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Chapter 1 Measurement Math 1201 1
Chapter 1 – Measurement
Sections 1.1-1.3:
Imperial units inch, foot, yard, mile
SI units metric system
(I) Converting between imperial units by unit analysis
Conversion Table:
(a) yards to feet
= yds x
Sometimes conversions may involve more than one scale
(b) miles to inches
= mi x x
1 ft. = 12 in.
1 yd. = 3 ft.
1 yd. = 36 in.
1 mi. = 1760 yd.
1 mi. = 5280 ft.
From the scale above, what units must be
placed in the numerator (top) and
denominator (bottom) so that yards cancel
and feet remain?
From the scale above, what units must be
placed in the numerators (top) and
denominators (bottom) so that miles
cancel and inches remain?
Chapter 1 Measurement Math 1201 2
Relationships between units:
Example 1:
(a) Convert 100 yd to ft.
(b) Convert 100 yd to inches
Example 2: A sunken ship is discovered by sonar to be in 13 200 ft of water.
Convert this depth to miles.
Example 3: Convert 19.75 yds to yards, feet and inches.
1 ft. = 12 in.
1 yd. = 3 ft.
1 yd. = 36 in.
1 mi. = 1760 yd.
1 mi. = 5280 ft.
Chapter 1 Measurement Math 1201 3
Example 4: The perimeter of a ceiling in a room is measured for the purpose of
installing crown molding. The perimeter of the ceiling is 272 in.
(a) What is the perimeter of the ceiling in feet?
(b) The cost of crown molding is $3.75/ft.. Determine the cost of ordering
crown molding to install in the room.
Example 5: There is 6 yd of material to be cut into strips that are 5in wide. How
many strips can be made?
Chapter 1 Measurement Math 1201 4
Math 1201 questions page 11-12 of Textbook
7. Convert: a) 3 ft. to inches, b) 63 yd. to feet, c) 48in. to feet
8. Convert a) 2 mi. to feet b) 574 in. to yards, feet and inches
10. Carolyn is building a pen for her dog.
The perimeter of the pen is 52 feet.
a) Covert the perimeter to yards and feet.
b) The fencing material is sold by the yard. It costs $10.99/yd. What is the cost of
material before taxes?
11. David has 10 yd. of material that he will cut into strips 15in. wide to make mats.
a) How many mats can David make?
Answers: 7a) 36in b) 189ft c) 4ft
8.a) 10560 ft b) 15yd 2 ft 10 in
10a) 17yd 1ft b) $197.82
11.a) 24 mats
15.a) $119.99 b) $18.59
16. 1062 ft
18. 28 tulip bulbs
Chapter 1 Measurement Math 1201 5
(II) Converting between SI units
The smallest metric measurement on the ruler below is the _____________.
How many divisions make up 1 cm? Answer:_______________
metric measurements for determining length is based on increments of 10.
SI units Abbreviation Relationship between
units
millimeter mm
centimetre cm 1 cm = 10 mm
metre m 1 m = 100 cm
kilometre km 1 km = 1000 m
Example 5: Convert each length to the indicated measurement.
1. 3 cm to mm 2. 15 m to km 3. 18 mm to cm
4. 360 m to cm 5. 6 km to m 6. 17 cm to m
Chapter 1 Measurement Math 1201 6
(III) Converting Imperial Units to SI units
1 foot = 12 inches 1 yard = 3 feet 1 mile = 1760 yards
1 inch = 2.54 centimetres ≐ 2.5 centimetres 1 mile ≐ 1.6 kilometres
Example 6: Convert each measurement to the nearest tenth.
