1.1
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1.2
 Measuring instruments
1.4
Imperial Units(in. yd. ft. mi.)
Referent
Abbreviation
Unit analysis
SI system measures(We are going to talk about it later.)
Proportional reasoning
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
Convert measurements between SI units and imperial units
SI units
1.4
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Right pyramid
Apex
Slant height
Polygon base
Lateral area
Right cone
1.5
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Cylinder
Right prism
Base area
Cone
Radius
1.6
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
1.7
Sphere
Surface area
Volume
Hemisphere


Substitute
Composite Objects

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Review how to:
 Convert the units
Imperial units with Imperial units——Jack
Imperial units with SI units——Ikaros
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 Calculate the surface area of 3-D shapes——Yoyo
Right Cone
Right Pyramid
Right Prism
Right Cylinder
Sphere
Hemisphere
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
Calculate the volumes of 3-D shapes(above-mentioned)——Yoyo
Solving Problems Involving Objects——Apple
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Example Questions
Quiz Time!10-15minutes/7 questions(multiple-choice)No written!

Imperial
Unit
Abbrevia tion inch foot yard mile in.
ft.
yd.
mi.
Referent Relations hip between
Units
Thumb length
Foot length
1ft.=12in.
Arm span 1yd.=36in
.=3ft.
Distance walked in
20 min
1mi.=
1760yd.=
5280ft.

SI UNIT
 Millimetre(mm)



Centimetre(cm)
Metre(m)
Kilometre(km)
IMPERIAL UNIT


Inch(in)
Foot(ft)


Yard(yd)
Mile(mi)

SI Units
1km=1000m
1m=100cm
1cm=10mm
1m=1000mm
1m=10dm
Imperial Units
1mi=1760yd
=5280ft
=63360 in
1yd=3ft=36in
1ft=12in
1in=0.083333ft
1ft=0.3333yd
SI Units to
Imperial Units
1mm ≈ 0.04in
Imperial Units to
SI units
1in=2.54cm
1cm≈0.4in
1m≈39in
1m≈3.25ft
1km≈0.6mi
1ft≈30cm
1ft≈0.3m
1yd=91.44cm
1yd≈0.9m
1mi≈1.6km

A lane is approximately
19m long.
What is this measurement to the nearest foot?
(1m≈3.25 ft.)



 From the table,1m≈3.25 ft.
So,19m≈19 x (3.25) ft.
19m≈62 ft.
A length of 19m is approximately 62 ft.

Convert 6 ft. 2 in. to inches




1 ft. = 12 in.
So, 6 ft. = 6
12 in.
6 ft. = 72 in
And, 6 ft. 2 in.=72 in. + 2 in. =74 in.

A truck driver knows that his truck
is 3.5m high.The support beams of a bridge are 11ft.9in. high. Can the truck cross the bridge smoothly?
(1cm≈0.4in)
 h truck
=3.5m=350cm
350cm × 0.4 in.=137.8 in
 h bridge
=11ft.9in=141in ＞ 137.8in
 h bridge
＞ h truck
 Yes! It can~



Area is the twodimensional (2-D) size of a surface.
Surface area (SA) of a solid is the total area of
the exposed surfaces of a three-dimensional (3-D) object.

 Right Cone
 A
Side
= πrs
 A
Base
= πr 2
 SA = πr 2 + πrs



Square-based Pyramid
 A triangle
= ½ bs
 A base
= b 2
 SA = 2bs + b 2
General Right Pyramid
 SA = sum of all the areas of all the faces
~Pyramid head~

 Rectangular Prism
 SA = 2( hl + lw + hw )

 Right Cylinder
 A top
=πr 2
 A bottom
=πr 2
 A side
=2πrh
 SA=2πr 2 + 2πrh

1. Which expression could be used to calculate the surface area of the right square-based pyramid with a base length of 10 cm and a
height of 12 cm?
*
SA = 2bs + b 2

2. Raj was asked to make a cylindrical tank with a lateral surface area of 2622 m 2 and a height of 23 m.
Which net diagram below would be correct for this cylinder?
*Lateral SA= A side
=2π rh
2πrh=114 × 23=2622


 is the space that a shape occupies often quantified numerically using the SI unit , the cubic meter .

 Right Cone
 A
Base
= πr 2
 V=1/3(area of base)h
=1/3πr 2 h

 General Right Pyramid
 V = 1/3(area of base) h
 Square-based Pyramid
 V = 1/3b 2 h
 Right Rectangular Pyramid
 V = 1/3lwh
~Pyramid head~



General Right Prism
 V=(area of base)h
Rectangular Right Prism
 V=lwh General Right Prism
Rectangular Prism

 Right Cylinder
 A base
=πr 2
 V=(area of base)h
=πr 2 h

3. Which of the following expressions represents the volume of the cylinder below?
(*
V cylinder = πr 2 h
)
 d=2x+4
 So, r=1x+2
 V= πr 2 h=π(1x+2) 2 (3x-1 )
 ……
 It’s “C”!

Definition of sphere:
A sphere is the set of points which are all the same distance from a fixed point which is the centre in space. A line segment that joins the centre to any point on the sphere is a radius. A
and passes through the centre is a diameter.

Surface Area of a Sphere
π
2

Surface Area of a Hemisphere
2

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

Solution:
Use the formula for the surface area of a sphere.
The radius is:
½(3 in.) = 1.5 in.
SA = 4 π r2
SA = 4 π (1.5)2
SA= 28.8
The surface area of a baseball is approximately 28 square inches.

V =4/3πr
3

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

Use the formula for the volume of a sphere.
The radius, r, is: r = ½ (870 000mi.) r = 435 000mi.
V = 4/3 πr 3
V = 4/3 π(435 000mi.) 3
V = 3.4479 * 10 17

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

First
The object comprises a right rectangular prism and a right rectangular pyramid.
Use the formula for the volume of a right rectangular prism.
V= lwh
V=(6.7)(2.9)(2.9)
V= 56.347
Then
Use the formula for the volume of a right rectangular pyramid.
V= 1/3 lwh
V= 1/3(6.7)(2.9)(2.1)
V= 13.601
Volume of the composite object is:
56.347 + 13.601= 69.948
The
So, the volume of the composite object is approximately 69.9 m3.

Let’s have a xiao quiz~