# GEOMETRY AND MEASUREMENT 3 CLAST MATHEMATICS COMPETENCIES

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GEOMETRY AND
MEASUREMENT
CLAST MATHEMATICS COMPETENCIES
IB1:
IB2a:
IB2b:
IB2c:
IIB1:
IIB2:
IIB3:
IIB4:
IIIBl:
IIIB2:
IVB1:
IVB2:
Round measurements to the nearest given unit of the measuring device used
Calculate distances
Calculate areas
Calculate volumes
Identify relationships between angle measures
Classify simple plane figures by recognizing their properties
Recognize similar triangles and their properties
Identify appropriate units of measurement for geometric objects
Infer formulas for measuring geometric figures
Identify applicable formulas for measuring geometric figures
Solve real-world problems involving perimeters, areas, volumes of
geometric figures
Solve real-world problems involving the Pythagorean theorem
132
3.1 MEASURING DISTANCE, AREA AND VOLUME
In this Chapter we shall discuss the measurement of distances, areas, volumes and their applications.
We shall then learn about the measurement of angles, the relationship between different types of
angles, the classification of plane figures and similar triangles.
A. Rounding Measurements
Objective IB1
1.
2.
3.
4.
CLAST SAMPLE PROBLEMS
Round 43.846 to the nearest hundredth of a centimeter
Round 19,283 tons to the nearest ten thousand tons
Round 9 pounds, 6 ounces to the nearest tenth of a pound
Round the measurement of the screw to the nearest 1/4 inch
In Section 1.3 we learned how to round numbers. We are now ready to round measurements in such
units as feet, yards, centimeters, meters, pounds, and liters.
1
ROUNDING MEASUREMENTS
RULE
EXAMPLES
1. Underline the place to which you
are rounding.
2. If the first number to the right of the
underlined place is 5 or more, add one to
the underlined number. Otherwise, do not
change the underlined number.
3. Change all the numbers to the right of the
underlined number to zeros.
Note: We are asking to round 3138 pounds
to the nearest hundred pounds. The idea and
procedure would be the same if you were asked
to round 3138 grams to the nearest hundred
grams or 3138 liters to the nearest hundred
liters.
1. 43.85 cm
Round 38.67 centimeters to the nearest
centimeter.
Since we want to round to the nearest
centimeter, we underline the units place.
1. Underline the 8. 38.67
2. The first number to the right of 8 is 6, so
we add one to 8. (8 + 1 = 9)
3. Change all the numbers to the right of 8 to
zeros obtaining 39.00 or 39.
Round 3138 pounds to the nearest hundred
pounds.
1. Underline the hundreds place. 3138
2. The first number to the right of 1 is 3. Do
not change the underlined number.
3. Change all the numbers to the right of 1 to
0's obtaining 3100. Thus, 3138 pounds
rounded to the nearest hundred pounds
becomes 3100 pounds.
2. 20,000 tons
3. 9.4 pounds
4. 3/4 inch
SECTION 3.1 Measuring Distance, Area and Volume
133
CLAST EXAMPLE
Example
1. Round the measurement of the length of the
1
paper clip to the nearest 4 inch.
A. 1 in.
1
B. 14 in.
3
C. 14 in.
D. 2 in.
Solution
Note: The abbreviation for inch is in.
1
If we are measuring (or counting) by 4 in., the
1
2
3
measures would be: 4 in., 4 in., 4 in.,
4
5
4 in., 4 in., and so on. The end of the paper
5
clip is closest to the 4 in. mark.
5
1
Since 4 in. = 14 in., the correct answer is B.
B. Calculating Distances, Areas and Volumes
Objectives IB2a,IB2b,IB2c CLAST SAMPLE PROBLEMS
1.
3.
5.
7.
8.
9.
Change 4 yards to feet
2. Change 8 1/2 ft to yards
Convert 9.7 kilometers to meters
4. Convert 90 centimeters to hectometers
Change 18 inches to yards
6. Change 2.2 miles to feet
Find the perimeter of a rectangle that has a length of 4 yards and a width of 4 feet
Find the distance around the polygon shown in kilometers
300 m
600 m
Find the distance around a regular pentagon with one side having a
400 m
length of 6 meters
500 m
10. What is the distance around a circular walk with a diameter of 6
ft.
11. What is the area in square centimeters of the triangle shown?
30 cm
12. What is the area of a circular region with a 6 yard diameter?
40 mm
13. Find the area of the 3 meter wide rectangle shown
13 cm
14. What is the surface area of a rectangular solid that is 6 feet long, 1
yard wide and 4 feet high?
15. Find the volume of a right circular cone 3 meters high and with a
5m
3m
16. What is the volume of a rectangular solid 5 feet long, 4 feet
wide and 9 inches high?
17. Find the volume of a sphere with a diameter of 2 yards
18. What is the volume in liters of a 850-milliliter bottle?
19. What is the volume in cubic centimeters of a 25.8-liter container?
20. What is the volume in cubic meters of a 675 -liter container?
1. 12 ft
5. 1/2 yd
9. 30 m
13. 12 m2
4
17. 3 π yd3
2. 2 5/6 yd
6. 11,616 ft
10. 6 π ft
14. 108 ft2
3. 9700 meters
7. 32 ft
11. 60 cm2
15. 40,000 π cm3
4. 0.009 hm
8. 1.8 km
12. 9p yd2
16. 15 ft3
18. 0.85 L
19. 25,800 cm3
20. 0.675 m3
134
CHAPTER 3
Geometry and Measurement
When we measure a distance we measure length and the result is a linear unit. In the U. S. system
we use inches, feet, yards and miles. In the metric system we use millimeters, centimeters, meters,
and kilometers. Here are some relationships you should know.
T
TERMINOLOGY -- LINEAR MEASURES
U. S. UNITS
EXAMPLES
12 inches (in.) = 1 foot (ft)
52
1
1
52 inches = 52
ft = 12 ft = 4 3 ft.
1
12
1 (in.) = 12 ft
3 ft = 1 yard (yd)
1
1 ft = 3 yd
8 ft
7920 ft
5280 ft = 1 mile (mi)
1
1 ft = 5280 mi
METRIC UNITS
kilometer hectometer dekameter meter
(km )
(hm)
(dam)
m
1000 m
100m
10 m
meter decimeter
m
(dm)
1
10 m
centimeter millimeter
(cm)
(mm)
1
1
100 m
1000 m
Chart: km hm dam m dm cm mm
= 8
8
2
1
yd = 3 yd = 2 3 yd.
3
= 7920
7920
1
mi = 5280 mi
5280
1
= 1 2 mi.
To convert from one metric unit to another is a
matter of moving the decimal point. For
example, to convert from km to meters, we
move three places right in the chart. Thus, to
convert 73 km to meters, we move the decimal
point in 73. three places right obtaining
73. km = 73,000 m.
To convert 3847 mm to meters, we have to
move the decimal point in 3847. three places
left. Thus, 3847. mm = 3.847 m.
You can recall the order in the metric chart if you remember that:
King Henry Died Monday Drinking Chocolate Milk!
2
FINDING PERIMETERS
RULE
A polygon is a closed plane figure having three
or more sides. (See figure at right.)
EXAMPLES
9 in
10 in
5 in.
15 in
The perimeter of (distance around) a polygon
is the sum of the lengths of its sides.
The perimeter of the polygon, in inches, is
(10 + 9 + 5 + 15) in. = 39 in. The perimeter
3
1
39
in feet is 12 = 3 12 = 3 4 ft.
SECTION 3.1 Measuring Distance, Area and Volume
3
135
FINDING CIRCUMFERENCES
The distance around a circle is called the
circumference C and given by the formula
C = π d = 2π r
diameter d
Note that the diameter d of the circle is twice
d
the radius, that is, d = 2r or r = 2 .
Note: The CLAST uses π as a factor in its
answers, you need not approximate π
If the radius of the circle is 7 ft, (r = 7) the
circumference C = 2 π r = 2 π (7) = 14 π ft.
CLAST EXAMPLE
Solution
Example
2. What is the distance around the polygon, in
meters?
75 cm
78 cm
The distance around the polygon is:
95 + 75 + 80 + 78 = 328 cm
Since 100 cm is 1 m, 328 cm = 3.28 m.
(If you look in the chart, to go from cm to m we
have to go two places left, so we move the
decimal point in 328. two places left obtaining
3.28 m.) The answer is C.
80 cm
95 cm
A. 328 m
B.
32.8 m
C. 3.28 m
D.
0.328 m
Hint: When adding the length of the sides, we
added 95 + 75 by writing 75 as 5 + 70 and then
adding 95 + 5 + 70 = 170. Then add 80 to 170,
obtaining 250 and finally, add 78 to get 328.
