Chapter 6 Practice Test -...

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) = 12 000; the area is 12 000 m2. Chapter 6 Prerequisite Skills Question 5 Page 294 P = 15 + 15 + 15 = 45; the perimeter is 45 cm.

A = ( )( )1 15 132

= 97.5; the area is 97.5 cm2.

262 MHR • Foundations for College Mathematics 11 Solutions

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 452.4 m2. =& Chapter 6 Prerequisite Skills Question 7 Page 295 a) = (8)(6) + 2(6)(2.5) + 2(8)(2.5) = 118

The area that was painted was 118 m2.

2 2SA lw wh lh= + +

b) V = lwh = (8)(6)(2.5) = 120

The volume of the room is 120 m3. Chapter 6 Prerequisite Skills Question 8 Page 295 a = 0.1, b = 0.6, and l = 0.9 Use the Pythagorean theorem.

61.037.0

37.06.01.0 22

222

===

+=+=

&c

bac

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 m2. 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 m2. V = π(5)2(12) = 942 The volume of the water is 942 m3. Chapter 6 Prerequisite Skills Question 10 Page 295 A pentagon has 5 sides. S = 180(5 – 2) = 540 The sum of the angles is 540°. 5405

= 108

The measure of each angle is 108°.

MHR • Foundations for College Mathematics 11 Solutions 263

Chapter 6 Prerequisite Skills Question 11 Page 295 Set up a proportion.

cm 182cm 36

cm 122cm 24

cm 36m 24m 2cm 1

==

==

==

y

x

yx

The diagram will be 12 cm by 18 cm. Chapter 6 Prerequisite Skills Question 12 Page 295 Set up a proportion.

40104ft. in. 10

ft. 4in. 1

=×=

=

xx

The sailboat is 40 ft. long.


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 302 32 ÷ 18 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 100–110 m long by 64–75 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 5331 yards.

• Professional (NBA) basketball courts are 94 ft. by 50 ft.


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: http://wwwbrr.cr.usgs.gov/projects/SW_corrosion/teachers-pupils/index.html b) The Taj Mahal, in Agra, India. One web image of this building can be found at: http://www.envocare.co.uk/taj_mahal_history.htm c) United Nations building in New York, NY. One web image of this building can be found at: http://www.inetours.com/New_York/Pages/United_Nations.html d) Buddhist Temple in Niagara Falls, ON. One web image of this building can be found at: http://groups.msn.com/BuddhistTemples/worldpeacetenthousandbuddhatemple.msnw 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 a2 + b2 = c2.

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.


http://wwwbrr.cr.usgs.gov/projects/SW_corrosion/teachers-pupils/index.html

http://www.envocare.co.uk/taj_mahal_history.htm

http://www.inetours.com/New_York/Pages/United_Nations.html

http://groups.msn.com/BuddhistTemples/worldpeacetenthousandbuddhatemple.msnw

Chapter 6 Section 1 Question 15 Page 306 a)

545.1618.1

5.25.2618.1

618.15.2

=

=

=

&x

x

xx

The length of the shorter side is about 1.545 m.

b) The remaining rectangle has dimensions 1.545 m by 0.955 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) 4143.10.217.29 =&

b) 4141.185.140.21 =&

c) 4143.15.10

84.14 =&

This is the same ratio as in part a). d) 4142.12 =& ; 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: http://tessellations.org/guestgallery1thumbs.htm


http://tessellations.org/guestgallery1thumbs.htm

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:

front view

side viewtop view


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) Diagrams may vary. For example:

This is the isometric view. This is the orthographic view.

top view

front view

side view

b) Diagrams may vary. For example:

top view

front viewside view

c) Answers may vary. For example:

They are congruent right triangles.


Chapter 6 Section 2 Question 9 Page 316 a) Answers may vary. b) Diagrams may vary. For example:

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 317 Diagrams may vary. For example:


Chapter 6 Section 2 Question 12 Page 317 Diagrams may vary. For example: A final drawing is shown below.

