262 MHR Foundations for College Mathematics 11 Solutions

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1 Chapter 6 Geometry in Design Chapter 6 Prerequisite Skills Chapter 6 Prerequisite Skills Question 1 Page 294 a) rectangle b) parallelogram c) circle d) trapezoid Chapter 6 Prerequisite Skills Question 2 Page 294 a) rectangular prism b) triangular prism c) square-based pyramid d) cylinder Chapter 6 Prerequisite Skills Question 3 Page 294 a) triangle b) square c) pentagon d) hexagon e) octagon Chapter 6 Prerequisite Skills Question 4 Page 294 P = 2(120) + 2(100) = 440; the perimeter is 440 m. A = (120)(100) = ; the area is m 2. Chapter 6 Prerequisite Skills Question 5 Page 294 P = = 45; the perimeter is 45 cm. 1 A = ( 15 )( 13 ) 2 = 97.5; the area is 97.5 cm MHR Foundations for College Mathematics 11 Solutions

2 Chapter 6 Prerequisite Skills Question 6 Page 294 P = 2π(12) =& 75.4; the perimeter is about 75.4 m. A = π(12) 2 =& 452.4; the area is about m 2. Chapter 6 Prerequisite Skills Question 7 Page 295 a) SA = lw + 2wh + 2lh = (8)(6) + 2(6)(2.5) + 2(8)(2.5) = 118 The area that was painted was 118 m 2. b) V = lwh = (8)(6)(2.5) = 120 The volume of the room is 120 m 3. Chapter 6 Prerequisite Skills Question 8 Page 295 a = 0.1, b = 0.6, and l = 0.9 Use the Pythagorean theorem. c2 = a 2 + b2 = = 0.37 c = 0.37 = & 0.61 SA = (0.1)(0.9) + (0.61)(0.9) + 2(0.5)(0.1)(0.6) =& 0.7 The surface area to be painted is about 0.7 m 2. Chapter 6 Prerequisite Skills Question 9 Page 295 r = 5 and h = 12 The surface area of the interior is the same as the surface area of the exterior. SA = 2π(5)(12) + 2π(5) 2 = 534 The area to be painted is 534 m 2. V = π(5) 2 (12) = 942 The volume of the water is 942 m 3. Chapter 6 Prerequisite Skills Question 10 Page 295 A pentagon has 5 sides. S = 180(5 2) = 540 The sum of the angles is = 108 The measure of each angle is 108. MHR Foundations for College Mathematics 11 Solutions 263

3 Chapter 6 Prerequisite Skills Question 11 Page 295 Set up a proportion. 1cm = x y = 2 m 24 m 36 cm x = 24 cm = 12 cm 2 y = 36 cm = 18 cm 2 The diagram will be 12 cm by 18 cm. Chapter 6 Prerequisite Skills Question 12 Page 295 Set up a proportion. 1in. = 10 in. 4 ft. x ft. x = 4 10 = 40 The sailboat is 40 ft. long. 264 MHR Foundations for College Mathematics 11 Solutions

4 Chapter 6 Section 1 Investigate Geometric Shapes and Figures Chapter 6 Section 1 Question 1 Page 302 D (Triangles, squares, and hexagons can be used to tile a plane.) Chapter 6 Section 1 Question 2 Page =& 1.78 No; the ratio of the two side lengths is about 1.78:1. Chapter 6 Section 1 Question 3 Page 302 Answers may vary. Chapter 6 Section 1 Question 4 Page 302 Answers may vary. Chapter 6 Section 1 Question 5 Page 302 Answers may vary. For example: three-dimensional: cone, cylinder, square-based pyramid, rectangular prism, triangular prism two-dimensional: triangle, rectangle, square, trapezoid Chapter 6 Section 1 Question 6 Page 302 Answers may vary. Chapter 6 Section 1 Question 7 Page 303 Olympic swimming pools are usually 50 m by 25 m. No; the ratio of the two side lengths is 2:1. Some other examples include: International soccer fields (pitches) are m long by m wide. Canadian (CFL) football field dimensions are 110 yards by 65 yards. (This is quite close to the golden ratio of 1.618:1) American (NFL) football field dimensions are 100 yards by yards. Professional (NBA) basketball courts are 94 ft. by 50 ft. MHR Foundations for College Mathematics 11 Solutions 265

