairs otriangles as being similar or congruent. Understanding these characteristics willallow or application o these concepts in real-world problems.

Similar Triangles

Similar figures must have the same shape, but their sizes may be different.

wo equal-sided polygons are said to be similar i all the corresponding anglesare equal in measure and the corresponding sides are proportional in length.

Te symbol or similar is ~

~ ~ ~Each pair o polygons shown are similar.

When writing the similarity relationship, the order in which the letters arewritten to represent the similar figures is very important.



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In similar figures, the corresponding angles are equal and the ratio ocorresponding sides are equal. Te ratio between the corresponding sides osimilar figures are expressed as a raction and is called the scale or scale actor.

For example, ABC is similar to DEF,

A

B C

2 3

4

100

50

30

D

E F

4 6

8

100

50 30

2

~ABC ~ DEF A =D, B =E, and C = F

Corresponding angles are equal.

(i)

ABC ~ DEF

Corresponding sides are proportional.

AB BC AC 2= =

DE EF DF 1=

(ii)

Note: B refers to the segment joining points Aand B.

ABrefers to the length, which is a number measurement.

Any triangle is defined by six measures: three sides and three angles. However,it is not necessary to know all o the six measures to demonstrate that the twotriangles are similar. I any one o the ollowing three conditions are met, thenthe triangles are similar:

1. AAA (angle, angle, angle)I all three pairs o corresponding angles are the same (equal), then thetriangles are similar.

.

A

B C

D

E F

. A=D

B=EC=F

Tis is the same as AA (angle, angle) because i any two angles othe two triangles are equal, then the third angle must be equal.



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2. SSS (side, side, side)I all three pairs o corresponding sides are in the same proportion, thenthe triangles are similar.

AB AC BC= =

DE DF EFA

C

B

54

6

A

BC

810

12

2 5 4 6 1

10 8 12 2= = =

3. SAS (side, angle, angle)I two pairs o sides are in the same proportion and the included angle isequal, then the triangles are similar.

AB BC=

DE EF

D

E F

A

B CB =E

4. RHS (Right angle, hypotenuse, side)In right-angled triangles, i the hypotenuses and lengths o one pair ocorresponding sides are proportional, then the triangles are similar.

A

B C

D

E F

BC

AC=EF

DF

ABC =

DEF = 90

Rules for Similar Triangles

State the property that will prove that the ollowing pairs o triangles are similar.

(i) (ii) (iii)

8

4

8

4

2

33

6

2

4

30

45 30

45

Solution (i) SSS Property

(ii) SAS Property

(iii) AAA Property

Example 11.3-a



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Using Similar Triangles to Find the Unknown Length

(i) I PQR ~ XYZ, find XY and XZ

X

Y Z8

P

Q R3

4

5

(ii) I ABC ~ DEF, find AB and DF

A

B

C10

6

E F

D

15

12

Solution (i) PQR ~ XYZ

Tereore,PQ QR PR

= =XY YZ XZ

3 4 5= =

XY 8 XZ

3 4=

XY 8 4(XY) = 3(8)

3 8X Y 6

4

= =

4 5=

8 XZ

4(XZ) = 5(8)

5 8XZ 10

4

= =

Tereore, XY = 6 and XZ = 10.

(ii) ABC ~ DEF

Tereore,AB BC AC

= =

DE EF DF

AB 6 EF= =15 12 DF

AB 6=

15 12

12(AB) = 6(15)

6 15AB 7.5

12

= =

6 10=

12 DF

6(DF) = 10(12)

10 12DF 20

6

= =

Tereore, AB = 7.5 and DF = 20.

Example 11.3-b

2



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Identifying Similar Triangles

(i)

120

P

RQ

12

18

40

20

Y

XZ

96

(ii) A

BC

50

12

10

D

EF

50

6 5

Solution (i) Y = 180 (40 + 20) = 120

Q = Y (Equal angles)PQ 12

= = 2YZ 6

, QR 18= = 2XY 9

i.e.,PQ QR

=

YZ XY(Lengths o two sides are in the same proportion)

PQR ~ ZYX (SAS Property)

(ii) B = E (Angles equal)2=

EF

BC, 2=DF

AC

i.e., =EF

BCDF

AC(Lengths o two sides are in the same proportion)

SSA is not one o the rules or identiying similar triangles. Tereore, wecannot conclude that ABC is similar to DEF.Note: SSA (side, side, angle) is not sufficient to conclude that two trianglesare similar.

