gr.9 pre-ib unit 10 measurement

2009-2010

George Li

MPM 1DE

Mr. Rocca

Measurement (I.S.U. Project)

(10.1) Areas of Composite Figures Pp.441-444 #1-3, 6-10, 15, 16

1. Calculate the perimeter of each figure.

a)

4 cm 3 cm

C= 5 cm

∵ this is a right-angled triangle, we

can use the Pythagorean Theorem

to find the value of c

c ² = 4 ² + 3 ² c ² = 16 + 9 c ² = 25

c = 25 c = 5 cm

p = a + b + c p = 3 + 4 + 5

p = 12 cm

∴, the perimeter of this triangle is 12 cm.

b) a

b c 10 cm

5 cm ∵ abc is right-angled, we can

use the Pythagorean Theorem to

find the length of ac

ac ² = ab ² + bc ² ac ² = 3 ² + 9 ² ac ² = 9 + 16 ac ² = 25

ac = 25 ac = 5 cm

p = 2l + 2w p = 2(10) + 2(5) p = 20 + 10

p = 30 cm

∴, the perimeter of this parallelogram is 30 cm.

3 cm

4 cm

#1 continued…

c)

a

b

8.5 m

4.0 m

p = 4a + 8b p = 4(8.5) + 8(4) p = 34 + 32

p = 66m

∴, the perimeter of this

cross shape is 66 m.

d)

12

km

a b c

10 km

∵ abd is a right-angled triangle, we can

use the Pythagorean Theorem to find the

length of ad

d

ad ² = ab ² + bd ² ad ² = 5 ² + 12 ² ad ² = 25 + 144 ad ² = 169

ad = 169 ad = 13 km

p = ac + cd + ad p = 10 + 13 + 13

p = 36 km

∴, the perimeter of this triangle is 36 km.

2. Calculate the area of each figure in exercise 1. a)

4 cm

3 cm

a = 𝑏ℎ

2

a = 3 (4)

2

a = 12

2

a = 6 cm²

∴, the area of this triangle is 6 cm².

b) 5 cm

10 cm

3 cm

4 cm

a = bh a = (10)(3)

a = 30 cm²

∴, the area of this parallelogram is 30 cm².

c) a

b

8.5 m

4.0 m

ALARGE = lw ALARGE = (4+4+8.5) (8.5) ALARGE = (16.5)(8.5)

ALARGE = 140.25 m² ASMALLx2 = 2lw ASMALLx2 = 2(4)(8.5) ASMALLx2 = 2(34)

ASMALLx2 = 68 m² A = ALARGE + ASMALLx2

A = 140.25 + 68

A = 208.25 m²

∴, the area of this cross is 208.25 m².

#2 continued…

d)

12

km

10 km

a = 𝑏ℎ

2

a = 10 (12)

2

a = 120

2

a = 60 km²

∴, the area of this triangle is 60 km².

3. Determine the shaded area of each figure.

4 cm

6 cm

ALARGE = 𝜋r² ALARGE = 𝜋6² ALARGE = 𝜋36

ALARGE = 113.1 𝑐𝑚²

ASMALL = 𝜋r² ASMALL = 𝜋4² ASMALL = 𝜋16

ASMALL = 50.3 𝑐𝑚²

A = ALARGE – ASMALL A = 113.1 – 50.3

A = 62.8 cm²

∴, the area of the shaded area is 62.8 cm².

a)

#3 continued…

b) 5 cm ARECTANGLE = lw ARECTANGLE = [8(5)][6(5)] ARECTANGLE = (40)(30)

ARECTANGLE = 1200 cm²

ACIRCLEx12 = 12πr²

ACIRLCEx12 = 12π5² ACIRCLEx12 = 12π25

ACIRCLEx12 = 942.5 cm²

A = ARECTANGLE – ACIRCLEx12 A = 1200 – 942.5

A = 257.5 cm²

∴, the area of the shaded area is 257.5 cm².

