Chapter 1 Measurement. §1.1 Imperial Measures of Length.

Chapter 1Measurement

§1.1 Imperial Measures of Length

Examples1. a) Convert 3 yards to feet. b) Convert 3 yards to inches.

c) Convert 47 inches to feet and inches. d) Convert 47 inches to yards.

Start with what you know:

3 yd

multiply by the conversion fraction (units you want on the top)

3 ft1 ydx = 9 ft

Start with what you know:

3 yd

multiply by the conversion fraction (units you want on the top)

36 in1 ydx = 108 in

Start with what you know:

47 in

multiply by the conversion fraction (units you want on the top)

1 ft12 inx = 3.92 ft

= 3 ft 11 in

47 in1 yd36 inx = 1.31 yd

2. Jessica is building a pen for the baby chickens. The perimeter of the pen will be 197 inches.a) What will the perimeter of the enclosure be in feet and inches?Start with what you know:

197 in

multiply by the conversion fraction (units you want on the top)

1 ft12 inx = 16.42 ft

= 16 ft 5 in

b) The wire mesh is sold by the foot. It costs $1.88/ft. What will be the cost of the materials before taxes?= 17 ($1.88) = $31.96

3. The school council has 6 yards of fabric that will be cut into strips 5 inches wide to make decorative banners for the school dance. How many banners can be made?

How many groups of 5 inches will go into 6 yards?

Change 6 yards into inches

6 yd 36 in1 ydx = 216 in

216 in ÷ 5 in = 43.2

43 banners can be made

4. A map of Alaska has a scale of 1: 4 750000. The distance on the map

between Paxson and the Canadian border is inches.

What is the distance to the nearest mile?

113

16

3 1 ft12 in

(4750000 in) x

= 276.4461 mi

The distance between Paxson and the Canadian border is approximately 276 miles.

1116

1 mi5280 ft

x

a b/c or

§1.2 Measuring Length and Distance

4 cm

610 = 4.6 cm

2 in 216 = 2 in

1 8

= 2.125 in

6 cm

8.510 = 6.85 cm

in 1516 = 0.9375 in

Using a Vernier Caliper: ***Question on Provincial***

The top scale gives the whole number: (where the zero on

the bottom scale points)

7 mm

To get the decimal portion of the measurement find the

first set of lines that match up

.4

Used to make very precise measurements

Using a Vernier Caliper: ***Question on Provincial***

The top scale gives the whole number: (where the zero on

the bottom scale points)

22 mm

To get the decimal portion of the measurement find the

first set of lines that match up

.7

Used to make very precise measurements

30 mm.0

9 mm.9

Item SI Measurement (Metric) Imperial Measurement CaliperPlastic Container Lid Diameter:Radius:Circumference:Depth of Lid:

Diameter:Radius:Circumference:Depth of Lid:Diameter:Depth of Lid:

Paper Clip Length:Width:Unravelled Length:Length:Width:Unravelled Length:

Length:Width:

Items you will need. - Ruler (showing both SI & Imperial Measurement) - String - Caliper - Plastic Container Lid - Paper Clip - Desk - Three other objects of your choice.

Item SI Measurement (Metric)

Imperial

the height of a person

the width of a television

the length of a staple

2. Determine the most suitable units in both the metric and imperial systems for measuring the following items.

cm or m in or ftcm inmm in

§1.3 Relating SI and Imperial

Measurements

Examples1. A bowling lane is approximately 19 meters long. What is this measurement to the nearest foot?

Start with what you know:

19 m

multiply by the conversion fraction (units you want on the top)

1 ft0.3048 mx = 62.34 ft

Bowling lane is approximately 62 ft.

Examples2. After meeting in Emerson, Manitoba, Hana drove 62 miles south and Famin drove 98 km north. Who drove farther? Round your answer to the nearest hundredth.

Convert so both are in same units (Km)

62 mi1.609 Km

1 mix = 99.76 Km

Hana drove 1.76 Km farther.

Examples3. Convert 6 feet 2 inches to centimetres. Round your answer to the nearest hundredth.

