1. 10B Apple Yoyo Jack Ikaros
Measurement
10B Apple Yoyo Jack Ikaros

2. Content 1.1 Imperial Measure of length
1.3 Relating SI and I mperial Units
1.4 SA of 3-D Shapes ~_~
1.5 Volumes of 3-D Shapes ~_~

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

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

5. READY?

6. 1.1 Imperial Measure of length
Develop personal referents to estimate imperial measures of length

7. Let’s figure it out. Imperial Unit Abbreviation Referent
Relationship between Units
inch
in.
Thumb length
foot
ft.
Foot length
1ft.=12in.
yard
yd.
Arm span
1yd.=36in.=3ft.
mile
mi.
Distance walked in 20 min
1mi.=
1760yd.=
5280ft.
Let’s figure it out.

8. Use the previous graph to solve the following problems
A)convert 5 yd. to inches and feet
B)convert 51 in. to (1)feet and inches
(2)yards, feet, and inches
Think about proportional reasoning(the relations between units)
e.g.12in.=12in.*(1/12)=1ft.
Use the previous graph to solve the following problems

9. Solution A)5yd.=5*3ft=15ft.=15* 12ft=180in.
B)51in.=51/12ft.=(4+ 3/12)ft.=4ft.3in.=1yd.1ft. 3in.
Solution

10. Let’s make it a bit more difficult.
Convert 12yd.32ft.144in. into yd. ft.
Let’s make it a bit more difficult.

11. The answer is……. 144in.=12ft. 12ft+32ft.=44ft.=14yd.2ft.
14yd.+12yd.=26yd.
So the answer is 26yd.2ft.
The answer is…….

12. 26yd.2ft.=960in.
12yd.32ft.144in.=960in.
correct
How can we verify it?

13. What do we call this? Unit analysis.
-Is one method of verifying that the units in a conversion are correct.
What do we call this?

14. 1.3 Relating SI and Imperial Units
SI UNIT
Millimetre(mm)
Centimetre(cm)
Metre(m)
Kilometre(km)
IMPERIAL UNIT
Inch(in)
Foot(ft)
Yard(yd)
Mile(mi)

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

16. Example I 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.

17. Example II Convert 6 ft. 2 in. to inches (1ft=12in) 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.

18. Example Ⅲ htruck =3.5m=350cm 350cm×0.4 in.=137.8 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)
htruck =3.5m=350cm
350cm×0.4 in.=137.8 in
hbridge =11ft.9in=141in＞137.8in
hbridge＞htruck
Yes! It can~

19. Surface Area 1.4 SA of 3-D Shapes ~_~
Area is the two-dimensional (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.

20. Surface Area Formulas
Right Cone
ASide= πrs
ABase= πr2
SA = πr2 + πrs

21. Surface Area Formulas Square-based Pyramid Atriangle = ½ bs Abase = b2
SA = 2bs + b2
General Right Pyramid
SA = sum of all the
areas of all the faces b S
~Pyramid head~

22. Surface Area Formulas
Rectangular Prism
SA = 2(hl + lw + hw)

23. Surface Area Formulas Right Cylinder Atop=πr2 Abottom=πr2 Aside=2πrh
SA=2πr2 + 2πrh

24. Example Questions
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 + b2
S= 13
h=12 5
b=10

25. Example Questions
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= Aside=2πrh
2πrh=114×23=2622

26. 1.5 Volumes of 3-D Shapes ~_~
is the space that a shape occupies
often quantified numerically using the SI unit , the cubic meter.

27. Volume Formulas
Right Cone
ABase= πr2
V=1/3(area of base)h
=1/3πr2h

28. Volume Formulas General Right Pyramid V = 1/3(area of base) h
Square-based Pyramid
V = 1/3b2h
Right Rectangular Pyramid
V = 1/3lwh
~Pyramid head~

29. Volume Formulas General Right Prism Rectangular Right Prism
V=(area of base)h
Rectangular Right Prism
V=lwh
General Right Prism
Rectangular Prism

30. Volume Formulas
Right Cylinder
Abase=πr2
V=(area of base)h
=πr2h

31. Example Questions (*Vcylinder= πr2 h)
3. Which of the following expressions represents the volume of the cylinder below?
(*Vcylinder= πr2 h)
d=2x+4
So, r=1x+2
V= πr2 h=π(1x+2) 2 (3x-1 ) ……
It’s “C”!

32. What is it ??? 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 line segment that joins two points on a sphere and passes through the centre is a diameter.
What is it ???

33. The surface area, SA, of a sphere with radius r is : SA = 4πr
Surface Area of a Sphere
The surface area, SA, of a sphere with radius r is :
SA = 4πr 2

34. The surface area, SA, of a hemisphere with radius r is : SA=3πr 2
Surface Area of a Hemisphere
The surface area, SA, of a hemisphere with radius r is :
SA=3πr 2

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

36. Solution: Use the formula for the surface area of a sphere.
The radius is:
½(3 in.) = 1.5 in.
SA = 4πr2
SA = 4π(1.5)2
SA= 28.8
The surface area of a baseball is approximately 28 square inches.

37. The volume, V, of a sphere with radius r is : V =4/3πr
Volume of a Sphere
The volume, V, of a sphere with radius r is :
V =4/3πr 3

38. Example:
The sun approximates a sphere with diameter mi. What is the approximate volume of the sun?

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

40. 1.7 Solving Problems Involving Objects

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

42. 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=
Solution:
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=
Volume of the composite object is:
= The So, the volume of the composite object is approximately 69.9 m3.

43. That’s all in the chapter 1
Easy Right?

44. NO MORE Q? So, that’s all we need to teach you today.
Let’s have a xiao quiz~

45. QUIZ TIME!
Remember to…
Choose “C”!