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Measurement 10B Apple Yoyo Jack Ikaros

Measurement 10B Apple Yoyo Jack Ikaros. 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 ~_~. Today’s objects. 1.1 Imperial Units(in. yd. ft. mi.) Referent Abbreviation Unit analysis

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Measurement 10B Apple Yoyo Jack Ikaros

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  1. 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.

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

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

  11. 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. 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

  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 A truck driver knows that histruck is 3.5m high.The support beams of a bridge are 11ft.9in. high. Can the truckcross 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. 1.4 SA of 3-D Shapes ~_~ Surface Area • Areais 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 S b b ~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 2and 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 ~_~ Volume • 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 • 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 3. Which of the following expressions represents the volume of the cylinder below? (*Vcylinder= πr2 h) • d=2x+4 • So, r=1x+2 • V= πr2h=π(1x+2) 2 (3x-1 ) • …… • It’s “C”!

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

  34. Surface Area of a Hemisphere • The surface area, SA, of a hemisphere with radius r is : • SA=3πr2

  35. 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. 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 870 000 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 (870 000mi.) r = 435 000mi. V = 4/3 πr3 V = 4/3 π(435 000mi.)3 V = 3.4479 * 1017

  40. 1.7 Solving Problems Involving Objects

  41. Example: Determinethe 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= 56.347 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= 13.601 Volume of the composite object is: 56.347 + 13.601= 69.948The So, the volume of the composite object is approximately 69.9 m3.

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

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

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

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