This diagram shows the construction lines used.

Chapter 6 Section 2 Question 13 Page 317


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 Diagrams may vary. For example:

Chapter 6 Section 3 Question 3 Page 322 Diagrams may vary. For example:

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.


Chapter 6 Section 3 Question 6 Page 323 a) Diagrams may vary. For example:

8 cm

8 cm

12 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) Diagrams may vary. For example:

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.



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 m2. c) Cost of metal: $6.5/m2 × 51.6 m2 =& $335.40

The cost of the metal is about $335. Chapter 6 Section 3 Question 9 Page 324 Diagrams may vary. For example:

20 cm

10 cm

10 cm


Chapter 6 Section 3 Question 10 Page 324 Diagrams may vary. For example: 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.

20 cm

10 cm

10 cm

30 cm

31.4 cm Chapter 6 Section 3 Question 11 Page 324 a) Diagrams may vary. For example: b) The folded tetrahedron should be similar to:

3 m

3 m 3 m

3 m

3 m

3 m


Chapter 6 Section 3 Question 12 Page 324 Diagrams may vary. For example: 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 4 Scale Models Chapter 6 Section 4 Question 1 Page 331 Set up a proportion to determine lengths.

5 102ft. 10

in. ft. 2in. 1

==

=

xx

x

6

122ft. 12in.

ft. 2in. 1

==

=

yy

y

The model will be 5 in. by 6 in. Chapter 6 Section 4 Question 2 Page 331 Set up a proportion to determine lengths.

1024 1288in. in. 281

81

in. in. 8 ft. 10

81

=×=

=

=

xx

x

1024 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 1.618 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.


Chapter 6 Section 4 Question 5 Page 332 2 ft.

2 ft.

2 ft. 2 ft.

2 ft.

2 ft.

Chapter 6 Section 4 Question 6 Page 332 Diagrams may vary. For example: The box will be a rectangular prism with dimensions 1 × 1 × 12, 1 × 2 × 6, 1 × 3 × 4, or 2 × 2 × 3 (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.).

Side viewTop viewFront view



Chapter 6 Section 4 Question 8 Page 332 a) Diagrams may vary. For example: 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 m2 – 6 m2 = 16 m2; the cost is $32. If the 1 × 3 face is the base, the area is 22 m2 – 3 m2 = 19 m2; the cost is $38. If the 1 × 2 face is the base, the area is 22 m2 – 2 m2 = 20 m2; the cost is $40.


Chapter 6 Section 4 Question 9 Page 332

a) The volume is: 42

322 dddhrV π=⎟

⎠⎞⎜

⎝⎛π=π=

Use trial-and-error to find d.

If d = 5, 17.984

)5( 3=π= &V

If d = 10, 40.7854

)10( 3=π= &V

If d = 9, 56.5724

)9( 3=π= &V

If d = 8, 12.4024

)8( 3=π= &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) Diagrams may vary. For example: Scale: 1 cm represents 4 cm of soup can


Chapter 6 Section 4 Question 10 Page 333 a) Diagrams may vary. For example: Scale: 1 cm represents 5 m

BC = 0.60 cm

BA = 2.00 cm

B A

C

c) Answers may vary. For example:

Area DEFGHC = 10.39 cm2

BC = 0.60 cm

BA = 2.00 cm

B A

H

G F

E

DC

Using The Geometer's Sketchpad®, the area of the base is found to be 10.39 cm2. The actual area is: 10.39 × 25 m2 = 259.75 m2 The volume of concrete is: 259.75 m2 × 0.1 m = 25.975 m3, or about 26 m3. d) Concrete cost: $75/m3 × 26 m3 = $1950


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) Diagrams may vary. For example: 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 m2 Cost of building: $1200/m2 × 144 m2 = $172 800

An estimate for the cost of the building is $172 800.


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 Diagrams may vary. For example:

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.