Be careful with the arithmetic!
Sometimes we have to use more than one formula to get the perimeter (distance around) of a figure.
For example, the next problem involves a rectangle and two semicircles (half circles). In order to
find the distance around the figure we need to know the formulas for the perimeter of a rectangle and
the perimeter (circumference) of a circle.
136
CHAPTER 3
Geometry and Measurement
CLAST EXAMPLES
Example
Solution
3. The figure below shows a running track in
the shape of a rectangle with semi-circles at
each end.
The total distance around the track is:
Straight
Distance
+
Distance around the
two semicircles
The straight distance is 3x + 3x or 6x and for
convenience, you can think of the two
semicircles as a complete circle of diameter
d = 2y as shown.
2y
3x
What is the distance around the track?
2y
A. 6x + 4y + 2 π y
B. 2 π y + 6x
C. π
D. π
y2 + 3x
The circumference C of this circle is
C = π d = π •2y = 2 π y
y2 + 6x
Note that the straight distance around the track is
3x + 3x = 6x. Thus, the answer must involve 6x.
Eliminate C.
Thus, the total distance around the track is:
Straight
Distance
+
6x
+
Distance around the
two semicircles
2π y
Since 6x + 2 π y = 2 π y + 6x, the answer is B.
To find the area of a plane figure such as a circle, square, rectangle, parallelogram or triangle, we
need the formulas that follow. Note: When we measure area or surface area the result is in
square units (abbreviated as sq. units or units2.)
4
FINDING AREAS
RULE
EXAMPLES
2
The
area
of
a
circle
A=π r
The area of a circle of radius r is:
2
A = π •5 sq. in. = 25 π sq. in.
The area of a square of side s is:
A = s2
The area of a square of side 3 m is:
A = 32 sq. m = 9 sq. m
The area of a rectangle of base b and height h
is: A = bh
The area of a rectangle of base 4 mm and
height 8 mm is:
A = 4•8 sq. mm = 32 sq. mm
The area of a triangle of base b and height h is:
1
A = 2 bh
The area of a triangle of base 2 yd and height 4
yd is
1
A = 2 •2•4 = 4 sq. yd
SECTION 3.1 Measuring Distance, Area and Volume
137
CLAST EXAMPLE
Example
Solution
4. What is the area of a circular region whose
diameter is 6 centimeters?
A.
36 π sq. cm
B.
6 π sq. cm
C.
12 π sq. cm
D.
9 π sq. cm
The formula to find the area of a circle is
A = π r2. Since the diameter is 6 cm, the
radius is half of that, or 3 cm. Thus, the area
is: A = π •(3)2 sq. cm = 9 π sq. cm.
(Some books use cm2 but the CLAST uses sq.
cm.)
In case the answer to the problem requires conversions (cm to meters or in. to ft), do the conversions
before you use the correct area formula, as shown in the next example.
CLAST EXAMPLE
Example
Solution
5. What is the area of a triangle with a base of
24 inches and a height of 12 inches?
A.
12 sq. ft
B.
1 sq. ft
C.
24 sq. ft
D.
144 sq. ft
Note that the answers are given in square feet.
5
First, convert the original dimensions to feet.
24 in. = 2 ft,
12 in. = 1 ft. The area of the
1
triangle is A = 2 bh, where b = 2 and h = 1.
1
Thus,
A = 2 •2•1 = 1 sq. ft.
FINDING SURFACE AREAS
The surface area of a rectangular solid of
length L, width W, and height H is:
A = 2HW + 2LH + 2WL
Note that to obtain the surface area of the
solid, we first need the area of the front and
the back (2HW), the two sides (2LH) and the
top and bottom (2WL).
The surface area of a rectangular solid 4 ft
long, 3 ft wide and 2 ft high is:
H
L
W
A = 2•(2•3) + 2•(4•2) + 2•(3•4)
= 12 + 16
+ 24
= 52 sq. ft
138
CHAPTER 3
Geometry and Measurement
CLAST EXAMPLE
Example
Solution
6. What is the surface area of a rectangular
solid that is 12 inches long, 5 inches
wide and 6 inches high?
Since we are looking for a surface area the
answer must be in square units (eliminate
choices A and C).
A.
B.
C.
D.
The area of the front is
The area of the back is
The area of the right side is
The area of the left side is
The area of the top is
The area of the bottom is
The total surface area is:
360 cubic inches
324 square inches
324 cubic inches
360 square inches
Note: Draw a picture whenever possible!
6
12
6•5 = 30 sq. in.
6•5 = 30 sq. in.
12•6 = 72 sq. in.
12•6 = 72 sq. in.
5•12 = 60 sq. in.
5•12 = 60 sq. in.
324 sq. in.
5
When we measure volume we measure the amount of space in a solid three dimensional object and
the result is in cubic units, sometimes written as units3. In the U.S. system we have cubic inches,
cubic feet and cubic yards. The metric system uses cubic centimeters, cubic meters and so on. The
volume of liquids in the metric system is measured in liters.
T
TERMINOLOGY -- METRIC CUBIC MEASURES
To convert from one metric unit to another
METRIC UNITS
is a matter of moving the decimal point. For
kiloliter hectoliter dekaliter
liter
example, to convert from kL to liters, we move
(kL)
(hL)
(daL)
L
three places right in the chart. Thus, to convert
1000 L
100L 10 L
37 kL to liters, we move the decimal point in
liter deciliter
centiliter
milliliter
37. three places right obtaining
L
(dL)
(cL)
(mL)
37. kL = 37,000 L.
1
1
1
To convert 2647 mL to liters, we have to move
10 L
100 L
1000 L
the decimal point in 2647. three places left.
Thus, 2647. mL = 2.647 L.
Chart: kL hL daL L dL cL mL
King Hendy Died Late Drinking Chocolate Milk
CLAST EXAMPLE
Example
7. Find the volume in centiliters (cl) of a 2.95
liter bottle.
A. 29.5 cl
C. 0.295 cl
B. 295 cl
D. 2950 cl
Solution
Since centi means 100, one liter = 100 cl.
Thus, 2.95 liter s= 2.95•100 cl = 295 cl. The
answer is B. You do not need any formulas but
remember your prefixes like milli, centi, deci
and kilo.
SECTION 3.1 Measuring Distance, Area and Volume
139
To find the volume of a rectangular solid (a box), right circular cylinder, circular cone or sphere we
need the formulas that follow.
6
H
FINDING VOLUMES
RULE
EXAMPLES
The volume of a rectangular solid
The volume of a rectangular solid
of length L, width W and height H is: 10 ft long, 9 ft wide and 8 ft high is:
V = 10•9•8 cubic ft = 720 cubic ft
V = LWH
L
W
r
The volume of a right circular
cylinder of radius r and height h is:
h
V = π r 2h
V = π r 2 h = π •52•7 cubic cm
= 175 π cubic cm
The volume of a right circular cone
of radius r and height h is:
h
1
V = π r 2h
3
r
The volume of a sphere of radius r
is:
4
V = π r3
3
r
The volume of a right circular
cylinder of radius 5 cm and height
7 cm is:
The volume of a right circular cone
of radius 6 in and height 12 in is:
1
1
V = 3 π r2h = 3 π •62•12 cubic in.
= 144 π cubic in.
The volume of a sphere of radius
3 m is
4
4
V = 3 π r3 = 3 π •33 cubic m
4
= 3 π •27 cubic m
= 36 π cubic m
CLAST EXAMPLE
Example
Solution
8. What is the volume of a sphere with a 12 in.
diameter?
Since the volume must be in cubic in.,
eliminate B and D. Since the diameter is 12 in,
4
Substituting 6 for r in
V = 3 π r3 we
obtain:
4
V = 3 π •216 cubic in = 288 π cubic in.
A. 48 π cubic in.
B.
48 π sq. in.
C. 288 π cubic in.
D.
288 π sq. in.
Caution: To use the formula for the volume of
a sphere you need the radius r.
140
CHAPTER 3
Geometry and Measurement
C. Identifying the Appropriate Unit of Measurement
Objective IIB4
CLAST SAMPLE PROBLEMS
1. Which unit of measurement would be used to give the amount of wall surface to be covered by
the contents of a can of paint?
A. gallons
B. liters
C. square feet
D. cubic feet
2. Which unit of measurement would be used to find the distance around a building?
A. liters
B. cubic feet
C. square meters
D. meters
3. What unit of measurement would be used to find the amount of paper needed for the label on the
can?