5 Chapter 6 Section 1 Question 8 Page 303 Answers may vary. For example: a) The Parthenon, in Athens, Greece. One web image of this building can be found at: b) The Taj Mahal, in Agra, India. One web image of this building can be found at: c) United Nations building in New York, NY. One web image of this building can be found at: d) Buddhist Temple in Niagara Falls, ON. One web image of this building can be found at: Chapter 6 Section 1 Question 9 Page 303 Answers may vary. Chapter 6 Section 1 Question 10 Page 303 Answers may vary. For example: The ratio is 1.2:1 in the photograph provided. Chapter 6 Section 1 Question 11 Page 304 a) 90 b) A triangle with side lengths of 3 units, 4 units, and 5 units is a right triangle. This fact is related to the Pythagorean theorem which states that a 90 angle is created for a triangle with sides a, b, and c such that a 2 + b 2 = c 2. Chapter 6 Section 1 Question 12 Page 304 a) Answers may vary. For example: The first and fourth sets of sixteenth notes form congruent triangles. These triangles are very close to being right-angled triangles. The remaining four triangles are congruent if you consider the notes at their extremities. These triangles are scalene and obtuse. b) Answers may vary. For example: the three notes at the beginning of bars 2 and 4 Chapter 6 Section 1 Question 13 Page 304 The red rectangle has the ratio of its sides as 1.6:1. Chapter 6 Section 1 Question 14 Page 305 Solutions for Achievement Checks are in the Teacher Resource. 266 MHR Foundations for College Mathematics 11 Solutions

6 Chapter 6 Section 1 Question 15 Page 306 a) 1.618x x x x = 2.5 = = & The length of the shorter side is about m. b) The remaining rectangle has dimensions m by m. The ratio for these dimensions is about 1.618:1. This is the golden ratio. Chapter 6 Section 1 Question 16 Page 306 a) 29.7 =& b) 21.0 =& c) =& This is the same ratio as in part a). d) 2 =& ; all the ratios calculated above match this standard to 3 decimal places. Chapter 6 Section 1 Question 17 Page 306 Answers may vary. Some examples from the Internet using themes such as motorbikes, elephants, turtles, and flowers can be found at: MHR Foundations for College Mathematics 11 Solutions 267

7 Chapter 6 Section 2 Perspective and Orthographic Drawings Chapter 6 Section 2 Question 1 Page 314 C (The blueprint suggested is missing a top view which is needed to detail the third dimension of the object.) Chapter 6 Section 2 Question 2 Page 314 Answers may vary. For example: The actual lengths are around 50 m. A scale model of a reasonable size would have dimensions around 25 cm. Therefore, use a scale: 1 cm represents 2 m. The dimensions for the model are length: 24.5 cm, width: 17.5 cm, height: 28.0 cm. Chapter 6 Section 2 Question 3 Page 314 B. (For the diameter to equal to the height, the side view must be a square.) Chapter 6 Section 2 Question 4 Page 314 The physical model will have length of 4 cubes, width of 2 cubes, and height of 3 cubes. Chapter 6 Section 2 Question 5 Page 315 Answers may vary. For example: The figure appears to be a rectangle but the sides are of lengths 4, 4, 4, and 3, which is impossible. It also appears that the diagram represents a figure on 1 level but there is a step up to create a second level, which is also impossible. Chapter 6 Section 2 Question 6 Page 315 a) Answers may vary. For example: a courtyard surrounded by 2 square and 2 rectangular rooms b) A possible isometric drawing: c) A possible orthographic drawing: top view side view front view 268 MHR Foundations for College Mathematics 11 Solutions

8 Chapter 6 Section 2 Question 7 Page 315 a) b) Answers may vary. For example: No; it is impossible to draw so that the farthest left and right tips of the base will be on dots. Chapter 6 Section 2 Question 8 Page 316 a) This is the isometric view. This is the orthographic view. top view front view side view b) top view side view front view c) Answers may vary. For example: They are congruent right triangles. MHR Foundations for College Mathematics 11 Solutions 269