Congruent Triangles

Congruent figures must have the same shape and size.

wo equal-sided polygons are said to be congruent i all the correspondingangles are equal in measure and the corresponding sides are equal in length (i.e.the polygons are similar and they have equal side lengths).

Te symbol or congruent is .

Example 11.3-c



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Each ofhe above pairs o polygons are congruent.

A

B

C F

E

D ABC DEF

Once again, it is important to note that when the congruent relationshipis written, the order o letters representing the figures must be consistent toillustrate the equal corresponding angles and sides. For example:

ABC DEF (i) A = D(ii) B = E(iii) C = F

(i) AB = DE

(ii) BC = EF

(iii) AC =DF

Corresponding angles and sides are equal.

I any o the ollowing our conditions are met, then the triangles are congruent.Te first three are or any triangles and the 4 this or right triangles.

1. SSS (side, side, side)I the lengths o all three pairs o corresponding sides are equal, then thetriangles are congruent.

A

B C

D

E F

AB = DE

BC = EF

AC = DF

2. SAS (side, angle, side)I the lengths o two pairs o corresponding sides are equal and the angleincluded between these sides are equal, then the triangles are congruent.

B

A

C

E

D

F

AB = DEBC = EF

B = E

4



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3. ASA (angle, side, angle)I two pairs o corresponding angles are equal and the lengths othe contained sides are equal, then the triangles are congruent.

A

B

CD

E

F B = E C = FBC = EF

4. RHS (Right angle, hypotenuse, side)In right-angled triangles, i the hypotenuses and lengths o one pairo corresponding sides are equal, then the triangles are congruent.

A

B C

D

E F

AC = DF

BC = EF

Note: In a non-right-angled triangle, if the congruent pair of angles is not inclusivewithin the two equal, corresponding sides of two triangles, then the SSA rulecannot be used to determine congruency.

Determining Similarity/Congruency of Triangles

In the ollowing examples, determine whether thepairs o triangles are

congruent, similar, or neither:(i) (ii)

A

B C

5 cm

D

E

F

18 cm

7.5 cm

O

MN

Q

P

(iii) (iv)

G

H I

J

K

L 4 in 8 in

T

U V

4 in8 in

3535

W

X Y

Example 11.3-d


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Solution(i) Since

AB BC=

DE DF

and the angle between those two sides ( B and D) areboth right angles, ABC and EDF are similar by the SAS Property.(i.e.,ABC ~ EDF). Since the corresponding lengths o

the sides are different, the triangles are not congruent.

(ii) Since MNand PQ are parallel, we know that alternate angles are equal(i.e., N = P and M = Q), and thereore, the triangles MNOand QPO are similar by AA (i.e., MNO ~ QPO). Since lengths othe sides are not known, we cannot determine congruency.However,judging by the scale, it appears that they are not congruent.

(iii) Since GH = JK, GI = JL, and HI = KL,the triangles are congruent by the SSSProperty.

(iv) Since U = XW, V = WY, and V = Y, it may be tempting to labelVU and WYX assimilar triangles. However, it is obvious that UV XY,and thereore,they are not in proportion with the lengths o the other twoside; thus, the triangles are not similar (this is an example to demonstratethat the SSA criterion is not sufficient to demonstrate similarity).

Calculating Angle and Side Measures of Similar Triangles

Calculate the lengths o the unknown sides and unknown angle measures o theollowing pairs o similar triangles:

(i)

A

B C 16 cm

7 cm 28 45

D

E F24 cm

18 cm 45 28

(ii)

O

M

N

Q

P

2.5

4.0

3.5

3.0

7857

..