A. inches
B. square inches
C. cubic inches
D. fluid ounces
7
RULE
PROPER MEASUREMENT UNITS
EXAMPLES
When measuring length use linear units
(kilometers, miles, meters, yards, centimeters,
inches, or millimeters)
The diameter of a coin may be measured in
inches or centimeters. The distance around a
circular track may be measured in meters or
yards.
When measuring area use square units.
(Square yards, square meters, square feet,
square centimeters)
The amount of paper needed to wrap a gift
may be in square inches. The amount of wall
surface covered by a gallon of paint may be in
square meters.
When measuring volumes use cubic units.
(Cubic centimeters, cubic inches, cubic feet,
cubic meters, gallons, liters)
The amount of gasoline in a fuel storage tank
may be measured in cubic feet, cubic meters,
gallons or liters.
CLAST EXAMPLE
Example
9. Which of the following would not be used to
measure the amount of water needed to fill a
swimming pool?
A. Cubic feet
C. Gallons
B. Liters
D. Meters
1. C
Solution
We are to measure the volume of the water.
Since volume is measured in cubic units,
meters is not an appropriate unit of measure.
2. D
3. B
SECTION 3.1 Measuring Distance, Area and Volume
141
Section 3.1 Exercises
WARM-UPS A
1.
Round the measurement of the length of the key to the nearest inch.
2.
1
Round the measurement of the length of the key to the nearest 2 inch.
CLAST PRACTICE A
PRACTICE PROBLEMS: Chapter 3, # 1-2
1
3. Round the measurement of the length of the key to the nearest 4 inch.
A.
2 in.
B.
1
24 in.
C.
1
2 2 in.
D.
3
24 in.
WARM-UPS B
4.
5.
6.
Find the distance around the polygon in meters.
150 m
200 m
250 m
Find the distance around the polygon in kilometers.
300 m
20 ft
Find the distance around the polygon in feet.
40 ft
7.
30 ft
Find the distance around the polygon in yards.
35 ft
8.
Find the circumference of a circle of radius 8 cm.
9
Find the circumference of a circle whose diameter is 10 in.
10. Find the area of a circle of radius 10 m.
142
CHAPTER 3
Geometry and Measurement
11. Find the area of a circle whose diameter is 12 ft.
12. Find the area of a rectangle of base 8 mm and height 10 mm.
13. Find the area of a rectangle of base 2 yards and height 4 yards.
14. Find the area of a triangle of base is 3 km and height 4 km.
15. Find the area of a triangle of base 2 ft and height 7 ft.
16. Find the surface area of a rectangular solid of length 8 cm, width 10 cm and height 7 cm.
17. Find the surface area of a shoe box that is 12 inches long, 8 inches wide and 4 inches
high.
18. Find the volume of the shoe box in Problem 17.
19. Find the volume of a rectangular solid of length 30 mm, width 20 mm and height
10 mm.
20. Find the volume of a right circular cylinder of radius 8 in and height 9 in.
21. Find the volume of a right circular cylinder whose diameter is 10 cm and whose height is 12
cm.
22. Find the volume in cubic feet of a right circular cylinder of radius 6 feet and height 30 inches.
23. Find the volume in cubic feet of a sphere whose diameter is 24 inches.
24. Find the volume in cubic yards of a sphere whose radius is 6 feet.
25. Find the volume in centiliters (cl) of a 1.95 liter bottle.
26. Find the volume in centiliters (cl) of a 3.90 liter bottle.
SECTION 3.1 Measuring Distance, Area and Volume
CLAST PRACTICE B
27.
143
PRACTICE PROBLEMS: Chapter 3, # 3-8
Find the distance around the polygon in kilometers
A.
2300 km
B.
23 km
C.
2.3 km
D.
230 km
600 m
300 m
400 m
400 m
600 m
28. Find the circumference of a circle of radius 12 in.
A.
24 π sq. in.
B. 144 π in.
C. 24 π in.
D. 22 π in.
29. What is the area of a circular region of diameter 10 centimeters?
A.
20 π sq. cm
B. 25 π sq. cm,
C, 10 π sq. cm.
D. 100 π sq. cm.
30. What is the area of a circular region of radius 6 ft?
A.
12 π sq. ft
B. 24 π sq. ft.
C. 36 π sq. ft
D. 144 π sq. ft.
31. What is the area of a square that has sides 6 centimeters in length?
A.
24 sq. cm
B. 12 sq. cm
C. 25 sq. cm.
D. 36 sq. cm
32. Find the area of a rectangle that is 6 inches long and 4 inches wide.
A.
20 sq. in.
B. 24 sq. in.
C. 10 sq. in.
D. 30 sq. in.
33. Find the area of a triangle with base 8 in. and height 10 in.
A.
40 sq. in.
B. 80 sq. in.
C. 40 in.
D. 80 in.
34. Find the surface area of a rectangular solid that is 14 inches long, 5 inches wide and
4 inches high.
A.
292 sq. in.
B. 650 in.
C. 900 sq. in.
D. 650 sq. in.
144
CHAPTER 3
Geometry and Measurement
35. Find the volume of a rectangular solid that is 7 cm long, 5 cm wide and 4 cm high.
A.
140 sq. cm
B. 168 cubic cm
C. 140 cubic cm
D. 166 cm
36. Find the volume of a right circular cylinder 7 inches high and with a 5 in. radius.
A.
35 π cubic in.
B. 42 π cubic in.
C. 175 π sq. in.
D. 175 π cubic in.
37. Find the volume of a right circular cone of radius 9 ft and height 10 ft.
A.
270 π sq. ft
B. 810 π cubic ft
C. 270 π cubic ft
D. 870 π sq. ft
38. Find the volume of a sphere with a radius of 9 mm.
A.
108 π sq. mm
B. 972 π cubic mm C. 18 π mm
D. 729 π cubic mm
WARM-UPS C
39. What type of measure is needed to express the circumference of a circle?
40. What type of measure is needed to express the diameter of a coin?
41. What type of measure is needed to express the area of a rectangle?
42. What type of measure is needed to express the area of a triangle?
43. What type of measure is needed to express the amount of water in a pool?
44. What type of measure is needed to express the amount of gas in your gas tank?
CLAST PRACTICE C
PRACTICE PROBLEMS: Chapter 3, # 9-11
45. What type of measure is needed to express the length of the line
segment AC of the given rectangle?
A.
Square
B.
Linear
C.
Cubic
D.
Equilateral
46. What type of measure is needed to express the volume
of the rectangular solid?
A.
Square
B.
Linear
C.
Cubic
D.
Equilateral
A
B
C
D
H
L
W
SECTION 3.1 Measuring Distance, Area and Volume
145
47. What type of measure is needed to express the surface
area of the rectangular solid?
A.
Square
B.
Linear
C.
Cubic
D.
Equilateral
H
L
W
48. What type of measure is needed to express the thickness of a coin?
A.
Square
B.
Linear
C.
Cubic
D.
Equilateral
49. What type of measure is needed to express the amount of paint in a can?
A.
Square
B.
Linear
C.
Cubic
D.
Equilateral
50. What type of measure is needed to express the amount of sand in a box?
A.
Square
B.
Linear
C.
Cubic
D.
Equilateral
EXTRA CLAST PRACTICE
51. What is the area of the figure in square meters?
A. 1400 square meters
B. 14 square meters
C. 14 square meters
D. 24 square meters
140 cm
10m
52. What is the area of a triangle whose base is 36 feet and whose height is 40 inches?
A. 60 square feet
B. 180 square feet
C. 360 square feet
D. 720 square feet
146
CHAPTER 3
Geometry and Measurement
3.2 APPLICATIONS AND PROBLEM SOLVING
The information we have learned in Section 3.1 will now be used to solve problems involving
perimeters, areas, and volumes. In addition, we shall learn how to select and create new formulas to
measure other geometric figures. If you have forgotten about the RSTUV procedure for solving
problems, review it before you go on.
A. Applications and the Pythagorean Theorem
Objectives IVB1, IVB2
CLAST SAMPLE PROBLEMS
1. A new street connecting A Street and B Avenue is to be
constructed. Construction costs are about \$100 per linear
foot. What is the estimated cost for constructing the new
street?
B Avenue
New
12 miles
A Street
5 miles
2. A football player runs 8 yards to the right as shown. He
then turns left, runs down field and catches a 10 yard pass
thrown from the spot from where he started. How far
down field did he run?
10 yd
d=?
8 yd
CLAST EXAMPLES
Example
Solution
1. What will be the cost of tiling a room
measuring 12 feet by 15 feet if square tiles
costing \$2 each and measuring
12 inches on each side are used?
A. \$180
B.
\$4320
C. \$360
D.
\$3600
Note: The measurements of the tile and of the
room are not in the same units. It is easier to
change the measurement of the tile to feet.