9 Chapter 6 Section 2 Question 9 Page 316 a) Answers may vary. b) Chapter 6 Section 2 Question 10 Page 316 All four drawings are valid because you cannot see behind the cubes shown. Chapter 6 Section 2 Question 11 Page MHR Foundations for College Mathematics 11 Solutions

10 Chapter 6 Section 2 Question 12 Page 317 A final drawing is shown below. This diagram shows the construction lines used. Chapter 6 Section 2 Question 13 Page 317 MHR Foundations for College Mathematics 11 Solutions 271

11 Chapter 6 Section 3 Create Nets, Plans, and Patterns Chapter 6 Section 3 Question 1 Page 322 A Chapter 6 Section 3 Question 2 Page 322 Chapter 6 Section 3 Question 3 Page 322 Chapter 6 Section 3 Question 4 Page 322 Answers may vary. For example: A pattern is the best choice because there are several pieces that need to be assembled to make a bookcase. A net would only include the outside of the bookcase and not any interior parts such as shelves. Similarly, a plan would only detail the bottom of the bookshelf, losing the details of the shelves and the parts higher up. Chapter 6 Section 3 Question 5 Page 322 a) Answers may vary. For example: The four small squares are intended to form the top of the cube. However, the top and bottom small squares are lined up vertically, which will overlap each other when the net is folded, and will not complete the cube-top. b) Move either the top square or the bottom square 1 unit to the left. 272 MHR Foundations for College Mathematics 11 Solutions

12 Chapter 6 Section 3 Question 6 Page 323 a) 8 cm 12 cm 8 cm c) Yes; it would be wise to make the net a bit larger to make room for taping the model together. Chapter 6 Section 3 Question 7 Page 323 a) 4 cm 12 cm c) Yes; it would be wise to make the net a bit larger to make room for taping the model together. MHR Foundations for College Mathematics 11 Solutions 273

13 Chapter 6 Section 3 Question 8 Page 323 a) 3 m 3 m 3 m b) The height of each triangle is: 3 m sin 60 =& 2.6 m SA = 4(3)(3) + 4(0.5)(3)(2.6) = 51.6 The surface area is about 51.6 m 2. c) Cost of metal: $6.5/m m 2 =& $ The cost of the metal is about $335. Chapter 6 Section 3 Question 9 Page cm 10 cm 20 cm 274 MHR Foundations for College Mathematics 11 Solutions

14 Chapter 6 Section 3 Question 10 Page 324 The cylinder when unfolded to make a net will be a rectangle with height 30 cm and length equal to the circumference of the circular hole, which is 10π cm, or 31.4 cm. 10 cm 10 cm 20 cm 30 cm 31.4 cm Chapter 6 Section 3 Question 11 Page 324 a) b) The folded tetrahedron should be similar to: 3 m 3 m 3 m 3 m 3 m 3 m MHR Foundations for College Mathematics 11 Solutions 275

15 Chapter 6 Section 3 Question 12 Page 324 The net below will fold to make a paper model of the square shape. Fold along all possible lines around the middle big square shape that has an empty centre. Chapter 6 Section 3 Question 13 Page MHR Foundations for College Mathematics 11 Solutions

16 Chapter 6 Section 4 Scale Models Chapter 6 Section 4 Question 1 Page 331 Set up a proportion to determine lengths. 1in. = x in. 2 ft. 10 ft. 2x = 10 x = 5 1in. y in. = 2 ft. 12 ft. 2y = 12 y = 6 The model will be 5 in. by 6 in. Chapter 6 Section 4 Question 2 Page 331 Set up a proportion to determine lengths. 1 = 10 ft. 8 in. 8 x in. 1 = 128 in. 8 x in. x = = in. = 85 ft. 4 in. The length of the full-size aircraft was 1024 in., or 85 ft. 4 in. Chapter 6 Section 4 Question 3 Page 331 Answers may vary. For example: A scale model is easier to visualize for most citizens. Chapter 6 Section 4 Question 4 Page 331 Answers may vary. For example: She could use half of a tennis ball placed on a roll of thick paper. First measure the diameter, d, of the tennis ball. Multiply this diameter by to get the height of the cylinder. To make the cylinder, choose a piece of thick paper that has dimensions 1.618d by πd. She can calculate the scale of her model by dividing the actual height of the silo (cylinder part) by the height of her model, 1.618d. MHR Foundations for College Mathematics 11 Solutions 277