Example 11.3-e

6



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Solution (i) Since ABC ~ DEF,Te ratio o the corresponding sides is equal.AB BC AC

= =

DE EF DF

i.e.,AB 16 7

= =

18 24 DF

AB 16=

18 24 16 18

AB24

=

= 12.0 cm

16 7=

24 DF 7 24

DF16

=

= 10.5 cm

C = F = 45, E = B = 28Tereore, A = D = 180 28 = 107.

(ii) Beore we start computing the lengths o the unknown sides and unknownangle measures, it is important to ensure that the correct sides and anglesare compared. We can see that N = P and M = Q because theyare alternate angles. Tereore, P = 78 and Q = 57.Tereore, MNO ~ QPO by AAA; hence, the lengths o the sides o the

two triangles are proportional:MN NO MO

= =

QP PO QO.

i.e.,

2.5 3 3.5= =

4 PO QO

2.5 3=

4 PO

4 3PO

2.5

=

= 4.8 in

2.5 3.5=

4 QO

4 3.5QO

2.5

=

= 5.6 in

Finally, we can compute MON = QOP = 180 78 57 = 45.Tey are opposite angles as well (notice we cannot simplyreer to the angles as O, as that creates ambiguity).

Solving Application Problems using Similar and

Congruent TrianglesWe can use similar and/or congruent triangles to solve a variety o real-lieapplication problems whenit is difficult or impossible to calculate certain anglesor lengths.

Determining the Height of a Building using Similar Triangles

Te new science building at a College is 6 stories tall. Arianna wishes to knowhow tall the building is. She devises a method whereby, she and a riend measurethe length o the shadow that the building casts at 3:00 in the afernoon. Te

Example 11.3-f



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8

length o this shadow is 6.24 m. She then has her riend measure her height 165 cm and the length o her shadow 44 cm. Using this inormation, howcan Arianna determine the height o the science building?

Solution Since the sun meets all points on the ground in a close vicinity at the same angle

at any given time, and it hits both the building and Arianna (standing vertically)at the same angle, the shadows created by the building and Arianna orm similartriangles as shown in the diagram.

E

A

h

B 6.24 m C

D

F165 cm

44 cm

Tereore, the measurements o the buildings height and length o its shadowareproportional to Ariannas respective measurements:

165 cm

6.24 m 44 cm

h=

Solving this ratio or h, h= (6.24 m)44 cm165 cmS X = 23.4 m .

Tereore, the science building has a height o approximately 23.4 m.

Calculating the Distance across a Lake using Congruent Triangles

A lake is situated on a property in the country. A couple looking to purchase

the property wishes to know how long the lake is. How can they determine this(without getting wet)?

Solution Tey can each mark a point on either end o the lake (denoted A and B onthe diagram below), each a fixed distance away rom the edge o the lake, andmeasure the distance to a common point on one o the adjacent sides o the lake(denoted C). Tey can then each continue to walk the same distance again inthe same direction to another set o points on their property (denoted E and Drespectively), creating congruent triangles (by SAS). Ten, they can measurethe distance rom D to E, which will be the same distance as A to B, since thetriangles are congruent. Finally, by subtracting the distance rom each o A and

B to the edge o the lake, and they will have determined the length o the lake.

A B

E D C

Example 11.3-g



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1

11.4 Perimeter and Area of Plane Figures

IntroductionIn Section 11.2, we introduced the concept of plane figuresthat is, geometric objects that can be

drawn in the 2-dimensional Cartesian plane. In this chapter, you will learn two very important

measurements of those figures, specifically of certain special convex polygons and circles:

perimeter and area.

The perimeter(P) of a plane figure is the total length of the boundary of the plane figure. In a

polygon, the perimeter is the sum of the lengths of the line segments (sides) that form the boundary

of the polygon.

The area (A) of a plane figure is the amount of 2-dimensional surface that is enclosed within the

figure. Area is measured using square unitse.g., the square centimeter (cm2), the square meter

(m2), the square inch (in2), or the square foot (ft2); that is, the amount of surface occupied by

squares with the respective side lengths.

Squares and RectanglesA squareis a quadrilateral whose sides are all equal in length and angles are all right anglesthis

makes it a regular polygon. We denote the length of each side by the letters.