Since 12 inches = 1 ft, the tile measures 1 ft on
each side.
If the tiles were 8 in. on each side, you would
need to convert 8 in. to ft.
1. \$6,864,000
1. Read the problem. You are given the
dimensions of the room and the cost of each
tile.
2. Select the unknown.
We want to find the cost of tiling the room.
3. Think of a plan.
Find how many tiles are needed and multiply
by the cost of each tile (\$2).
4. Use the fact that we need to cover a
rectangle measuring 12 tiles by 15 tiles. Thus,
we need 12 × 15 = 180 tiles.
Total cost = Cost of each tile × 180
=
\$2
×180 or \$360
2. 6 yards
SECTION 3.2 Applications and Problem Solving
Example
2. Find the cost of building a rectangular
driveway measuring 12 feet by 30 feet if
concrete costs \$12.50 per square yard.
A. \$4500
B.
\$400
C. \$1500
D.
\$500
147
Solution
Since the cost is per sq. yd, we change
12 ft to 4 yd and 30 ft to 10 yd. In square yards,
the area of the driveway is:
4 yd × 10 yd = 40 sq. yd.
Total Cost = Cost per sq. yd × sq. yd needed
=
\$12.50 × 40 = \$500
Some applications in the CLAST require the use of the Pythagorean theorem.
1
PYTHAGOREAN THEOREM
RULE
EXAMPLES
For any right triangle, the square of the length
A jogger runs 4 miles east and then 3 miles
of the hypotenuse is equal to the sum of the
north. How far is the jogger from his starting
squares of the lengths of the other two sides.
point?
In symbols,
If we let a = 4 and b = 3 in the diagram, we
have to find the distance c. By the
c2 = a2 + b2
Pythagorean theorem,
Note: In a right triangle (a triangle with a 90
c2 = a2 + b2
degree angle) the hypotenuse c is the side
2
=
2
Thus, c 3 + 42
opposite the 90 degree angle.
c2 = 9 + 16
c
c2 = 25
b
c= 5
a
Thus, he is 5 miles from the starting point.
CLAST EXAMPLE
Example
3. A television antenna 12 ft high is to be
anchored by three wires each attached to the
top of the antenna and to points on the roof
5 ft from the base of the antenna. If wire
costs \$0.75 per foot, what will be the cost of
the wire needed to anchor the antenna?
A. \$27
B.
\$29.25
C. \$9.75
D.
\$38.25
Hint: After you read the problem, see if you
can draw a picture of the situation.
c
12 ft
5 ft
Solution
1. Read the problem. The antenna is 12 ft
high and each wire is 5 ft from the base. The
cost of each foot of wire is \$0.75.
2. Select the unknown.
We need to find the length of each wire,
multiply by 3 (we need 3 wires) and then by
\$0.75 (each foot of wire is \$0.75)
3. Think of a plan.
First, find the length of each wire. Draw a
picture like the one at left.
4. Use the Pythagorean theorem
c2 = 52 + 122
c2 = 25 + 144
c2 = 169
c = 13
Thus, each wire is 13 ft long. Since we need
three wires, we need 3 ×13 ft = 39 ft. The total
cost would be \$0.75 ×39 = \$29.25.
148
CHAPTER 3
Geometry and Measurement
B. Inferring and Selecting Geometric Formulas
Objectives IIIB1, IIIB2
CLAST SAMPLE PROBLEMS
1. In the figure, S represents the sum of the measures of the interior angles. Study the figure and
calculate the sum S of the measures of the interior angles of an eight-sided convex polygon.
3 sides
1 triangle
S = 180o
2.
4 sides
2 triangles
S = 360o
6 sides
4 triangles
S = 720o
The figure consists of a rectangle and a semi-circle of
radius r. Determine the formula for finding the area of
Many geometric measurements can be generalized to be used in more complicated situations.
Look at the three spheres with radii 1, 2 and 3 units and surface areas (SA) 4 π ,
16 π and 36 π , respectively. The CLAST asks
us to generalize this information and find the
surface area of a sphere with a radius of 4 units.
To solve this problem set up a table and see if
there is a pattern.
1
2
3
4
Surface Area
4 π = 1•4 π
16 π = 2•8 π
36 π = 3•12 π
= 4•?
r=1
SA = 4 π
r=2
SA = 16 π
r=3
SA = 36 π
The underlined numbers are 4 π , 8 π and 12 π . The next number in the pattern is 16 π so the
surface area of a sphere with radius 4 should be 4•16 π or 64 π . (You can also reason that since the
surface area is measured in square units, the pattern should involve r2, the square of the radius, that
is 12•? = 4 π , 22•? = 16 π , 32•? = 36 π , where we multiply the square of the radius by 4 π . For a
radius of 4 units, the answer should be 42 • 4 π = 64 π .)
1. S = 1080o
2. 2r2 -
π r2
2
SECTION 3.2 Applications and Problem Solving
149
CLAST EXAMPLE
Example
Solution
4. In each of the figures, S represents the
shaded area when a triangle is removed from a
square
Side = 4
S=8
Side = 6
S = 18
Side = 8
S = 32
B. 75
C. 50
The area of the first square is
42 = 16
The area of the second square is
62 = 36
The area of the third square is
82 = 64
Now, from the diagram, S = 8, 18 and 32.
What is the relationship between the area of the
squares and the area S?
Area of squares
Area of S
Find S if the side of the square is 10
A. 100
First find the areas of each of the squares.
D. 40
16
8
36
18
64
32
In each case, S is half of the area of the
corresponding square. If the side of a square is
10, its area is 100 and S is half of that or 50.
CLAST EXAMPLE
Example
5. Study the information given with the regular
hexagons. Then calculate the height h of the
corresponding triangle in a regular hexagon
with each side 4 units long.
h
h
b
h
b
b
s = 1 unit
s = 2 units
3
h = 2 units h = 3 units
s = 3 units
3 3
h = 2 units
A. 2 3 units B.
3 3 units
C. 4 3 units D.
5 3
2 units
Solution
Look at the pattern involving the length s of the
side of the hexagon and the height h.
Length of side
Height
3
1
1• 2
3
2
2• 2 = 3
3 3
3
3
3• 2 = 2
In each case multiply the length s of the side by
3
2 . For a hexagon whose side measures 4
3
units, the height is 4• 2 = 2 3 . The answer is
A.
Note: We were looking for a linear pattern, so
we used a linear factor.
150
CHAPTER 3
Geometry and Measurement
Some of the patterns appearing in the CLAST involve the measure of an angle which is in degrees.
You should know that the sum of the measures of the angles of any triangle is 180o.
CLAST EXAMPLE
Example
Solution
6. Study each figure below and then calculate
S, the sum of the measures of the interior
angles of a 9-sided convex polygon.
4 sides
2 triangles
S = 360o
5 sides
3 triangles
S = 540o
6 sides
4 triangles
S = 720o
A. 14400 B. 1260o C. 1080o D. 1000o
We want to discover a pattern involving the
number of sides, the number of triangles and S,
the sum of the measures of the angles.
No. of sides No. of triangles
S
o
4
2
2•180 = 360o
5
3
3•180o = 540o
6
4
4•180o = 720o
9
?
?•180o
Note that the number of triangles is 2 less than
the number of sides. If the number of sides is
9, the number of triangles is 9 - 2, or 7, and S =
7•180o = 1260o. The answer is B.
So far we have inferred new formulas for different geometric figures. A different CLAST skill
requires you to calculate the measure of a geometric figure composed of several figures for which
the formulas are known. Here is the rule you need.
2
FORMULAS FOR AREAS AND VOLUMES
RULE
EXAMPLES
Study the figure of a right circular cylinder
1. Write the formulas necessary to find the
with a hemisphere (half a sphere) mounted on
area or volume of the individual figures.
top. Find the volume.
2. Use the figure to find the variables to be
r
used in the formula.
r
r
3. Simplify the area or volume formula.
4. Find the sum of the areas or volumes.
5. If the area (or volume) to be found is
shaded, you may be able to find it by
calculating the difference between the areas or
volumes. (See Problems 16, 30 and 31.)
h
h
4
The volume of the sphere is 3 π r3. One half
2
1 4
of this is 2 × 3 π r3 = 3 π r3 The volume of
the cylinder is π r2h. Thus the total volume is
2
the indicated sum 3 π r3 + π r2h.
SECTION 3.2 Applications and Problem Solving
151
CLAST EXAMPLE
Example
Solution
7. The figure shows a regular hexagon. Select
the formula for computing the total area of
the hexagon.
The total area of the hexagon is composed of 6
identical triangles as shown. The area
h
h
b
b
A. Area = 3h + b
B.
6(h + b)
C. 6hb
D.