17 Chapter 6 Section 4 Question 5 Page ft. 2 ft. 2 ft. 2 ft. 2 ft. 2 ft. Chapter 6 Section 4 Question 6 Page 332 The box will be a rectangular prism with dimensions , 1 2 6, 1 3 4, or (units are in ball diameters). The last option is likely chosen since it uses the least packaging material. (From previous learning, the optimal shape is the one closest to that of a cube.). Front view Top view Side view Chapter 6 Section 4 Question 7 Page MHR Foundations for College Mathematics 11 Solutions

18 Chapter 6 Section 4 Question 8 Page 332 a) The distance between a pair of dots represents 1 m in the actual sculpture. c) The area of the complete net as shown in the diagram is 22 square units. If the 2 3 face is the base, the area is 22 m 2 6 m 2 = 16 m 2 ; the cost is $32. If the 1 3 face is the base, the area is 22 m 2 3 m 2 = 19 m 2 ; the cost is $38. If the 1 2 face is the base, the area is 22 m 2 2 m 2 = 20 m 2 ; the cost is $40. MHR Foundations for College Mathematics 11 Solutions 279

19 Chapter 6 Section 4 Question 9 Page a) The volume is: V = πr 2h = π d d = πd 2 4 Use trial-and-error to find d. π( 5) 3 If d = 5, V = = & π( 10) 3 If d = 10, V = = & π( 9) 3 If d = 9, V = = & π( 8) 3 If d = 8, V = = & The smallest integer that gives a volume of at least 500 is 9. The minimum height and width of the can is 9 cm. b) Scale: 1 cm represents 4 cm of soup can MHR Foundations for College Mathematics 11 Solutions

20 Chapter 6 Section 4 Question 10 Page 333 a) Scale: 1 cm represents 5 m BA = 2.00 cm BC = 0.60 cm B A C c) Answers may vary. For example: BA = 2.00 cm BC = 0.60 cm Area DEFGHC = cm 2 B A C D H E G F Using The Geometer's Sketchpad, the area of the base is found to be cm 2. The actual area is: m 2 = m 2 The volume of concrete is: m m = m 3, or about 26 m 3. d) Concrete cost: $75/m 3 26 m 3 = $1950 MHR Foundations for College Mathematics 11 Solutions 281

21 Chapter 6 Section 4 Question 11 Page 333 a) The roof of the net will now be made of 6 isosceles triangles instead of a hexagon. b) No change; the base remains the same. Chapter 6 Section 4 Question 12 Page 333 a) Note that the net for the roof needs to be wider than the orthographic top view because two parts of the roof are slanted. c) Area of the base of the building: (6 2 m)(6 2 m) = 144 m 2 Cost of building: $1200/m m 2 = $ An estimate for the cost of the building is $ MHR Foundations for College Mathematics 11 Solutions

22 Chapter 6 Section 4 Question 13 Page 334 B (The tank's cylinder is 7 m long and the hemisphere has a radius of 2 m. The overall length is 11 m and the diameter of the hemisphere is 4 m.) Chapter 6 Section 4 Question 14 Page 334 Chapter 6 Section 4 Question 15 Page 334 The cone has a radius of 5 cm and a height of 3 cm. The slant height, s, can be found using the Pythagorean theorem. s 2 = = 34 s = 34 = & 5.8 The slant height is the radius of the circle of which a sector forms the net for the cone. The arc length of the sector equals the circumference of the base of the cone. Arc length = 2 πr = 2π(5) = & The central angle that the arc subtends is: = & 310 2π(5.8) Diagram for the net may vary. For example: 5.8 cm 50 MHR Foundations for College Mathematics 11 Solutions 283