A rectangleis a quadrilateral whose angles are all right angles and opposite sides are equal in

lengthit is differentiated from a square by the property that the sides need not all be of the same

length. We denote the longer side by the letter l(for length), and the shorter side by the letter w

(for width).

Square P = 4s A =s2

RectangleP = 2l + 2wP = 2(l+ w)

A = l w

Perimeters and Areas of PlaneGeometric Figures

11.4

s

l

s


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2

Chapter 11

Calculating the Perimeter and Area of Squares and Rectangles

Find the perimeter and area of the following figures:

(i) (ii)

Solution

(i) The figure is a square.

Using P = 4s

P = 4(12) = 48 cm.

Using A =s2

A = (12)

2

= 144 cm

2

Therefore, the perimeter is 48 cm and the area is 144 cm2.

(ii) The figure is a rectangle.

Using P = 2l + 2w

P = 2(23.4) + 2(7.5) = 61.8 m.

Using A = lw

A = (23.4)(7.5)= 175.5 m2

Therefore, the perimeter is 61.8 m and the area is 175.5 m2.

Determining the Cost to Lay a Rectangular Garden

A rectangular garden is being built to be 6.5 m long and 3.2 m wide. The

fencing for the garden costs $2.95/m and the soil costs $6.25/m2. Calculate

the cost to lay the garden.

Solution

Using P = 2l + 2w

P = 2(6.5) + 2(3.2) = 19.4 m

Fencing cost =

2.95

(19.4)( ) $57.231

Using A = lw

A = (6.5)(3.2) = 20.8 m2

Soil cost =6.25

(20.8)( ) $130.001

Therefore, the total cost to lay the garden = $57.23 + $130.00 = $187.23.

Example 11.4-a

Example 11.4-b

12 cm 7.5 m

23.4 m

l= 6.5 m

w= 3.2 m


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Finding the Area of a Square Given the Perimeter

A square picture is being framed around its border with 180 cm of wood.

What area of glass is needed to frame the picture?

Solution

Perimeter of the square frameP= 180 cm.

Rearranging the formulaP= 4s, we get4

Ps

i.e.,180

45 cm4

s

UsingA=s2,A= (45)2= 2,025 cm2.

Therefore, the area of glass needed to frame the picture is 2,025 cm2.

Rhombuses and ParallelogramsA rhombusis a quadrilateral whose sides are all equal in lengthit is differentiated from a square

by the property that the angles are not right angles. We denote the length of each side by the letter

b, and the perpendicular height by the letter h.

Rhombuses, like squares, have four equal side lengths, which makes the calculation of the

perimeter of a rhombus is equal to that of a square.

The area of a rhombus is determined as follows:

Draw a perpendicular line from the top corner of the rhombus to its base. This is the height, h,

of the rhombus. Cutthe resulting triangle that is created and paste it on the opposite side. The

result is a rectangle with length band width h, as shown in the diagram below:

A parallelogramis a quadrilateral whose opposite sides are equal and parallel. It is differentiated

from a rectangle by the property that the angles are not right angles. We denote the length of thebaseby the letter b, the length of the slantside by the letter a, and the perpendicular heightby the

letter h.

The calculation of the perimeter of a parallelogram is equal to that of a rectangle, replacing the

letters l and w with aand b.

The area of a parallelogram is determined using the same procedure as that of a rhombus.

Example 11.4-c

b

bh

b

h


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Chapter 11

Rhombus P = 4b A =b h

ParallelogramP = 2a + 2bP = 2(a + b)

A = b h

Calculating the Perimeter and Area of Rhombuses and Parallelograms

Find the perimeter and area of the following figures:

(i) (ii)

Solution

(i) The figure is a rhombus.

Using P = 4band A = bhP = 4(25) = 100 cm

A = (25)(22) = 550 cm2

Therefore, the perimeter is 100 cm and the area is 550 cm2.

(ii) The figure is a parallelogram.