8.534

3435 222

===

+=

&s

s

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 4.31)5(22 =π=π= &r

The central angle that the arc subtends is: °=°×π

310360)8.5(2

4.31 &

Diagram for the net may vary. For example:

50° 5.8 cm


Chapter 6 Section 5 Solve Problems With Given Constraints Chapter 6 Section 5 Question 1 Page 340 1 L = 1000 cm3 If each side of the cube has a length of 10 cm, the volume will be (10 cm)3, or 1000 cm3. The side length of the box is 10 cm. Chapter 6 Section 5 Question 2 Page 340 For a circle, the area is: A = πr2

6.12

500

500500

2

2

=π

=

π=

π=

&

r

r

r

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)



Hallway

Bedroom #2

10 by 15

Bedroom # 3

8 by 15

Bathroom

7 by 15

Bedroom #1

15 by 15

Kitchen 10 by 20

Living Room

10 by 20

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.

8.3565.0000200

5.0000200

)5.0(0002002

2

2

=π

=

π=

π=π=

&

r

r

rhrV

The radius of the circle is about 356.8 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.



front view side view

top view


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 m2.

Cost of metal: $15/m2 × 55.4 m2 = $831


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.

ED = 5.00 cm

m DA = 8.00 cm

m AB = 6.00 cm

E

D C

BA

b) Method 1 (Assume that the base for the cylinder already exists). The surface area of a cylinder = 2πrh + πr2

9.329105

2580)5()8)(5(2 2

=π=

π+π=π+π=

&

SA

The area of the walls and roof is about 329.9 m2. ; the approximate volume of concrete required is 264 m3. 2649.3298.0 =× & Method 2 (Assume that the base for the cylinder needs to be constructed). The surface area of a cylinder = 2πrh + 2πr2

4.408130

5080)5(2)8)(5(2 2

=π=

π+π=π+π=

&

SA

The area of the walls, base, and roof is about 408.4 m2. ; the approximate volume of concrete required is 327 m3. 3274.4088.0 =× & c) Method 1 (from above) Cost: $90/m3 × 264 m3 = $23 760 The estimated cost for the concrete is $23 760. Method 2 (from above) Cost: $90/m3 × 327 m3 = $29 430 The estimated cost for the concrete is $29 430.


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,

75.075.01

75.0175.01

)3()5.0(1

3

3

3

2

2

=π

=

π=

π=π=π=

&

d

d

ddd

hrV

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 2 × 0.5 in. = 1 in. = 112

ft. =& 0.083 ft.

The outside diameter and the height will be 0.83 ft. and 2.33 ft. respectively. Diagrams may vary. For example:

top view

side view front view 2.33 ft.

2.33 ft.

0.83 ft. 0.83 ft.

0.83 ft.


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 m2 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 m2 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 m2 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 m2 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 m2 The heat required is 11 250 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 Diagrams may vary. For example: Each small square in the diagram represents a square with side length 50 m.

GateCluster

GateCluster

GateCluster

Check-in Area


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:

625.180

130 =

The half blanket has dimensions 80 cm by 65 cm. The ratio is:

231.16580 =&

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) Diagrams may vary. For example:

front view side view

top view

Chapter 6 Review Question 4 Page 346 Diagrams may vary. For example:


Chapter 6 Review Question 5 Page 346 Diagrams may vary. For example: Scale: 1 cm represents 2 ft.

Bedroom 2

Washroom 2

Washroom 1

Bedroom 1

CD = 4.00 cm

AB = 12.00 cm

C D

BA

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. Diagrams may vary. For example:

radius 3 cm

18 cm

18.8 cm


Chapter 6 Review Question 7 Page 347 Diagrams may vary. For example:

Fold line

1 m

1 m

5 m

Chapter 6 Review Question 8 Page 347 Divide each of the actual measurements by 72.

846.072

9.60 =&

The wingspan is about 0.846 m, or 84.6 cm.