23 Chapter 6 Section 5 Solve Problems With Given Constraints Chapter 6 Section 5 Question 1 Page L = 1000 cm 3 If each side of the cube has a length of 10 cm, the volume will be (10 cm) 3, or 1000 cm 3. The side length of the box is 10 cm. Chapter 6 Section 5 Question 2 Page 340 For a circle, the area is: A = πr = πr 2 r 2 = 500 π r = 500 π = & 12.6 The radius of the pond is about 12.6 m. The minimum length of berm required to enclose the pond is: 2πr = 2π(12.6 m) =& 79 m Chapter 6 Section 5 Question 3 Page 340 Answers may vary. For example: Some possible reasons for choosing an octagonal shape: Less material is needed to make the box,. The box takes up less space. A box with straight sides is stronger than one with curved sides. The shape is more interesting from a marketing viewpoint. Chapter 6 Section 5 Question 4 Page 340 Answers may vary. For example: Some possible constraints: minimum and maximum space requirements size of plots of land for installation available sizes of sheet metal cost of materials (and hence price of finished home) 284 MHR Foundations for College Mathematics 11 Solutions

24 Chapter 6 Section 5 Question 5 Page 341 Bedroom #2 10 by 15 Bedroom #1 15 by 15 Living Room 10 by 20 Kitchen 10 by 20 Hallway Bathroom 7 by 15 Bedroom # 3 8 by 15 dot spacing Chapter 6 Section 5 Question 6 Page 341 The pipe forms a shallow cylinder of depth 50 cm, or 0.5 m, to contain the oil. V = πr 2h = πr 2 (0.5) r 2 = 0.5π r = 0.5π = & The radius of the circle is about m. Circumference = 2πr = 2π(356.8) =& 2242 The length of pipe needed is at least 2242 m. Chapter 6 Section 5 Question 7 Page 341 Answers may vary. Chapter 6 Section 5 Question 8 Page 341 Answers may vary. MHR Foundations for College Mathematics 11 Solutions 285

25 Chapter 6 Section 5 Question 9 Page 342 a) front view side view top view b) Note the triangles on the ends are 1.7 units high so that the slant height is 2 units. c) SA = 2(4) + 3(4) + 3(4) +2(4) + 2(3) + 2(3) + 2(0.5)(2)(1.7) = 55.4 The area of sheet metal needed is 55.4 m 2. Cost of metal: $15/m m 2 = $ MHR Foundations for College Mathematics 11 Solutions

26 Chapter 6 Section 5 Question 10 Page 342 a) In an orthographic top view of the uranium fuel cylinder and containment building, the cylinder is a rectangle (6 m by 8 m) and the containment building is in the shape of a circle. The diagram details this situation. An accurate measurement gives the diameter of the containment building, 10 m. Since the fuel cylinder has diameter 8 m, the containment building must be at least 8 m high. D C ED = 5.00 cm m DA = 8.00 cm E A m AB = 6.00 cm B b) Method 1 (Assume that the base for the cylinder already exists). The surface area of a cylinder = 2πrh + πr 2 SA = 2π(5)(8) + π(5) 2 = 80π + 25π = 105π = & The area of the walls and roof is about m = & 264 ; the approximate volume of concrete required is 264 m 3. Method 2 (Assume that the base for the cylinder needs to be constructed). The surface area of a cylinder = 2πrh + 2πr 2 SA = 2π(5)(8) + 2π(5) 2 = 80π + 50π = 130π = & The area of the walls, base, and roof is about m = & 327 ; the approximate volume of concrete required is 327 m 3. c) Method 1 (from above) Cost: $90/m m 3 = $ The estimated cost for the concrete is $ Method 2 (from above) Cost: $90/m m 3 = $ The estimated cost for the concrete is $ MHR Foundations for College Mathematics 11 Solutions 287

27 Chapter 6 Section 5 Question 11 Page 343 Solutions for Achievement Checks are in the Teacher Resource. Chapter 6 Section 5 Question 12 Page 343 Let the inside diameter be d. The inside height will be 3d. For this cylinder, V = πr 2h 1 = π(0.5d ) 2 (3d ) 1 = 0.75πd 3 d 3 = π d = π = & 0.75 The inside diameter is 0.75 ft. and the inside height will be 2.25 ft. The walls need to be 0.5 in. thick. The diameter and the height will be increased by in. = 1 in. = 1 ft. =& ft. 12 The outside diameter and the height will be 0.83 ft. and 2.33 ft. respectively. top view front view 2.33 ft. side view 2.33 ft ft ft ft. 288 MHR Foundations for College Mathematics 11 Solutions