Using P = 2a+ 2band A = bh

P = 2(3.12) + 2 (0.96) = 8.16 m

A = (3.12)(0.75) = 2.34 m2

Therefore, the perimeter is 8.16 m and the area is 2.34 m2.

TrapezoidsA trapezoidis a quadrilateral with one pair of opposite sides that are parallelit is differentiated

from a parallelogram by the property that the other pair of opposite sides need not be parallel.

Since all four sides may have different lengths, we denote the length of the smaller parallel side by

the letter a, the length of the larger parallel side by the letter b, and the lengths of the other two

sides by the letters cand d. Again, we denote the perpendicular height by the letter h.

Example 11.4-d

25 cm22 cm

h

b

b

0.96 m

3.12 m

0.75 m

ha

b


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The perimeter of a trapezoid is the sum of the four side lengths, a, b, c, and d.

To calculate the area of a trapezoid, copy the trapezoid, rotate the image 180, and paste it to the

original trapezoid, as shown below. The result will be a parallelogram with an area of (a+ b)h.

The area of the trapezoid is half the area of the parallelogram =1

( )2

a b h .

Trapezoid P = a + b+ c+ d A = 1 ( )2

a b h

Calculating the Perimeter and Area of a Trapezoid

Find the perimeter and area of the following trapezoid:

Solution

Using P = a + b+ c+ d

P = 32 + 48 + 15 + 17 = 112 cm.

Using A =1

( )

2

a b h

A = 2560)14)(40()14(2

4832cm

Therefore, the perimeter of the trapezoid is 112 cm and the area of the

trapezoid is 560 cm2.

Determining the Cost of Fencing and Sodding a Trapezoidal Lawn

Example 11.4-e

Example 11.4-f

a

b

c d h

b

a b

a

48 cm

32 cm

17 cm15 cm 14 cm

a

b

c dh


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Chapter 11

A house on the corner of a crescent has a backyard in the shape of a

trapezoid, with the dimensions given on the figure below. If fencing costs

$25 per linear foot and sod costs $0.40 per square foot, how much will it

cost to fence and sod the backyard?

Solution

Using P = a + b+ c+ d

P = 85 + 120 + 50 + 61 = 316 ftFencing Cost = (316)($25) $7900

Using A =1

( )2

a b h

A = 25125)50)(5.102()50(2

12085ft

Sod cost = (5125)($0.40) $2050

Therefore, the total cost to fence and sod the backyard = $7,900 + $2,050 =

$9,950.

TrianglesA triangle is a closed figure formed by three sides and three internal angles. We use the letters a, b,

and cto denote the side lengths, and hto denote the height.

The perimeter of a triangle, regardless of whether it is acute, right, or obtuse, is the sum of the

three side lengths a, b, and c. P = a+ b+ c.

To calculate the area of a triangle (regardless of whether it is acute, right, or obtuse), copy the

triangle, rotate the image 180, and paste it to the original triangle as shown below. The result in all

120 ft

85 ft

50 ft 61 ft

b

cah

(i) Acute Triangle

b

ca = h

(ii) Right Triangle

b

cah

(iii) Obtuse Triangle


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three cases will be a parallelogram, with base band height h. Thus, the area of a triangle is half

that of a parallelogram:1

A2

b h .

Triangle P = a + b+ c A = 1

( )2

b h

Calculating the Perimeter and Area of a Triangle

Find the perimeter and area of the following triangles:

(i) (ii)

Solution

Using P = a+ b+ cand A =1

( )2

b h

(i) P = 16 + 38.4 + 41.6 = 96 cm

A = 21

(16)(38.4) 307.2 cm2

Therefore, the perimeter of the triangle is 96 cm and the area of the

triangle is 307.2 cm2.

(ii) P = 2(27.5) + 18 = 73 in

A = 21

(18)(26) 234 in2

b

h

b

ca h

Example 11.4-g

38.4 cm

16 cm41.6 cm

18 in

27.5 in

26 in

i

b

h

ii

b

h

iii

b

c

ah

b

ca


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8

Chapter 11

Therefore, the perimeter of the triangle is 73 in and the area of the

triangle is 234 in2.