Convert 6 ft 2 in to inches.

74 in2.54 cm

1 inx = 187.96 cm

6 feet 2 inches = 187.96 cm

6(12) + 2 = 74 in

Examples4. A truck driver knows that her semitrailer is 3.5 meters high. The support beams of a bridge are 11ft. 9in. high. Will the vehicle fit under the bridge?

Convert metric into imperial, or imperial into metric

141 in2.54 cm

1 inx

= 3.5814 m

Yes the semitrailer will fit

11(12) + 9 = 141 in

Imperial to Metric (convert 11 ft 9 in to inches)

1 m100 cmx

§1.4 Surface Areas of Right Pyramids and

Right Cones

* All formulae will be on the Provincial Exam Formula Sheet *

Examples1. Jeanne-Marie measured then recorded the lengths of the edges and slant height of this regular tetrahedron. What is its surface area to the nearest square centimetre?

bh2

= 35.1 cm2

SA = 4(A Triangle)

A Triangle = (7.8)(9.0)

2=

= 4(35.1) = 140.4 cm2

SA = 140 cm2

Examples2. A right rectangular pyramid has base dimensions 8 ft. by 10 ft., and a height of 16ft. Calculate the surface area of the pyramid to the nearest square foot.

(length)(width)

= 80 ft2Need to find slant height (s)

A Base = = (10)(8)

a2 + b2 = c2

SA = 80 + 2(82.45) + 2(67.04)

A Front/Back = 42 + 162 = s2

16 + 256 = s2

272 = s2

√272 = s16.49 ft = s

bh2

(10)(16.49)2=

= 82.45 ft2 Blue

a2 + b2 = c2

A Side = 52 + 162 = s2

25 + 256 = s2

281 = s2

√281 = s16.76 ft = s

bh2

(8)(16.76)2=

= 67.04 ft2 Red

= 378.98 ft2

Examples3. A right cone has a base radius of 2 ft. and a height of 7 ft. Calculate the surface area of this cone to the nearest square foot.

Need to find slant height (s)

SA = 58 ft2

SACone = πr2 + πrs

= π(2)2 + π(2)(7.28)

a2 + b2 = c2

22 + 72 = s2

4 + 49 = s2

53 = s2

√53= s7.28 ft = s = 4π + 14.56π

= 18.56π= 58.31

Examples4. The lateral area of a cone is 220 cm2. The diameter of the cone is 10cm. Determine the height of the cone to the nearest tenth of a centimetre.

Lateral Area: the surface area of an object not including the area of its base

h = 13.1 cm

Lateral Area = πrs

220 = a2 + b2 = c2

r2 + h2 = s2

25 + h2 = 196.28

h2 = 171.2801h = √171.2801h = 13.087 cm

π(5)s

14.01 = s5π 5π 52 + h2 = 14.012

§1.5 Volumes of Right Pyramids & Right Cones

1(area of base) h

3V

1(area of base) h

3V

Volume of Right Pyramid Volume of Right Cone

Examples1. Calculate the volume of this right square pyramid to the nearest cubic inch.

Need to find height (h)

V = 30 in3

V = ⅓(4)(4)(h)

= ⅓(4)(4)(√32)

a2 + b2 = c2

h2 + 22 = 62

h2 + 4 = 36h2 = 32

h= √32

1(area of base) h

3V

= 30.17

Examples2. Determine the volume of a right rectangular pyramid with base dimensions 5.4cm by 3.2cm and height 8.1cm. Answer to the nearest tenth of a cubic centimetre.

V = 46.7 cm3

V = ⅓(5.4)(3.2)(8.1)

= 46.656

1(area of base) h

3V

Examples3. Determine the volume of this cone to the nearest cubic inch.