28 Chapter 6 Section 5 Question 13 Page 343 a) From previous learning, the most efficient building (least surface area for a given volume) occurs when the prism is in the shape of a cube. Try a building in the shape of a cube, given that the height must be a multiple of 3 m. If the building has 1 story, the dimensions are 3 m by 3 m by 3 m. The surface area is: 5 m 9 m = 45 m 2 The heat required is 450 W. If the building has 2 stories, the dimensions are 6 m by 6 m by 6 m. The surface area is 5 m 36 m = 180 m 2 The heat required is 1800 W. If the building has three stories, the dimensions are 9 m by 9 m by 9 m. The surface area is 5 m 81 m = 405 m 2 The heat required is 4050 W. If the building has four stories, the dimensions are 12 m by 12 m by 12 m. The surface area is 5 m 144 m = 720 m 2 The heat required is 7200 W. If the building has five stories, the dimensions are 15 m by 15 m by 15 m. The surface area is 5 m 225 m = 1125 m 2 The heat required is W. Clearly, the 5-story building is the most efficient, as the furnace can produce sufficient heat for this building. b) Designs and scale models may vary. Chapter 6 Section 5 Question 14 Page 343 Each small square in the diagram represents a square with side length 50 m. Gate Cluster Check-in Area Gate Cluster Gate Cluster MHR Foundations for College Mathematics 11 Solutions 289

29 Chapter 6 Review Chapter 6 Review Question 1 Page 346 Answers may vary. For example: They measured to see if the diagonals were equal. This was a way to check the accuracy of their initial layout since every rectangle and square has equal diagonals. Chapter 6 Review Question 2 Page 346 For the full blanket, the ratio is: 130 = The half blanket has dimensions 80 cm by 65 cm. The ratio is: 80 =& The full blanket is close to a golden rectangle, but not when it is folded in half. Chapter 6 Review Question 3 Page 346 b) side view front view top view Chapter 6 Review Question 4 Page MHR Foundations for College Mathematics 11 Solutions

30 Chapter 6 Review Question 5 Page 346 Scale: 1 cm represents 2 ft. Bedroom 1 C CD = 4.00 cm D A AB = cm Washroom 1 B Washroom 2 Bedroom 2 Chapter 6 Review Question 6 Page 346 The bottom of the container will be a circle with radius 3 cm. The sides of the cylinder will be a rectangle with width 18 cm and length 6π = 18.8 cm. radius 3 cm 18 cm 18.8 cm MHR Foundations for College Mathematics 11 Solutions 291

31 Chapter 6 Review Question 7 Page m 1 m Fold line 1 m Chapter 6 Review Question 8 Page 347 Divide each of the actual measurements by =& The wingspan is about m, or 84.6 cm =& The length is about m, or cm. Chapter 6 Review Question 9 Page 347 Scale: 1 cm represents 1 m radius 5 cm 15.7 cm 10 cm 292 MHR Foundations for College Mathematics 11 Solutions

32 Chapter 6 Review Question 10 Page 347 The net for the building and the restaurant are separated to aid construction of the model. Scale: one space between dots represents 20 m Chapter 6 Review Question 11 Page 347 a) The six rooms in the hexagon are all equilateral triangles. Suppose each triangle has side length x. Using a scale drawing or trigonometry, the height is 0.866x. Total area of the 6 rooms is: 6(0.5)( x )(0.866x) = 2.598x 2 Thickness of concrete: 10 cm = 0.10 m Volume of concrete required for the base is: 2.598x = x Cost, in dollars, of the concrete is x 75 = x Since the budget is $600, x 2 = 600 x 2 = x = x = & 5.55 The side length of the hexagon is about 5.55 m. b) Area of one wall is: (5.55 m) 2 = m 2 Area of the roof = area of the base: 2.598(5.55) 2 = m 2 Area of wood required is: 12(30.80) = m 2 Cost of the wood is: (20) = The cost of wood is about $8992. The total cost of the project is: = $959 MHR Foundations for College Mathematics 11 Solutions 293

33 Chapter 6 Practice Test Chapter 6 Practice Test Question 1 Page 348 A (Length to width ratios: 8 = 1.6; 5 = 2.5; 6 = 3 ; 6 = The first ratio is closest to 1.618, the golden ratio.) Chapter 6 Practice Test Question 2 Page 348 B (The angle must divide evenly into 360. Only 120 divides evenly.) Chapter 6 Practice Test Question 3 Page 348 B (The diameter of the sphere is larger than the side of the square base.) Chapter 6 Practice Test Question 4 Page 348 front view side view top view 6 cm Chapter 6 Practice Test Question 5 Page MHR Foundations for College Mathematics 11 Solutions