3hb
1
of each triangle is 2 bh. Since there are 6 of
1
them, the total area is 6•2 bh = 3bh.
Since 3bh = 3hb, the answer is D.
Section 3.2 Exercises
WARM-UPS A
1.
Find the cost of tiling a room measuring 12 feet by 18 feet if square tiles costing \$2 each and
measuring 8 in. by 8 in. are used.
2.
A gardener wants to prepare a flower bed measuring 10 feet by 9 feet and 6 inches deep. How
many cubic yards of soil are needed to do this?
3.
Roofing material come in 100 sq. ft. bundles costing \$90 each. How much would the material
cost to repair a roof measuring 15 ft by 20 ft?
4.
How many square yards of wallpaper costing \$10 per roll are needed to cover an area 10 ft
long and 8 ft high?
5.
A baking pan in the shape of a rectangular solid is 20 cm long, 10 cm wide and 4 cm deep.
How many liters will it hold? Hint: 1 liter = 1000 cc.
6.
What will be the cost of carpeting an office measuring 12 ft by 18 ft if carpeting costs \$15 per
square yard?
7.
How much would it cost to fence a 30 ft by 60 ft yard if fencing material costs \$12 per linear
foot?
8.
A paper cup is in the shape of a right circular cone 10 cm high and 6 cm in diameter. How
many cubic centimeters does the cup hold?
152
CHAPTER 3
Geometry and Measurement
9.
A radio station is going to construct a 16 foot antenna anchored by three cables, each attached
to the top of the antenna and to points on the roof of the building that are 12 feet from the base
of the antenna. What is the total length of the three cables?
10.
An airplane leaves the airport and flies 40 miles north. It then flies 30 miles west. How far
from the airport is the plane?
11.
The door in a warehouse is 9 ft tall and 48 inches wide. If a sheet of paneling is 10 ft long,
what is the widest it can be and still fit through the door?
12. A contractor wants to put a sidewalk in front of the
triangular garden shown. If it costs \$50 per linear foot
to build the sidewalk, how much will the job cost?
CLAST PRACTICE A
Front
8 yd
10 yd
PRACTICE PROBLEMS: Chapter 3, # 12-15
13.
A farmer wants to build a fence to enclose a rectangular area 20 ft long by 12 ft wide.
many feet of fencing will be needed?
A.
30 ft
B. 23 ft
C. 64 ft
How
D. 112 ft
14. Find the cost of carpeting an office that measures 21 ft by 24 ft if carpet costs \$13 per square
yard.
A.
\$6552
B. \$2184
C. \$1092
D. \$728
15. The outside dimensions of a picture frame are 4 ft by 30 inches. If its inside dimensions are
3
24 ft by 21 inches, find the area of the frame.
A.
759 sq. in.
B. 747 sq. in.
C. 67.50 sq. ft
D. 182.25 sq. ft
16. A gardener uses a six-inch mulch border around a garden
measuring 6 ft in diameter. What is the area of the border?
A.
C.
3.2 sq. in.
2.75 π sq. ft
B.
D.
6 inches
6 feet
5 π sq. ft
2.75 sq. ft
17. The city commission wants to construct a new street that
connects Main Street and North Boulevard as shown in the
diagram. If the construction cost is \$90 per linear foot, find the
estimated cost for constructing the street. Hint: 1 mile = 5280
ft.
N
New
4 mi
North Blvd
3 miles
Main St
A.
\$33,808
B.
\$3690
C. \$2,376,000
D.
\$475,200
E
S
SECTION 3.2 Applications and Problem Solving
153
WARM-UPS B
18. Study the information given with the right triangles then calculate the area of a right
triangle with hypotenuse 10 and base 5 3 .
4
6
8
3 3
4 3
2
2 3
3
1
A = 2 •2 3 •2
1
A = 2 • 3 •1
19.
1
A = 2 •4 3 •
Study the given figures then find the area of a the regular 6- sided polygon.
h
h
h
b
4 sides
A = 2 bh
20.
1
A = 2 •3 3 •3
b
b
5 sides
5
A = 2 bh
6 sides
A=?
Study the figure formed from a semicircle and a triangle.
Then give the formula for calculating the total area of the
figure.
r
h
21.
Study the figure showing a square with a semicircle attached
to each side. Find the formula for computing the perimeter of
the figure.
22.
Find the formula for the total area of the figure in Problem 21.
s
154
CHAPTER 3
Geometry and Measurement
CLAST PRACTICE B
23.
Study the information given with
the regular hexagons, then calculate
the area A of a regular hexagon with
a side equal to 8.
A. 96 3
C.
24
25
48 3
B.
72 3
D.
36 3
h
h
h
b
b
Side 2
A=6 3
Side 4
A = 24 3
b
Side 6
A = 54 3
Study the information given for the
3, 4 and 6 sided figures. If S
represents the sum of the measures
of the interior angles, find S for a 10
sided polygon.
A.
1080o
B.
1800o
C.
1440o
D.
1260o
3 sides
1 triangle
S = 180o
4 sides
2 triangles
S = 360o
The given figure is formed by
joining a triangle and a square as
shown. Study the figure then select
the formula for computing its total
area.
A. A = s4
3s2
C. A = 4 + s2
26.
PRACTICE PROBLEMS: Chapter 3, #16-20
s
A.
SA = 2(2 π r) + rh
B.
SA = π r2h
C.
SA = 2 π r2 + 2 π rh
D.
SA = 2 π r2 + 2rh
s
s
B. A = s3
5 s2
D. A = 4
A cylindrical container can be formed
from two tops and a rectangle as
shown. Based on the figure, what is
the formula for calculating the surface
area SA of the right circular cylinder?
6 sides
4 triangles
S = 720o
s
r
r
h
r
h
SECTION 3.2 Applications and Problem Solving
27.
The figure shows a rectangular box
without a top formed by cutting out
squares x units on each side from a
rectangular piece of length L and
width W. Select the formula for
calculating the surface area SA of the
box.
A.
B.
C.
D.
28.
29.
30.
155
x
x
W
SA = LW - 4x2
SA = LW - x2
SA = (L - 2x)W - 2x) - 4x2
SA = (L - 2x)(W - 2x)
L
Study the figure showing a regular
pentagon. Then select the formula for
computing the total area A of the
pentagon.
A.
5
A = 2 hb
B. A = 5(h + b)
C.
A = 5h + b
D. A = 5hb
Study the figure showing a trapezoid,
then select the formula for computing
the total area A of the trapezoid, if
B = x + b + y.
A. A = b(B - x - y) + xh2
B. A = h(B + b)
C. A = b(B - x - y)h
1
D. A = 2 h(B + b)
Study the figure showing a square
inscribed in a circle. Select the
formula for computing the area of the
A. A = π r2 - s2
s2
B. A = π 2 - s2
s2
C. A = 2
s2
D. A = π 2
h
b
h
h
b
y
x
s
156
31.
CHAPTER 3
Geometry and Measurement
Study the figure showing a triangle
inscribed in a circle. Select the
formula for computing the area of the
A.
A = π r2 - 2r2
B.
A= π -2
C.
A= π
D.
A = π r2 - r2
r
EXTRA CLAST PRACTICE
32.
Study the information given in the
figures and then calculate A, the
exterior angle in the last figure.
A.
65o
B.
C. 110o
o
60
135o
D. 115o
o
70
o
130
o
20
100
45
o
25
90
o
o
115
o
65
120
o
o
o
A
33. The base of a tool shed is to be a slab of concrete 15 feet long by 12 feet wide by 6 inches
thick. If one cubic yard of concrete costs \$39, how much will the concrete for the
base of the tool shed cost?
A. \$3510
B. \$1560
C. \$130
D. \$65
1
34. A rectangular flower bed measures 22 feet by 50 inches. The outside dimensions of a path
around the bed are 4 ft by 60 inches. What is the area of the path?
A. 115 square feet
B. 200.50 square feet
C. 1200 square inches
D. 1380 square inches
35. The diameter of a tree trunk is 2.4 meters. What is its circumference?
A.
0.72 π square meters
B. 1.2 π meters
C.
2.4 π meters
D. 2.88 π square meters
SECTION 3.3
Lines, Angles and Triangles
157
3.3 LINES, ANGLES AND TRIANGLES
Before we proceed to discuss the different types of lines, angles and triangles we need to learn some
of the basic terminology.
T
TERMINOLOGY--TYPES OF ANGLES
PLANE ANGLES
EXAMPLES
A plane angle is a figure formed by two rays
C
with a common endpoint called the vertex.
vertex
The rays AB and AC are the sides of the angle
1
A
and the angle can be named ∠ 1 (read "angle
B
1"), ∠ CAB , ∠ BAC or ∠ A
The vertex of angle BAC is A.