Finding the Area of a Kite

Calculate the area of the kite shown in the figure below:

Solution

The kite consists of two identical triangles, each with a base of 35 cm and a

height of 9 cm.

Using A =1

( )2

b h

A = 21

35 9 157.5 cm2

Therefore, the area of the kite is 2 157.5 = 315 cm2.

Circles and Sectors

CirclesA circleis a closed plane curve such that any point on the curve lies within a fixed distance (called

the radius) from a fixed point (called the centre).

Example 11.4-h

35 cm

9 cm

9 cm

Exhibit 11.4-a A circle and its components.

Chord

Radius (r)

Diameter dCentre O

The radius (r) is the distance from the centre

point of the circle to the boundary of the circle.

A chord is the line segment that connects any two

points on the boundary of the circle.

The diameter (d)of the circle is the length of the

largest chord on the circlethe one that passes

through the centre point. Notice that the diameter

is exactly twice the radius: d= 2r.


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To describe the boundary length of the circle, the word circumference, rather than perimeter, is

used.. For any circle, the ratio of the circumference, c,to its diameter, d, is a constant special

irrational number discovered by the ancient Greeks, known as (pi, pronounced pie).

i.e.,c

d , which gives c d

(2 )c r

2c r

Note:Though is an irrational number, which means we cannot express its exact value as a

decimal number, we can write down a decent approximation: 3.14159 , or more simply,

3.14 . When doing small, approximate calculations, we use 3.14 in place of . However, when

performing large or precise calculations, we use the button in the calculator.

The area of a circle is calculated, as follows:

1. Cut a circle into an even number of equal slices (for example, 16).

2. Take half of the slices and arrange them end-to-end in the shape of teeth. Do the samewith the other half and place it on each end to make the interlocking shape symmetrical.

3. The result is approximately a parallelogram. The length of the parallelogram is half of the

circumference of the circle; i.e., 2

2

rb r

. The height of the parallelogram is the

distance from the boundary of the circle to the centre, which is the radius, r. The area of the

circle, therefore, is approximately equal to the area of the parallelogram:2A ( )b h r r r

Note: The more slices used in the circle, the closer the approximation gets.Therefore, the

formula in step 3 is indeed the exact formula for the area of a circle: 2A r .

r

b =

h


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Chapter 11

CircleC = d

C = 2 r 2A r

Calculating the Circumference and Area of a Circle

Find the circumference and area of the following circles (use 3.14 and

round the answer to 2 decimal places, as needed):

(i) (ii)

Solution

Using C = 2 r and 2A r

(i) C 2 2(3.14)(15) 94.20 cmr = 94.2 cm2 2 2A = (3.14)(15) 706.50 cmr = 706.5 cm2

Therefore, the circumference of the circle is 94.2 cm and the area is

706.5 cm

2

.(ii) C 3.14(1.24) 3.89 md

1.240.62 m

2 2

dr

2 2 2A = 3.14(0.62) 1.21 mr

Therefore, the circumference of the circle is 3.89 m and the area is

1.21 m2.

Calculating the Distance Travelled on a Bicycle

A road bike has a wheel with a 622 mm diameter. If the wheel spins at 192

rpm (revolutions per minute), determine the distance the cyclist travels in 1

hour and 20 minutes, rounded to the nearest tenth of a km.

Solution

Example 11.4-i

Example 11.4-j

r

radius

diameter

15 cm

1.24 m


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11


The distance travelled in one revolution of the wheel, is equivalent to the

circumference of the wheel (since we are doing a large calculation, we will

use the exact value of ).

Distance travelled in one revolution = C = d = (622)

= 1954 mm = 1.954 m

Since it spins at 192 rpm (revolution in one minute), the distance travelled in

1 minute = 192(1.954) 375.2 m.

The total distance travelled in 1 hour and 20 minutes (80 minutes) =

80(375.2) = 30016 m = 30.0 km

Therefore, the cyclist travelled approximately 30.0 km in 1 hour and 20

minutes.