34 Chapter 6 Practice Test Question 6 Page 348 Scale: one space between 2 dots represents 10 cm Chapter 6 Practice Test Question 7 Page 349 front view side view top view MHR Foundations for College Mathematics 11 Solutions 295

35 Chapter 6 Practice Test Question 8 Page 349 Scale: one space between 2 dots represents 1 in. Chapter 6 Practice Test Question 9 Page 349 a) Scale: one space between 2 dots represents 2 cm b) The height of the can is: cm = cm V = πr 2h = π( 6) 2 (19.416) = & 2196 The volume is about 2196 ml. 296 MHR Foundations for College Mathematics 11 Solutions

36 Chapters 4 to 6 Review Chapters 4 to 6 Review Question 1 Page 350 a) i) x y First Differences Second Differences Answers may vary. For example: Not quadratic; the first differences are constant ( 4) and the graph is a straight line. ii) x y First Differences Second Differences Answers may vary. For example: Quadratic; the second differences are constant (1) and the graph is a parabola. iii) Answers may vary. For example: Not quadratic; the graph is a straight line since there is no x 2 -term. iv) Answers may vary. For example: Quadratic; the graph is a parabola since there is an x 2 -term. MHR Foundations for College Mathematics 11 Solutions 297

37 b) i) ii) iii) 10 iv) Chapters 4 to 6 Review Question 2 Page 350 a) The parabola has not been stretched, it opens upward, and the vertex has been translated 4 units to the left of the y-axis. b) The parabola has not been stretched, it opens downward, and the vertex has been translated 1 unit to the right of the y-axis. c) The parabola has been vertically compressed, it opens downward, and the vertex has been translated 7 units to the left of the y-axis. d) The parabola is vertically stretched, it opens upward, and the vertex has been translated 9 units to the left of the y-axis. e) The parabola is vertically compressed, it opens upward, and the vertex has been translated 32 units to the left of the y-axis. f) The parabola is vertically stretched, it opens downward, and the vertex has been translated 18 units to the right of the y-axis. 298 MHR Foundations for College Mathematics 11 Solutions

38 Chapters 4 to 6 Review Question 3 Page 350 a) i) vertex at (1, 9) ii) parabola opens upward iii) not stretched iv) b) i) vertex at ( 8, 5) ii) parabola opens downward iii) vertically stretched iv) c) i) vertex at (0, 1) ii) parabola opens upward iii) vertically compressed iv) d) i) vertex at (0, 0) ii) parabola opens upward iii) vertically stretched iv) MHR Foundations for College Mathematics 11 Solutions 299

39 e) i) vertex at (2, 2) ii) parabola opens upward iii) vertically compressed iv) f) i) vertex at ( 1, 13) ii) parabola opens upward iii) vertically stretched iv) Chapters 4 to 6 Review Question 4 Page 350 a) The vertex is (6, 177.4) and the curve opens downward. Therefore, it takes 6 s to reach a maximum height of m. b) Substitute t = 7 into the equation. 2 h = 4.9(7 6) = = The height of the flare 7 seconds after launch will be m. c) 200 d) From the graph, the flare will hit the water after about 12 s MHR Foundations for College Mathematics 11 Solutions