Angles are measured by the amount of rotation
needed to turn one side of an angle so that it
coincides with (falls exactly on top of) the
other side. Note that ∠ YXZ. is "greater than"
∠ BAC.
The rotation of an angle is measured in
degrees and is denoted by m ∠ BAC. One
complete revolution is 360o (read "360
1
degrees") and 360 of a complete revolution is
1o. The CLAST sometimes states that the
measure of ∠ A, for example, is represented by
"x".
C
Z
A
X
B
A complete revolution is 360o
C
C
A
30°
x
A
B
B
m ∠ BAC = 30o
One-half of a complete revolution is 180o and
results in an angle called a straight angle.
One-quarter of a complete revolution is 90o
and results in an angle called a right angle
(denoted by using the little square ).
Y
x = 30o
180°
90°
An acute angle is an angle whose measure is
less than 90o.
An obtuse angle is an angle whose measure is
greater than 90o and less than 180o.
158
CHAPTER 3
Geometry and Measurement
We now give several relationships between angles.
T
TERMINOLOGY--COMPLEMENTARY, SUPPLEMENTARY, VERTICAL ANGLES
COMPLEMENTARY ANGLES
EXAMPLES
o
Two angles whose measures add to 90 are
called complementary angles.
B
Note that
A
m ∠ A + m ∠ B = 90o or
m ∠ A = 90o - m ∠ B
SUPPLEMENTARY ANGLES
Two angles whose measures add to 180o are
B
A
supplementary angles. Note that
o
o
m ∠ A + m ∠ B = 180 or m ∠ A =180 - m ∠ B
VERTICAL ANGLES
When two lines intersect, the opposite angles
are called vertical angles and they have the
u
x
y
same measure. If x, y, u and v are the
v
measures then x = y and u = v.
We are now ready to study the properties existing between different plane figures such as
angles and triangles. Before we do that, let us see what type of questions we will
encounter.
A. Properties of Plane Figures
Objective IIB2
CLAST SAMPLE PROBLEMS
1. What type of triangle is Δ ABC below?
A
65 o
B
45
o
C
In the figure to the right:
2.
3.
4.
5.
Find two right angles
Find two pairs of complementary angles
Find two pairs of vertical angles that are not right angles
Which geometric figure has all of the following
characteristics?
i. quadrilateral ii. opposite sides are parallel
iii. diagonals are not equal
1. Scalene
4. 3 and 6; 2 and 5
2. 1 and 4
5. Rhombus
3
2
4
1
5
6
3. 5 and 6; 2 and 3
SECTION 3.3
Lines, Angles and Triangles
159
CLAST EXAMPLE
Example
Solution
Complementary angles are angles whose sum is
90o, so 1 can not be complementary to any other
angle. (Eliminate answers A and D.) Now, 5
and 6 are complementary, but this is not one of
the choices. However, angles 6 and 3 are
vertical angles, so they must have the same
measure. Since angles 5 and 6 are
complementary and angles 3 and 6 have equal
measures, ∠ 5 and ∠ 3 are complementary.
The answer is C. If you are not convinced,
here is the proof:
1.
6
1
5
4
2
3
Which of the following pairs of angles are
complementary?
A. 1 and 2
B.
5 and 2
C. 3 and 5
D.
4 and 1
m ∠ 5 + m ∠ 6 = 90o
m∠ 6 = m∠ 3
m ∠ 5 + m ∠ 3 = 90o,
so ∠ 3 and ∠ 5 are complementary.
Triangles can be classified according to their angles (right, acute or obtuse) or according to the
number of equal sides (scalene, isosceles, or equilateral).
T
TERMINOLOGY -- CLASSIFYING TRIANGLES BY ANGLES AND SIDES
CLASSIFICATION
EXAMPLES
A right triangle is a triangle containing a right
angle.
An acute triangle is a triangle in which all the
angles are acute.
An obtuse triangle is a triangle containing an
obtuse angle.
A scalene triangle is a triangle with no equal
sides. Note that the sides are labeled |, || and |||
to indicate that the lengths of the sides are
different.
160
T
CHAPTER 3
Geometry and Measurement
TERMINOLOGY--CLASSIFYING TRIANGLES BY ANGLES AND SIDES (CONT.)
An isosceles triangle is a triangle with two
equal sides and two equal base angles. Note
that the equal angles are indicated by arcs, the
equal sides by slashes.
An equilateral triangle is a triangle with three
equal sides and three 60o angles. As before
the equal angles are indicated by arcs and the
equal sides by slashes.
Next, we give the rules used to name specific triangles, and a very important rule that
applies to the measures of the sum of the angles of any type of triangle. Do you remember
what that is?
T
TERMINOLOGY--NAMING TRIANGLES
NAMING TRIANGLES
EXAMPLES
Triangles are named using the letters at the
A
vertices and the symbol Δ . The given triangle
is Δ ABC (read "triangle ABC").
C
B
1
SUM OF THE ANGLES OF A TRIANGLE
RULE
60o
35 o
The sum of the measures of the angles of a
triangle is 180o. You can use this fact to find
The measure of the other angle, call it x, is
the measure of the third angle when the
measures of the other two angles are given.
such that: 60o + 35o + x = 180o
or
95o + x = 180o
x = 85o
CLAST EXAMPLE
Example
2. What type of triangle is Δ ABC?
A
55
o
70
o
B
C
A. Isosceles
C. Equilateral
B.
D.
Right
Scalene
Solution
First, we have to determine the measure "x" of
the missing angle. Since the sum of the three
angles in a triangle is 180o, we have:
55o + 70o + x = 180o
125o + x = 180o
x = 180o - 125o
x = 55o
Since two of the angles are of equal measure,
the triangle is an isosceles triangle. Thus, A.
SECTION 3.3
Lines, Angles and Triangles
161
In the preceding sections we have mentioned geometric figures such as squares, circles, rectangles
and polygons. These figures have definite properties that should be familiar to you. We shall give
the definitions and examples for several polygons and quadrilaterals.
T
TERMINOLOGY--TYPES OF POLYGONS
TYPES OF POLYGONS
EXAMPLES
A polygon is a closed figure whose sides are
line segments that do not cross each other.
A regular polygon is a polygon with all sides
of equal length and all angles of equal
measure. They are usually named by using the
number of sides. For example: Pentagon is a
5-sided polygon.
Hexagon is a 6-sided polygon.
Octagon is an 8-sided polygon.
A
F
E
A
B
B
E
D
C
D
C
A quadrilateral is a four-sided polygon
A trapezoid is a quadrilateral with one pair of
parallel sides.
A parallelogram is a quadrilateral in which
the opposite sides are parallel.
A rhombus is a parallelogram with all sides
equal.
A rectangle is a parallelogram with a right
angle.
A square is a rectangle with all sides equal.
A diagonal is a line segment joining two nonconsecutive vertices of a polygon.
162
CHAPTER 3
Geometry and Measurement
The CLAST emphasizes the characteristics of these polygons as shown in the example.
CLAST EXAMPLE
Example
3. Select the geometric figure that possesses all
of the following characteristics:
i.
polygon
ii.
iii.
two and only two sides are
parallel.
A. Parallelogram
B.
Rectangle
C. Rhombus
D.
Trapezoid
Solution
First note that all of the responses give figures
that are polygons and quadrilaterals, so we have
to concentrate on the figure having two and
only two parallel sides. The parallelogram, the
rectangle and the rhombus have more than two
parallel sides. The only quadrilateral with
exactly one pair of parallel sides is the
B. Relationships Between Angle Measures
Objective IIB1
CLAST SAMPLE PROBLEMS
Referring to the diagram on the right m ∠ g = 130o:
1. Which angles are supplements of g?
2. Find m ∠ f, m ∠ h and m ∠ j
3. Find three angles with the same measure as m ∠ c
4. Find two pairs of supplementary angles that are not right
angles.
a
b
d
c f
g
e
h j
The material we have studied can be used to find additional relations between angles. One of these
relations is congruence as detailed next.
T
TERMINOLOGY--CONGRUENT ANGLES AND SIDES
CONGRUENT ANGLES
EXAMPLES
B
Angles ∠ 1 and ∠ 2 are congruent, denoted
by ∠ 1 ≅ ∠ 2, means that m ∠ 1 = m ∠ 2.
2
The congruence symbol should remind you of
"=".
1
3
A
C
The angles of an equilateral triangle are
congruent, that is, ∠ 1 ≅ ∠ 2 ≅ ∠ 3
CONGRUENT SIDES
In the equilateral triangle above
Two sides of a triangle are congruent, denoted
___
___
___
___
___
___
AB ≅ B C ≅ AC
by AB ≅ AC means that side AB is the
___
same length as side AC .
