1. Chapter 1 Measurement

2. §1.1 Imperial Measures of Length

4. Examples 1. 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 ft 1 yd x = 9 ft Start with what you know: 3 yd multiply by the conversion fraction (units you want on the top) 36 in 1 yd x = 108 in Start with what you know: 47 in multiply by the conversion fraction (units you want on the top) 1 ft 12 in x = 3.92 ft = 3 ft11 in 47 in 1 yd 36 in x = 1.31 yd

5. 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 ft 12 in x = 16.42 ft = 16 ft5 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

6. 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 in 1 yd x = 216 in 216 in ÷ 5 in = 43.2 43 banners can be made

7. 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? 3 1 ft 12 in (4750000 in) x = 276.4461 mi The distance between Paxson and the Canadian border is approximately 276 miles. 11 16 1 mi 5280 ft x a b / c or

8. §1.2 Measuring Length and Distance

9. 4 cm 6 10 = 4.6 cm 2 in 2 16 = 2 in 1 8 = 2.125 in

10. 6 cm 8.5 10 = 6.85 cm in 15 16 = 0.9375 in

11. 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

12. 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

13. 30 mm.0 9 mm.9

14. Item SI Measurement (Metric) Imperial Measurement Caliper Plastic Container Lid Diameter: Radius: Circumference: Depth of Lid: Diameter: Radius: Circumference: Depth of Lid: Diameter: Depth of Lid: Paper ClipLength: 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.

15. 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 min or ft cm in mmin

16. §1.3 Relating SI and Imperial Measurements

17. Examples 1. 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 ft 0.3048 m x = 62.34 ft Bowling lane is approximately 62 ft.

18. Examples 2. 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 mi 1.609 Km 1 mi x = 99.76 Km Hana drove 1.76 Km farther.

19. Examples 3. Convert 6 feet 2 inches to centimetres. Round your answer to the nearest hundredth. Convert 6 ft 2 in to inches. 74 in 2.54 cm 1 in x = 187.96 cm 6 feet 2 inches = 187.96 cm 6(12) + 2 = 74 in

20. Examples 4. 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 in 2.54 cm 1 in x = 3.5814 m Yes the semitrailer will fit 11(12) + 9 =141 in Imperial to Metric (convert 11 ft 9 in to inches) 1 m 100 cm x

21. §1.4 Surface Areas of Right Pyramids and Right Cones

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

23. Examples 1. 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? bh 2 = 35.1 cm 2 SA = 4(A Triangle ) A Triangle = (7.8)(9.0) 2 = = 4(35.1) = 140.4 cm 2 SA = 140 cm 2

24. Examples 2. 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 ft 2 Need to find slant height (s) A Base = = (10)(8) a 2 + b 2 = c 2 SA = 80 + 2(82.45) + 2(67.04) A Front/Back = 4 2 + 16 2 = s 2 16 + 256 = s 2 272 = s 2 √ 272 = s 16.49 ft = s bh 2 (10)(16.49) 2 = = 82.45 ft 2 Blue a 2 + b 2 = c 2 A Side = 5 2 + 16 2 = s 2 25 + 256 = s 2 281 = s 2 √ 281 = s 16.76 ft = s bh 2 (8)(16.76) 2 = = 67.04 ft 2 Red = 378.98 ft 2

25. Examples 3. 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 ft 2 SA Cone = πr 2 + πrs = π(2) 2 + π(2)(7.28) a 2 + b 2 = c 2 2 2 + 7 2 = s 2 4 + 49 = s 2 53 = s 2 √ 53= s 7.28 ft = s = 4π + 14.56π = 18.56π = 58.31

26. Examples 4. The lateral area of a cone is 220 cm 2. 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 = πrsπrs 220 = a 2 + b 2 = c 2 r 2 + h 2 = s 2 25 + h 2 = 196.28 h 2 = 171.2801 h = √ 171.2801 h = 13.087 cm π(5)s 14.01 = s 5π5π 5π5π 5 2 + h 2 = 14.01 2

27. §1.5 Volumes of Right Pyramids & Right Cones

28. Volume of Right Pyramid Volume of Right Cone

29. Examples 1. Calculate the volume of this right square pyramid to the nearest cubic inch. Need to find height (h) V = 30 in 3 V = ⅓(4)(4)(h) = ⅓(4)(4)( √ 32) a 2 + b 2 = c 2 h 2 + 2 2 = 6 2 h 2 + 4 = 36 h 2 = 32 h= √ 32 = 30.17