40 Chapters 4 to 6 Review Question 5 Page 350 a) (3x + 4)(10x + 1) b) (x 2)(4x + 15) = 30x 2 + 3x + 40x + 4 = 4x x 8x 30 = 30x x + 4 = 4x 2 + 7x 30 c) (12x 8)(2x + 0.5) d) (6 + 2x)(6 2x) = 24x 2 + 6x 16x 4 = 36 12x + 12x 4x 2 = 24x 2 10x 4 = 36 4x 2 Chapters 4 to 6 Review Question 6 Page 350 a) (s 1)(4s 7) = 4s 2 7s 4s +7 = 4s 2 11s + 7 b) 4(15) 2 11(15) + 7 = = 742 The actual area of the apartment is 742 m 2. Chapters 4 to 6 Review Question 7 Page 350 a) b) y = ( x 4) y = ( x 4)( x 4) y = x 2 4x 4x y = x 2 8x c) d) y = 8( x + 2) y = 8( x + 2)( x + 2) + 27 y = 8( x 2 + 2x + 2x + 4) + 27 y = 8( x 2 + 4x + 4) + 27 y = 8x x y = 8x x + 59 y = ( x + 10) 2 3 y = ( x + 10)( x + 10) 3 y = x x + 10x y = x x + 97 y = 3.2( x 4) y = 3.2( x 4)( x 4) 0.8 y = 3.2( x 2 4x 4x + 16) 0.8 y = 3.2( x 2 8x + 16) 0.8 y = 3.2x x y = 3.2x x 52 Chapters 4 to 6 Review Question 8 Page 350 To find the y-intercept, substitute x = 0 into the equation. y = (7 3(0))(2(0) 6) = (7)( 6) = 42 The y-intercept is 42. MHR Foundations for College Mathematics 11 Solutions 301

41 Chapters 4 to 6 Review Question 9 Page 351 a) x x + 24 (Find 2 numbers with product 24 and sum 11.) = (x + 8)(x + 3) b) x 2 + x 30 (Find 2 numbers with product 30 and sum 1.) = (x + 6)(x 5) c) x 2 8x + 7 (Find 2 numbers with product 7 and sum 8.) = (x 7)(x 1) d) x 2 + 8x + 16 (Find 2 numbers with product 16 and sum 8.) = (x + 4)(x + 4) e) 3x x (Factor out the GCF) = 3(x x + 36) (Find 2 numbers with product 36 and sum 13.) = 3(x + 9)(x + 4) f) 10x 2 110x 100 (Factor out the GCF) = 10(x x + 10) (Find 2 numbers with product 10 and sum 11.) = 10(x + 10)(x + 1) Chapters 4 to 6 Review Question 10 Page x x 1 = 5x 5+ x x= x + 4x 5 a) ( )( ) 2 2 Equivalent; explanations may vary. x 2 x 1.5 = x 1.5x 2x+ 3 = x 3.5x+ 3 x + 3.5x+ 3 Not equivalent; explanations may vary. b) ( )( ) x 1 x+ 2 = x + 2x 1x 2= x + x 2 Equivalent; explanations may vary. c) ( )( ) MHR Foundations for College Mathematics 11 Solutions

42 Chapters 4 to 6 Review Question 11 Page 351 a) y = x 2 x 20) (Find 2 numbers with product 20 and sum 1.) y = (x 5)(x + 4) The zeros are x = 5 and x = 4. b) y = 10x (Factor out the GCF) y = 10(x 2 36) (Find 2 numbers with product 36 and sum 0.) y = 10(x + 6)(x 6) The zeros are x = 6 and x = 6. c) y = 2x 2 + 4x + 70 (Factor out the GCF) y = 2(x 2 2x 35) (Find 2 numbers with product 35 and sum 2.) y = 2(x + 5)(x 7) The zeros are x = 7 and x = 5. d) y = 3.5x x (Factor out the GCF) y = 3.5(x 2 + 6x + 9) (Find 2 numbers with product 9 and sum 6.) y = 3(x + 3)(x +3) The zero is x = 3. Chapters 4 to 6 Review Question 12 Page 351 a) x + 16 b) x x 2 SA = 2x x 2x( x 16) 2 x ( x + 16) 2 2 SA = x 2 + 2x x + x x SA = 4x x c) Since the width of the base is 5, x = 5 2 x = 10 2 ( ) ( ) SA = = = 880 The surface area is 880 cm 2. MHR Foundations for College Mathematics 11 Solutions 303

43 Chapters 4 to 6 Review Question 13 Page 351 Yes, the shape can tile the plane. Chapters 4 to 6 Review Question 14 Page 351 b) c) top side front 304 MHR Foundations for College Mathematics 11 Solutions

44 Chapters 4 to 6 Review Question 15 Page 351 a) front side top b) c) The top of the platform is a rectangle with width 6 m and length 8 m. Its area is 48 m 2. Cost: $40/m 2 48 m 2 = $1920 The cost of the plywood is $1920. MHR Foundations for College Mathematics 11 Solutions 305