1. f and j
2. 50o, 130o, 50o
3. b, d, e
4. f and g, g and j, j and h, f and h
SECTION 3.3
Lines, Angles and Triangles
163
CLAST EXAMPLE
Example
___
Solution
___
4. In Δ ABC, AB ≅ CB . Which of the
following statements is true for the figure
shown? (The measures of the angles are
represented by x, y, z as indicated.)
___
___
___
AB ≅ CB means that the line segments AB
___
and CB are the same length. This makes the
triangle an isosceles triangle, so y = x.
B
Now, x + 100o = 180o
and
x = 80o
z
y
x
C
A
Since y = x = 80o, y = 80o
100°
w
A. z = 80o
B.
z=x
C. y = 80o
D.
z = 10o
The CLAST combines the idea of perpendicular and parallel lines to establish other relationships
between the measures of angles. Here are the concepts we need.
T
TERMINOLOGY--PARALLEL AND PERPENDICULAR LINES
PARALLEL LINES
EXAMPLES
___
___
A
B
Lines AB and CD are parallel, denoted by
___ ___
___
___
E
AB || CD , if the lines AB and CD are in the
same plane and do not intersect..
C
D
F
These pairs of parallel lines are denoted by
___
___
___
AB || CD and
___
EF || GH .
PERPENDICULAR LINES
___
___
D
Lines AB and CD are perpendicular,
___
___
denoted by AB ⊥ CD , if they intersect at
right angles.
A
___
C
B
___
Lines AB and CD are perpendicular.
G
H
164
CHAPTER 3
Geometry and Measurement
CLAST EXAMPLE
Example
___
___
Solution
___
___
5. Given that CD || BE and AE ⊥ CD
which of the following statements is true for
the figure at the right? (The measure of
angle DCB is represented by "v".)
C
B
A. w = x
D. ∠ DCB and ∠ DEB are supplementary
angles
w
o
u E
y
z
135
x
B. y = x
C. ∠ EDC and ∠ DEB are vertical angles
D
v
A
Since y and the 135o angle are supplementary,
y = 45o. Now, z = 90o and x + y + z = 180o.
For y = 45o, z = 90o,
x + 45o + 90o = 180o
x + 135o = 180o
x = 45o
o
Since both x and y are 45 , y = x and the
There is one more important fact concerning parallel lines and the measures of related angles. Here
it is:
T
TERMINOLOGY--CORRESPONDING, ALTERNATE AND VERTICAL ANGLES
ANGLE RELATIONSHIPS
EXAMPLES
If L1 and L2 are parallel lines crossed by a
1
2
transversal as shown in the Example, the
L1
3
4
following relationships hold:
Corresponding angles are congruent.
∠ 1 ≅ ∠ 5, ∠ 3 ≅ ∠ 7, ∠ 2 ≅ ∠ 6 and
∠4 ≅ ∠8
These are the pairs of angles that lie on the
same side of the transversal.
Alternate interior angles are congruent.
∠ 3 ≅ ∠ 6, and ∠ 4 ≅ ∠ 5. These are the
pairs of angles between the parallel lines and
on opposite sides of the transversal
Vertical angles are congruent.
∠ 1 ≅ ∠ 4, ∠ 2 ≅ ∠ 3, ∠ 5 ≅ ∠ 8, and
∠6≅ ∠7
L2
5
6
7
8
Here is a quick way of remembering all these
facts: Look at angles 1 and 2. Angle 1 is
acute and 2 is obtuse. Think of
∠ 1 as “small" and ∠ 2 as "big".
All acute ("small" ) angles are congruent.
Thus, ∠ 1, ∠ 4, ∠ 5, and ∠ 8 are congruent.
All obtuse ("big") angles are congruent
Thus, ∠ 2, ∠ 3, ∠ 6 and ∠ 7 are congruent.
SECTION 3.3
Lines, Angles and Triangles
165
CLAST EXAMPLES
Example
Solution
6. Which statement is true for the figure shown
at right given that L1 and L2 are parallel
lines?
A. Since m ∠ T = 750, m ∠ S = 60o
L2
B. Since m ∠ T = 750, m ∠ S = 105o
P
L1
Q
R
75°
S
U
T
W
X
Y
V
45°
Look at the transversal on the left
m ∠ T = 75o
(Remember, all the "small" angles are
congruent). Now, ∠ S and ∠ T are
suppplementary, so m ∠ S = 180o - m ∠ T
= 180o - 75o = 105o.
C. m ∠ V = m ∠ R
D. None of the statements is true.
Example
Solution
x = 40o because x and the 40o angle are
corresponding angles.
7. If L1 || L2 and L3 || L4, which of the
following statements is true?
y = 180o - 40o = 140o, since x and y are
supplementary.
s
o
40
t
r
z
x
L
y
3
L
L
L
1
z = y = 140o, since z and y are alternate interior
angles (remember "big" angles are congruent)
A. z = 40o
B.
r = 140o
C. z = 140o
D.
s=t
4
2
166
CHAPTER 3
Geometry and Measurement
C. Similar Triangles
Objective IIB3
CLAST SAMPLE PROBLEMS
A
In the figure to the right, the measure of angle A is x
CB
?
BA
?
1. x = ? .
3. BD = ?
4. If the measure of angle A is z, what is AC ?
w
o
30
x
y
C
o
30
4
E
C
y
120 o
E
120o
D
z
B
A
z
D
5
x
7.5
B
Geometric figures with exactly the same shape, but not necessarily the same size, are called
similar figures. If you know that two triangles are similar, there are several relationships
you should know. We list them next.
T
TERMINOLOGY--SIMILAR TRIANGLES
DEFINITION OF SIMILAR TRIANGLES
EXAMPLES
E
When two triangles are similar, denoted by
Δ ABC ~ Δ DEF
(1) Corresponding angles are equal and
8
B
(2) Corresponding sides are proportional.
4
To remember the information in (1) note:
A
Δ ABC ~ Δ DEF
∠A ≅ ∠D, ∠B ≅ ∠E, ∠C ≅ ∠ F.
Match first and first (A and D), second and
second (B and E), third and third (C and F).
To remember the information in (2), note:
Δ ABC ~ Δ DEF
___
___
match AB and DE .
___
___
___
___
Δ ABC ~ Δ DEF match BC and EF .
Δ ABC ~ Δ DEF match AC and DF .
1. y
4
2
3
C
D
6
F
The two triangles above are similar.
(1) ∠ A ≅ ∠ D, ∠ B ≅ ∠ E, ∠ C ≅ ∠ F,
___
___
___
AB
BC
CA
2
4
3
= ___ = ___ = 4 = 8 = 6
DE
EF
FD
A
In the figure at right,
if BC is parallel to DE
C
B
D
E
(2)
___
2. CE
BC
3. BE
4. AC = 6
SECTION 3.3
Lines, Angles and Triangles
167
1. From a group of triangles, select the ones that are similar
2. Find specific measures for angles or sides of one of two similar triangles.
ÜCLAST EXAMPLES
Example
Solution
To find the correct answer, we have to check
each of the responses. The first triangle in A is
an equilateral triangle. For the second triangle
to be similar, it has to be equilateral and the
third angle must also be 55o.
But 55o + 55o + 55o = 165o and not 180o, so
the second triangle is not equilateral.
8. Which triangles are similar:
A.
5
5
55°
55°
5
B.
If the triangles are similar, the sides must be
6 5
proportional. Thus, 9 = 7 . But this is not
true, so the triangles are not similar.
9
5
30°
7
6
The two triangles in C are obviously not similar
since corresponding angles are not equal.
C.
A
o
25
B
D
C
60 o
E
D.
Q
A
80°
80°
45°
55°
B
C
R
S
The correct answer must be D. Let us see
why. The measure of the angles in the first
triangle are 55o and 80o, so the measure of the
third angle must be
180o - 80o - 55o = 45o.
But the measures of the angle in the second
triangle are 80o and 45o, so the third angle must
be 55o, making the corresponding angles equal
and the triangles similar.
168
CHAPTER 3
Geometry and Measurement
Example
Solution
Since ∠ 1 and ∠ 2 are vertical angles, they have
the same measure, so i is true.
9. Which statements are true for the pictured
triangles?
B
C
1
A
5
Triangle BAC and triangle DEC have equal
corresponding angles (the 90o angle and angles
1 and 2, which make the third angles equal).
Thus, Δ BAC ~ Δ DEC which makes statement
ii true.
E
2
3
D
___
___
___
AB
AC
___
i.
m∠ 1 = m∠ 2
ii. ___
ED
___
___
iii.