V = 679 in3

V =

= ⅓π(6)2(18)

1(area of base) h

3V

⅓(πr2h)

= 678.584

Examples4. A cone has a height of 4yd. and a volume of 205 cubic yards. Determine the radius of the base of the cone to the nearest yard.

r = 7 yards

205 =

205 =

1(area of base) h

3V

⅓πr2(4)

3 x x3

4 ydV = ⅓πr2h

4πr2

33(205) = 4πr2

4π4π3(205) = r2

4π√ √r = 6.996

Examples5. Find the top area, bottom area and lateral area of the cylinder.ATop = πr2 = ABottom

= π(6)2 = 36π

Top Area = _____ Bottom Area = ______ Lateral Area = _____

Total Surface Area = _________

= 113.1 cm2

113.1 cm2 113.1 cm2

ASide = ALateral

Circumference= 2πr H

eight

Alateral = 2πrh= 2π(6)(25)= 300π= 942.5 cm2

942.5 cm2

1168.7 cm2

§1.6 Surface Area and Volume of a Sphere

Examples1. The diameter of a baseball is approximately 3in. Determine the surface area of a baseball to the nearest square inch.

SA = πd2

= π(3)2

= 9π= 28.27= 28 in2

Examples2. The surface area of a lacrosse ball is approximately 20 square inches. What is the diameter of the lacrosse ball to the nearest tenth of an inch? SA = πd2

20 = πd2

20 = d2

2.52 = d

π π

√ √π

2.5 in = d

Examples3. The sun approximates a sphere with diameter 870 000 mi. What is the approximate volume of the sun?

V = πr3

V = 3.45 x 1017 cubic miles

34

2870 000= π( )3

34

= π(435 000)3

34

(345 000 000 000 000 000 cubic miles)

4. A hemisphere has radius 8.0cm.a) What is the surface area of the hemisphere to the nearest tenth of a square centimetre?

SAHEMISPHERE = SASPHERE = 4πr2

2 = 2πr2

2

SA = 2πr2 + πr2

HemisphereCircle on Bottom

= 3πr2 = 3π(8)2 = 3π(64) = 192π = 603.2 cm2

4. A hemisphere has radius 8.0cm.b) What is the volume of the hemisphere to the nearest tenth of a cubic centimetre? VHEMISPHERE =

4πr3

321 ( )

= πr3

64

= π(8)3

64

= π(512)64

= π 6

2048

= 1072.3 cm3

§1.7 Solving Problems Involving Objects

Examples1. Determine the volume of this composite object to the nearest tenth of a cubic metre.

VTOP = (area of base)(height)31

= (6.7)(2.9)(2.1)31

= 13.601Don’t round

until the end.

VBOTTOM = L x W x H= (6.7)(2.9)(2.9)= 56.347

VTOTAL = 13.601 + 56.347 = 69.948VTOTAL = 69.9 m3

Examples2. Determine the surface area of this composite object to the nearest square foot.

SATOP = (4πr2)21

= 2π(2)2

Leave like this

SABOTTOM = 2πr2 + 2πrh

= π(2)2 + 2π(2)(4)

= 20π

SATOTAL = 8π + 20π = 87.965SATOTAL = 88 ft2

= 2πr2Does not include

bottom of hemisphere

= 8π

Minus top circle of cylinder

= πr2 + 2πrh

= 4π + 16π

= 28π

Examples3. A cabane a sucre is a composite object formed by a rectangular prism with a right triangular prism as its roof. Determine the surface area

SATOP: 2bh = 3=

Two ∆ so A∆ = 2(3) = 6 yd2

A∆= 2

(3)(2)

LxW = 12.5

Two so A = 2(12.5) = 25 yd2

A = = (5)(2.5)a2 + b2 = c2

a

b

c

Need to find slant

22 + (1.5)2 = c2

4 + 2.25 = c2

6.25 = c2

√6.25 = √c2

2.5= c

SABOTTOM: w∙h = 6 Front & Back so AFRONT/BACK= 2(6) = 12 yd2AFRONT= = (3)(2)

w

L

h

w

L∙h = 10 Two sides so ASIDES = 2(10) = 20 yd2ASIDE = = (5)(2)

w∙L = 15 yd2ABOTTOM = = (3)(5)

**

**

*SATOTAL= 6 + 25 + 12 + 20 + 15 SA TOTAL = 78 yd2