(a) 8 in to cm (b) 264 mi to km (c) 7 ft to cm
(d) 1 yd = ______ mm
(e) 6 ft. 7 in to cm (f) 4 yd. 2 ft. 2 in to m
Chapter 1 Measurement Math 1201 7
Questions Page 22-23 of testbook
Answers: 4a) 40.6cm b) 1.2mt c) 4.6m d) 1.5km
e) 9.7km f) 50.8mm
5.a) 1 in b) 8 ft c) 11yd d) 93mi
6a) 55.9 cm b) 256.5cm c) 9.6m
7.a) i) 2ft 6in ii) 3 yd iii) 6mi
10. the odometer is accurate; 142 km is close to 87 mi
12. a) Michael
14. 144 sections of casing
15. 28 in
Chapter 1 Measurement Math 1201 8
REVIEW OF SURFACE AREA
Area formulas: (𝒖𝒏𝒊𝒕𝒔)𝟐
Rectangle: 𝑨 = 𝒍 × 𝒘 Square: 𝑨 = 𝒍 × 𝒍 or 𝑨 = 𝒍𝟐
Circle: 𝑨 = 𝝅 ∙ 𝒓𝟐
Triangle: 𝑨 =𝒃𝒉
𝟐 or 𝑨 =
𝟏
𝟐𝒃𝒉 Pythagorean Theorem: 𝑎2 + 𝑏2 = 𝑐2
Net Diagram the diagram formed if all sides were unfolded
Find the surface area of the following:
1. Square/Rectangular Prism
(A)
(B)
10cm
10cm
10cm
10cm
6cm
8cm
Chapter 1 Measurement Math 1201 9
NOTE:
2.
12 cm
8 cm
NOTE: the top/bottom are circles and
the curved surface is a rectangle
whose length is the circumference of
the circle
Chapter 1 Measurement Math 1201 10
Section 1.4: Surface Area of Right Pyramids and Right Cones
Pyramids: are named or described by the shape of its base.
SA of pyramids = SA of base + SA of lateral sides (triangles)
NOTE if the base polygon of a pyramid is regular (all sides =) then all
lateral triangles are congruent
Example 1: Determine the surface area of the regular triangular pyramid (regular
tetrahedron)
Sketch the tetrahedron net diagram
13 yd
Chapter 1 Measurement Math 1201 11
The Right Pyramid
A three dimensional object that has
Triangular faces (lateral faces)
Base is a polygon (many sides)
Point on top is called the apex
S represents slant height (the height
Of the triangular face)
h is the height of the pyramid (the
Perpendicular distance from the
Apex to the center of the base)
The Lateral Area, AL, is the area
Of the exposed triangles (surface area without the base)
Example 2: Determine the lateral area and the surface area of the square pyramid
Sketch the net diagram
Chapter 1 Measurement Math 1201 12
Example 3: Determine the slant height of a square pyramid given the surface area.
Example 4: Determine the surface area of a square pyramid given the height of the pyramid.
Height is 8m and base is 12m
Chapter 1 Measurement Math 1201 13
Example 5: Determine the surface area of a rectangular pyramid.
Net Diagram:
Questions: Page 34-35 #4,5,8a,10,16b,13b,18
Chapter 1 Measurement Math 1201 14
Extra practice: Surface Area of a pyramid
1. If the surface area of the square pyramid is 160.8 cm2 then determine the slant height.
2. Determine the surface area for each pyramid.
(A)
6 cm
6 cm
S
10 cm
10 cm
12 cm
Chapter 1 Measurement Math 1201 15
(B)
12 cm
18 cm
15 cm
Chapter 1 Measurement Math 1201 16
(C)
(D)
10 cm
13 cm
22 cm
15 cm
16 cm
13 cm
Answers: 1. 10.4 cm
2.(A) 360 cm2 (B) 717.6 cm2
(C) 652.8 cm2 (D) 709.5 cm2
Chapter 1 Measurement Math 1201 17
Right Cones:
SA = lateral area + base
SA =
Example 6: Determine the surface area of the right cone.
8 cm
18cm
Chapter 1 Measurement Math 1201 18
Example 7: Determining an unknown measure of a cone
SA = 2325 cm2 and radius r = 10 cm.