026.172

9.73 =&

The length is about 1.026 m, or 102.6 cm. Chapter 6 Review Question 9 Page 347 Diagrams may vary. For example: Scale: 1 cm represents 1 m

radius 5 cm

10 cm

15.7 cm


Chapter 6 Review Question 10 Page 347 Diagrams may vary. For example: 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: 2598.2)866.0)()(5.0(6 xxx = Thickness of concrete: 10 cm = 0.10 m Volume of concrete required for the base is: 22 2598.001.0598.2 xx =× Cost, in dollars, of the concrete is 2 20.2598 75 19.485x x× = Since the budget is $600,

55.5485.19

600485.19

600600485.19

2

2

=

=

=

=

&x

x

x

x

The side length of the hexagon is about 5.55 m.

b) Area of one wall is: (5.55 m)2 = 30.80 m2 Area of the roof = area of the base: 2.598(5.55)2 = 80.02 m2 Area of wood required is: 12(30.80) + 80.02 = 449.62 m2 Cost of the wood is: 449.62(20) = 8992.4 The cost of wood is about $8992.

The total cost of the project is: 8992 + 600 = $959


Chapter 6 Practice Test Chapter 6 Practice Test Question 1 Page 348 A

(Length to width ratios: 236;3

26;5.2

25;6.1

58 ==== .

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 Diagrams may vary. For example:

side view front view

top view 6 cm

Chapter 6 Practice Test Question 5 Page 348 Diagrams may vary. For example:


Chapter 6 Practice Test Question 6 Page 348 Diagrams may vary. For example: Scale: one space between 2 dots represents 10 cm

Chapter 6 Practice Test Question 7 Page 349

side view front view

top view


Chapter 6 Practice Test Question 8 Page 349 Diagrams may vary. For example: Scale: one space between 2 dots represents 1 in.

Chapter 6 Practice Test Question 9 Page 349 a) Diagrams may vary. For example: Scale: one space between 2 dots represents 2 cm

b) The height of the can is: 1.618 × 12 cm = 19.416 cm

The volume is about 2196 mL.

2196)416.19()6( 2

2

=π=π=

&

hrV


Chapters 4 to 6 Review Chapters 4 to 6 Review Question 1 Page 350 a) i)

x y First Differences

Second Differences

0 17 –4 1 13 –4 2 9 –4 3 5 –4 4 1 –4 5 –3 –4 6 –7 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

1 0 1 3 1 1 2 5 3 1 3 7 6 1 4 9 10 1 5 11 15 1 6 13 21 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 x2-term. iv) Answers may vary. For example:

Quadratic; the graph is a parabola since there is an x2-term.


b) i) ii)

20 25

6

14 −10 iii) iv)

15

−10 10

10

−5 5

−15 −10 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.


Chapters 4 to 6 Review Question 3 Page 350 a) i) vertex at (1, 9) ii) parabola opens upward iii) not stretched iv)

−25 25

60

b) i) vertex at (–8, –5) ii) parabola opens downward iii) vertically stretched iv)

−10

−25 25

40

c) i) vertex at (0, –1) ii) parabola opens upward iii) vertically compressed iv)

40

−10

−25 25

d) i) vertex at (0, 0) ii) parabola opens upward iii) vertically stretched iv)

−10 10

20


e) i) vertex at (2, 2) ii) parabola opens upward iii) vertically compressed iv)

−25 25

50

f) i) vertex at (–1, 13) ii) parabola opens upward iii) vertically stretched iv)

−25 25

50

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 177.4 m. b) Substitute t = 7 into the equation.

24.9(7 6) 177.44.9 177.4

172.5

h = − − += − +=

The height of the flare 7 seconds after launch will be 172.5 m.

c)

200

13 d) From the graph, the flare will hit the water after about 12 s.


Chapters 4 to 6 Review Question 5 Page 350 a) (3x + 4)(10x + 1) b) (x − 2)(4x + 15) = 30x2 + 3x + 40x + 4 = 4x2 + 15x − 8x − 30 = 30x2 + 43x + 4 = 4x2 + 7x − 30 c) (12x − 8)(2x + 0.5) d) (6 + 2x)(6 − 2x) = 24x2 + 6x − 16x − 4 = 36 − 12x + 12x − 4x2 = 24x2 − 10x − 4 = 36 − 4x2 Chapters 4 to 6 Review Question 6 Page 350 a) (s − 1)(4s − 7) = 4s2 − 7s − 4s +7 = 4s2 − 11s + 7 b) 4(15)2 − 11(15) + 7 = 900 − 165 + 7 = 742

The actual area of the apartment is 742 m2.