30. Examples 2. 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 cm 3 V = ⅓(5.4)(3.2)(8.1) = 46.656

31. Examples 3. Determine the volume of this cone to the nearest cubic inch. V = 679 in 3 V = = ⅓π(6) 2 (18) ⅓(πr 2 h) = 678.584

32. Examples 4. 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 = ⅓πr 2 (4) 3 x x3 4 yd V = ⅓πr 2 h 4πr24πr2 3 3(205) = 4πr 2 4π4π 4π4π 3(205) = r 2 4π4π √ √ r = 6.996

33. Examples 5. Find the top area, bottom area and lateral area of the cylinder. A Top = πr2πr2 = A Bottom = π(6) 2 = 36π Top Area = _____ Bottom Area = ______ Lateral Area = _____ Total Surface Area = _________ = 113.1 cm 2 113.1 cm 2 A Side = A Lateral Circumference = 2πr Height A lateral = 2πrh = 2π(6)(25) = 300π = 942.5 cm 2 942.5 cm 2 1168.7 cm 2

34. §1.6 Surface Area and Volume of a Sphere

35. Examples 1. The diameter of a baseball is approximately 3in. Determine the surface area of a baseball to the nearest square inch. SA = πd 2 = π(3) 2 = 9π = 28.27 = 28 in 2

36. Examples 2. 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 = πd 2 20 = πd 2 20 = d 2 2.52 = d π π √ √ π 2.5 in = d

37. Examples 3. The sun approximates a sphere with diameter 870 000 mi. What is the approximate volume of the sun? V = πr 3 V = 3.45 x 10 17 cubic miles 3 4 2 870 000 = π ( ) 3 3 4 = π(435 000) 3 3 4 (345 000 000 000 000 000 cubic miles)

38. 4. A hemisphere has radius 8.0cm. a) What is the surface area of the hemisphere to the nearest tenth of a square centimetre? SA HEMISPHERE = SA SPHERE = 4πr 2 2 = 2πr 2 2 SA = 2πr 2 + πr 2 Hemisphere Circle on Bottom = 3πr 2 = 3π(8) 2 = 3π(64) = 192π = 603.2 cm 2

39. 4. A hemisphere has radius 8.0cm. b) What is the volume of the hemisphere to the nearest tenth of a cubic centimetre? V HEMISPHERE = 4πr34πr3 3 2 1 ( ) = πr 3 6 4 = π(8) 3 6 4 = π(512) 6 4 = π 6 2048 = 1072.3 cm 3

40. §1.7 Solving Problems Involving Objects

41. Examples 1. Determine the volume of this composite object to the nearest tenth of a cubic metre. V TOP =( area of base )( height ) 3 1 = (6.7)(2.9)(2.1) 3 1 = 13.601 Don’t round until the end. V BOTTOM = L x W x H = (6.7)(2.9)(2.9) = 56.347 V TOTAL = 13.601 + 56.347 = 69.948 V TOTAL = 69.9 m 3

42. Examples 2. Determine the surface area of this composite object to the nearest square foot. SA TOP = (4πr 2 ) 2 1 = 2π(2) 2 Leave like this SA BOTTOM = 2πr 2 + 2πrh = π(2) 2 + 2π(2)(4) = 20π SA TOTAL =8π + 20π = 87.965 SA TOTAL = 88 ft 2 = 2πr 2 Does not include bottom of hemisphere = 8π Minus top circle of cylinder = πr 2 + 2πrh = 4π + 16π = 28π

43. Examples 3. 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 SA TOP : 2 bh = 3 = Two ∆ so A ∆ = 2(3) = 6 yd 2 A∆=A∆= 2 (3)(2) LxWLxW= 12.5 Two so A = 2(12.5) = 25 yd 2 A = = (5)(2.5) a 2 + b 2 = c 2 a b c Need to find slant 2 2 + (1.5) 2 = c 2 4 + 2.25 = c 2 6.25 = c 2 √ 6.25 = √ c 2 2.5= c SA BOTTOM : w∙hw∙h = 6 Front & Back so A FRONT/BACK = 2(6) = 12 yd 2 A FRONT = = (3)(2) w L h w L∙hL∙h = 10 Two sides so A SIDES = 2(10) = 20 yd 2 A SIDE = = (5)(2) w∙Lw∙L = 15 yd 2 A BOTTOM = = (3)(5) * * * * * SA TOTAL = 6 + 25 + 12 + 20 + 15 SA TOTAL = 78 yd 2