Determine the height h of the cone
Questions: Page 34-35 #6a7b,8b,11,15,16a
10cm
h
Chapter 1 Measurement Math 1201 19
Section 1.5: Volume of Right Pyramids and Right Cones
Volume – the amount of space an object occupies
Capacity – the amount of material a container holds
--volume and capacity are measured in cubic units. (𝑢𝑛𝑖𝑡𝑠)3
(A) Right Square Prisms, Right Rectangular Prisms 𝑉 = 𝐴𝐵 × 𝐻
The Volume of a Rectangular Pyramid is one-third the volume of the Right
Rectangular Prism with same dimensions
2 cm
3 cm
5 cm
2 cm
3 cm
5 cm
Chapter 1 Measurement Math 1201 20
(B) Right Cylinder
Similarly, the volume of a right cone is one third the volume of a right
cylinder with the same base and height.
5 cm
16cm
5 cm
16cm
Questions: Page 42-43 #4a,5b,6a,7b,8,9
Chapter 1 Measurement Math 1201 21
Volume Examples where you may need to find a missing measure first:
Example 1: A right square pyramid has a base side length of 6.4 cm. Each
triangular face has two equal sides of length 8 cm. Determine the height and volume of the pyramid.
6.4 cm
6.4 cm
8cm
8cm
Chapter 1 Measurement Math 1201 22
Example 2: Determine the volume for:
Example 3: The height of the right cone below is 10 cm and its volume is 377 cm3
Determine the diameter.
4 cm
5 cm
10 cm
Questions: Page 42-43 #,11,15,18bd
Chapter 1 Measurement Math 1201 23
Section 1.6: Surface Area and Volume of a Sphere:
Sphere: Hemisphere:
Surface Area: 𝑆𝐴 = 4𝜋𝑟2 Surface Area: 𝑆𝐴 = 2𝜋𝑟2 + 𝜋𝑟2 = 3𝜋𝑟2
Volume: 𝑣 =4
3𝜋𝑟3 or 𝑣 =
4𝜋𝑟3
3 Volume: 𝑣 =
2
3𝜋𝑟3 or 𝑣 =
2𝜋𝑟3
3
Example 1: An official basketball has a radius of 12.5 cm and usually
has a leather covering. Approximately how much leather,
in cm2, is required to cover 12 official basketballs?
Example 2: The surface area of a softball is approximately 50.24 in2.
Determine the diameter of the softball.
A sphere is the set of points in
space that are the same fixed
distance from a fixed point,
which is the center.
Chapter 1 Measurement Math 1201 24
Example 3: Find the surface area of the hemisphere that has a radius of 8.0cm
Example 4: A fitness ball when inflated with air has a circumference of
198 cm. Determine the volume of the fitness ball to the nearest tenth of a cm3.
Questions: Page 51 #3d,4a,5,8,11,18,19,20
Chapter 1 Measurement Math 1201 25
Section 1.7 Solving Problems Involving Objects
Example 1:
(A) Find Surface Area (B) Find Volume
2 ft
4 ft
Chapter 1 Measurement Math 1201 26
Example 2: Below is a sketch of a grain bin. If a farmer's grain truck can hold 560 cubic feet of barley, how many truckloads of barley are required to fill the bin?
Chapter 1 Measurement Math 1201 27
Example 3: Determine the surface area and volume of a right cylinder with a right cone of the same height removed.
(A) Find Surface Area (B) Find Volume
Questions: Page 59-60 #3-11 (omit #4)
Chapter 1 Measurement Math 1201 28
Formulas
Surface Area of a
Cylinder
SA = 2𝜋𝑟2 + 2𝜋𝑟ℎ
Surface Area of a
Cone
SA = 𝜋𝑟2 + 𝜋𝑟𝑠
Surface Area of a
Sphere
SA = 4𝜋𝑟2
Volume of a Sphere V =
4
3 𝜋𝑟3 or V =
4 𝜋𝑟3
3
Volume of a Cone V =
1
3 𝜋𝑟2ℎ or V =
𝜋𝑟2ℎ
3
Volume of a Pyramid V = 1
3 Ah or V =
𝐿𝑤ℎ
3
Imperial 1 ft. = 12 in. 1 yd. = 3 ft.
1 mi. = 1760 yd. Imperial to SI Units
1 in. = 2.54 cm ~ 2.5 cm 1 mi. ~ 1.6 km