Chapters 4 to 6 Review Question 7 Page 350 a) b)

5.1685.01644

5.0)4)(4(5.0)4(

2

2

2

+−=++−−=

+−−=+−=

xxyxxxy

xxyxy

972031001010

3)10)(10(3)10(

2

2

2

++=−+++=

−++=−+=

xxyxxxy

xxyxy

c) d)

593282732328

27)44(827)422(8

27)2)(2(827)2(8

2

2

2

2

2

++=+++=+++=

++++=+++=

++=

xxyxxyxxy

xxxyxxy

xy

526.252.38.02.516.252.3

8.0)168(2.38.0)1644(2.3

8.0)4)(4(2.38.0)4(2.3

2

2

2

2

2

−+−=−−+−=

−+−−=−+−−−=

−−−−=−−−=

xxyxxy

xxyxxxy

xxyxy

Chapters 4 to 6 Review Question 8 Page 350 To find the y-intercept, substitute x = 0 into the equation.

42)6)(7(

)6)0(2))(0(37(

−=−=

−−=y

The y-intercept is –42.


Chapters 4 to 6 Review Question 9 Page 351 a) x2 + 11x + 24 (Find 2 numbers with product 24 and sum 11.) = (x + 8)(x + 3) b) x2 + x – 30 (Find 2 numbers with product –30 and sum 1.) = (x + 6)(x – 5) c) x2 – 8x + 7 (Find 2 numbers with product 7 and sum –8.) = (x – 7)(x – 1) d) x2 + 8x + 16 (Find 2 numbers with product 16 and sum 8.) = (x + 4)(x + 4) e) 3x2 + 39x + 108 (Factor out the GCF) = 3(x2 + 13x + 36) (Find 2 numbers with product 36 and sum 13.) = 3(x + 9)(x + 4) f) –10x2 – 110x – 100 (Factor out the GCF) = –10(x2 + 11x + 10) (Find 2 numbers with product 10 and sum 11.) = –10(x + 10)(x + 1) Chapters 4 to 6 Review Question 10 Page 351 a) ( )( ) 2 25 1 5 5 4x x x x x x x+ − = − + − = + − 5 Equivalent; explanations may vary. b) ( )( ) 2 2 22 1.5 1.5 2 3 3.5 3 3.5 3x x x x x x x x x− − = − − + = − + ≠ + + Not equivalent; explanations may vary. c) ( )( ) 2 21 2 2 1 2x x x x x x x− + = + − − = + − 2 Equivalent; explanations may vary.


Chapters 4 to 6 Review Question 11 Page 351 a) y = x2 – 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 = 10x2– 360 (Factor out the GCF) y = 10(x2 – 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 = –2x2 + 4x + 70 (Factor out the GCF) y = –2(x2 – 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 = 3.5x2 + 21x + 31.5 (Factor out the GCF) y = 3.5(x2 + 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

x + 16

2x

b)

xxSA

xxxxxSA

xxxxxxSA

48416322

)16(2

2)16(22

2

2

222

+=++++=

+⎟⎠⎞⎜

⎝⎛+++⎟

⎠⎞⎜

⎝⎛=

c) Since the width of the base is 5,

10

52=

=

x

x

( ) ( )24 10 48 10

400 480880

SA = +

= +=

The surface area is 880 cm2.


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)

front side

top


Chapters 4 to 6 Review Question 15 Page 351 a)

side

top

front


c) The top of the platform is a rectangle with width 6 m and length 8 m. Its area is 48 m2. Cost: $40/m2 × 48 m2 = $1920

The cost of the plywood is $1920.


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