AB is parallel to ED
= ___
EC
A. i only
B.
ii only
C. i and ii only
D.
i, ii, and iii
___
Solution
Since m ∠ G = m ∠ H (vertical angles) and
m ∠ D = m ∠ B = 40o, the third angles must be
equal.
Thus, m ∠ A = m ∠ E, and i is true.
A
7.5
5
B
C
4
E
i.
iii.
m∠ A = m∠ E
___
___
CE
CB
= ___
CD
___
CA
Since Δ CDE ~ Δ ABC, corresponding sides
are proportional. Thus,
40°
H
G
___
This makes iii a true statement and the correct
10. Which statements are true for the pictured
triangles?
40°
___
AB ⊥ BD and DE ⊥ BD .
Example
D
___
Finally, AB is parallel to ED because
___
ii.
AC = 6
___
___
4 AC
4•7.5 30
5 = 7.5 or AC = 5 = 5 = 6
hence ii is true.
Since the two triangles are similar,
___
___
CE
CD
___
CA
A. i only
B.
ii only
C. i and ii only
D.
i, ii, and iii
= ___ , so iii is not true.
CB
The correct answer is C, i and ii are the only
true ones.
SECTION 3.3
Lines, Angles and Triangles
169
Section 3.3 Exercises
WARM-UPS A
1.
Name two pairs of complementary angles in the figure.
2.
Name a pair of supplementary angles in the figure.
3.
Name two pairs of vertical angles in the figure.
4.
Name two right angles in the figure.
5.
Name two acute angles in the figure.
6.
Is there an obtuse angle in the figure?
7.
Find m ∠ ABC
B
8. Find m ∠ 3
6
5
4
1
2
3
9. Find x
10. Find y and z
y
A
o
o
60 60
Triangle 1
C
x
o
100
50o
o
30
3
Triangle 2
Triangle 3
60
o
o
75 z
Triangle 4
In Exercises 11-14, classify the triangles according to their angles and sides.
11.
What type of triangle is triangle 1?
12.
What type of triangle is triangle 2?
13.
What type of triangle is triangle 3?
14.
What type of triangle is triangle 4?
15.
Name the geometric figure that is a parallelogram in which the diagonals are perpendicular and
of the same length.
16.
Name a quadrilateral in which only two sides are parallel.
170
CHAPTER 3
Geometry and Measurement
CLAST PRACTICE A
17.
PRACTICE PROBLEMS: Chapter 3, # 21-22
Which of the following pairs of angles
are supplementary, given that L1 and L2
are parallel lines?
A.
∠ P and ∠ S
B. ∠ Y and ∠ Q
D.
∠ R and ∠ Q
D. ∠ R and ∠ X
18.
What type of triangle is Δ ABC?
A.
Isosceles
B.
Equilateral
C.
Right
D.
Obtuse
19.
i.
X
L2
A
40
S
Y
50 o
o
B
C
B.
C.
D.
Select the geometric figure that possesses all of the following characteristics:
Triangle
A.
C.
21.
R
Q
Which of the angles is an obtuse angle?
A.
20.
P
L1
ii.
Scalene triangle
Acute triangle
All acute angles
iii.
All sides are equal
B. Equilateral triangle
D. Isosceles triangle
The words "right," "acute," "obtuse," "scalene," "isosceles," and "equilateral" are used
individually to describe triangles. Which of the following combinations of words are
impossible to describe a triangle?
A.
C.
Right; scalene
Right; obtuse
B. Isosceles; obtuse
D. Isosceles; right
SECTION 3.3
Lines, Angles and Triangles
171
WARM-UPS B
22. Find m ∠ 1.
23. Find m ∠ 4.
1 2
L
4
L
3
24. Find m ∠ 2.
5
L
1
___
115 o
1
2
and L are parallel
2
___
25. In D ABC, AB ≅ CB . Find m ∠ y.
B
z
___
___
26. In D ABC, AB ≅ CB . Find m ∠ x.
o
110
y
A
___
x
C
w
___
27. In D ABC, AB ≅ CB . Find m ∠ z.
___
___
___
___
Given that CD || BE and AD ⊥ CD .
28. Find m ∠ x.
29. Find m ∠ y.
30. Find m ∠ u.
31. Find m ∠ z.
C
D
45o
w
B v
u
E
y z
x
A
172
CHAPTER 3
Geometry and Measurement
Given that L1 and L2 are parallel
32. Find m ∠ Q.
S
L2
CLAST PRACTICE B
R
30
75°
33. Find m ∠ T.
34. Find m ∠ W.
Q
P
L1
U
o
W
T
X
Y
V
PRACTICE PROBLEMS: Chapter 3, # 23-24
35.
Which statement is true for the figure?
A.
m ∠ B = 100o
B. m ∠ E = 20o
C.
m∠ C ≠ m∠ H
D. ∠ A and ∠ B are
complementary
36.
Which statement is true for the figure?
E
A.
m∠ E π m∠ C
B. m ∠ I = 120o
C.
m∠ B = m∠ G
D. ∠ A and ∠ I are
supplementary
8
A
i only
B.
ii only
C.
iii only
D.
i and iii only
38.
Which of the statements is true for the
figure shown, given that L1 and L2 are
parallel lines?
A.
m∠ P = m∠ Y
B. m ∠ P = 100o
C.
∠ X and ∠ P are supplementary
D.
∠ Q and ∠ X are complementary
B
C
F
4
D
C
A B
D
100°
G H
E
6
37. Which statements are true for the figure
if M1 and M2 are parallel?
i.
m ∠ 3 = 50o ii. m ∠ 4 = 70o
iii. ∠ 1 and ∠ 2 are supplementary
A.
8
6
F I
G H
6
70°
1
3
L1
L2
2
M1
4
50°
M2
P
Q
80°
X
Y
SECTION 3.3
39.
i.
ii.
iii.
The diagram to the right shows lines in
the same plane. If line L1 is a horizontal
line, which statement is true?
L2 and L3 are the only parallel lines
L2 and L4 are perpendicular lines
Lines L2 and L3 are vertical lines.
A.
i only
B.
ii only
C.
iii only
D.
i and iii only
Lines, Angles and Triangles
L2
173
L3
L1
60°
60°
L4
L5
40.
Given that L1 || L2, which of the following
statements is true?
A.
u + v = 180o
C.
u = 40o
___
B.
D.
___
C
AC ⊥ BC
u
y = 50o
o
40
y
B
130 o A
L
v
x
z
1
L
2
WARM-UPS C
41.
The two triangles to the right are similar.
Find the lengths marked x and y.
5
y
4
x
3
4
42. The parallelograms to the right are
similar.
__
Find the length of the diagonal PR .
S
D
R
C
10
10
A
8
B
P
10
Q
174
CHAPTER 3
Geometry and Measurement
__
In the figure to the right, the line PQ is
parallel to the line AB In each problem,
find the missing lengths.
__
__ ___
__
AP
PC BQ
BC
43.
3
4
6
?
44.
5
4
?
6
45.
2
?
3
5
46.
?
4
4
8
C
Q
P
B
A
CLAST PRACTICE C
PRACTICE PROBLEMS: Chapter 3, # 25-27
47. Study figures A, B, C, D, then select the figure in which all triangles are similar.
B.
A.
80°
80°
75°
70°
C..
48.
D.
Study figures A, B, C, D, then select the figure in which all triangles are similar.
A.
B.
C.
D.
5
60°
5
60°
5
5
58°
9
30°
7
58°
6
SECTION 3.3
49.
A.
Which statement is true for the triangles
to the right?
m∠ A = m∠ E
B. m ∠ x = 30o
___
C.
AC = 3.5
___
___
CE
CB
D. ___
CA
C.
m∠ X = m∠ Z
___
___
CE
AB
___
= ___
CB
z
D
30°
x
4
30°
3
E
A
X
C
Y
130°
E
D
Z
EXTRA CLAST PRACTICE
___
51.
___
Given that AB || CD and
___
___
BD ⊥ CD which of the following
statements is true for the figure
shown. (The measure of angle ACD
is represented by x)
52.
E
A
A.
w=v
B.
∠ EAB and ∠ EBA are
supplementary
C.
∠ AEB and ∠ ECD are
complementary angles
D.
z=x
C
x
v
u z B
y
D
w
Select the geometric figure that
possesses all of the following
characteristics
i. four sided ii. right angle
opposite sides parallel
A. rectangle
B. square
C. trapezoid
D. rhombus
B
C
130°
None is true
5
y
B. m ∠ X = m ∠ Y
D.
175
A
= ___
CD
50. Which statement is true for the triangles
to the right?
A.
Lines, Angles and Triangles
B
```