Determining the Amount of Pizza Sauce Needed

An extra-large pizza is circular with a diameter of 16 inches. Pizza sauce is

spread on the pizza dough at a rate of 1.5 mL of pizza sauce per square inch

(in2) of crust. How much pizza sauce (rounded to the nearest 10 mL) is

required to cover the entire pizza, if a 1-inch crust is to be left around the

edge of the entire pizza?

Solution

Radius of pizza dough surface (area) =16

8 in2 2

dr .

Since there is a 1-inch crust to be left at the edge of the pizza, the radius of

the surface to be covered with pizza sauce is 7 inch.

The area of the pizza dough to be covered with the sauce =2 2 2A = (7) (49) 154 inr

The quantity of pizza sauce needed to cover the pizza = (1.5)(154) 230 ml .

Therefore, approximately 230 ml of pizza sauce is needed to cover the pizza.

SectorsA sector(denoted by a capital S) is a portion of a circle that is bounded by two radii from the

centre of the circle to the boundary of the circle, as shown in the table below. The section of the

circumference that bounds the sector is known as the arc(denoted by a capital L). The internal

angle of the sector inscribed by the two radii is known as the angle subtended by the arc(denoted

by the Greek letter ).

Example 11.4-k

16 in

8 in

1 in


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12

Chapter 11

In this sector with a sector angle, in a circle with 360o

:The arc length, L, of the sector is proportional to the

circumference of the circle, d .

(i) The area, As, of the sector is proportional to the

area of the circle,

2

r

.

i.e., (i)L

360 d

==> L

360d

Perimeter of sector = r+ r+ L = 2r+ L

(ii) s2

A

360 r

==> 2

SA360

r

SectorL 360d

P = 2r+ L

2

SA360

r

Calculating the Perimeter and Area of a Sector

Calculate the perimeter and area (rounded to the nearest mm and mm2,

respectively) of a sector of a circle with a radius of 75 mm and an inscribed

angle of 75.Solution

Using P = 2r+ L and L360

d

,

P =75

2(75) (2)(75)( ) 150 98 248 mm360

Using 2SA

360r

,

2 275

75 3682 mm360

Determining the Speed of a Gondola on a Ferris Wheel

The Niagara SkyWheel is a giant ferris wheel in Niagara Falls that has a

diameter of 50.5 m. If the SkyWheel rotates at a maximum speed of 9per

second, determine the speed at which the gondolas on the rim of the wheel

are moving (in km/h, rounded to the nearest tenth of a km).

Example 11.4-l

Example 11.4-m

r

r

L

AS= Sector Angle

Or

r

L


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13


Solution

The speed of the gondolas per second is the distance they travel in one

second, which is the arc length of a sector of the wheel with an inscribed

angle of 9.

Using L 360d

9(50.5)( ) 3.966 m

360L

Hence, the gondolas travel at a maximum speed of 3.966 m/s.

Converting this speed in km/h,

Speed =m s min m km

(3.966 )(60 )(60 ) = 14277.6 14.3s min hr hr hr

Therefore, the gondolas travel at a maximum speed of approximately 14.3

km/h.

Composite FiguresIt is quite common, when solving application problems, to see a complex geometric figure

constructed out of two or more simple geometric figures that we have previously studied. Such

figures are called composite figures.

To determine the perimeter of a composite figure, simply calculate the length of the boundary, by

adding up all the straight lengths and sector lengths along the boundary.

To determine the area of a composite figure, break the figure up into simple figures and add up all

the areas.

Calculating the Perimeter and Area of a Parking Lot

A new parking lot is to be created around a commercial building (see image

below). The edge of the parking lot is to be enclosed with concrete curbs

and the surface of the parking lot is to be paved with asphalt. Determinehow many linear metres of concrete curbing and square metres of asphalt are

required to create the parking lot.

Example 11.4-n


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14

Chapter 11

Solution

Letx,y,zbe the unknown lengths as marked in the diagram below:

x= 53 + 55 + 42 = 150 ft

y= 7543 = 32 ft

z= 50y =5032 = 18 ft

P = 50 + 53 +z+ 55 + 43 + 42 + 75 +x= 486 ft

To calculate the amount of asphalt neede

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