c) 3 m

3 m

7 m

1 m

1 m

ATRIANGLE = 𝑏ℎ

2

ATRIANGLE = 7 (3)

2

ATRIANGLE = 21

2

ATRIANGLE = 10.5 m²

ARECTANGLE = lw ARECTANGLE = (7)(3)

ARECTANGLE = 21 m²

ADOOR= lw ADOOR = (1)(2.5)

ADOOR = 2.5 m²

2.5 m

ACIRCLE = πr²

ACIRLCE = π0.5² ACIRCLE = π0.25

ACIRCLE = 0.79 m²

A = ATRIANGLE + ARECTANGLE – ADOOR – ACIRCLE A = 10.5 + 21 – 2.5 – 0.79

A = 28.21 m²

∴, the area of the shaded area is 28.21 m².

#3 continued…

d)

3 cm

4 cm

1 cm

ASQUARE = lw ASQUARE = (4)(4)

ASQUARE = 16 cm²

APARALLELOGRAMx2 = 2bh APARALLELOGRAMx2 = 2(3)(1)

APARALLELOGRAMx2 = 6 cm²

ASHADED = ASQUARE - APARALLELOGRAMx2

ASHADED = 16 – 6

ASHADED = 10 cm²

∴, the area of the shaded area is 10 cm².

6. The circumference of a circle is 14.5 cm.

a) Calculate the diameter of the circle.

C = πd 14.5 = πd 14.5/π = d 4.62 = d

∴, the diameter of the circle is 4.62 cm.

b) Calculate the radius of the circle.

r = d/2 r = 4.62/2 r = 2.31

∴, the radius of the circle is 2.31 cm.

#6 continued…

c) Calculate the area of the circle.

a = πr²

a = π2.31² a = π5.34 a = 16.78

∴, the area of the circle is 16.78 cm².

7. The area of a circle is 63.6 m².

20.24 = 𝑟

a) Calculate the radius of the circle.

a = πr² 63.6 = πr² 63.6/π = r²

20.24 = r²

4.5 = r

∴, the radius of the circle is 4.5 m.

b) Calculate the diameter of the circle.

d = 2r d = 2(4.5) d = 9

∴, the diameter of the circle is 9 m.

c) Calculate the circumference of the circle.

c = 2πr c = 2π(4.5) c = 2(14.14) c = 28.28

∴, the area of the circle is 16.78 m.

8. A Roman window comprises of rectangle and a semicircle (below left).

The dimensions of the rectangle are 60 cm by 100 cm. Determine the

perimeter of this window.

100 cm

60 cm

60 cm

Circumference = 2πr c = 2π(60/2) c = 2π30 c = 2(94.25) c = 188.5

p = 100 + 100 + 60 + (188.5/2) p = 260 + 94.25 p = 354.25

∴, the perimeter of the window is 354.25 cm.

9. A picture frame measures 30 cm by 20 cm (above right). The four

trapezoids that comprise this frame are cut from one piece of wood 2

cm wide. Each cut removes 0.3 cm from that piece of wood.

20 cm

30 cm

2 cm

16 cm

26 cm

a) Determine the area of a picture that fills this frame.

a = lw a = (26)(16)

a = 416 cm² ∴, the area of a picture that will fill this frame is 416 cm².

#9 continued…

b) Determine the minimum length of wood needed to make this frame.

w = 2[30+2(0.3)] + 2[20+2(0.3)] w = 2(30+0.6) + 2(20+0.6) w = 2(30.6) + 2(20.6) w = 61.2 + 21.2 w = 102.4 ∴, the minimum length of wood needed to make this frame is 102.4 cm.

c) How would your answer to part b change if the framing material was

wider? Explain.

The answer would not have changed, because the size of the frame would

not become bigger, just the inside of the frame becoming smaller, so the

length of would needed would stay the same.

10. The field inside a 400-m running track is to be seeded. Each straight

portion of the track is 100 m. Each curved part if the track is a semicircle.

One 1.5-kg bag of grass seed will seed an area of 80 m².

a) What is the length of each curved part of the track? 400 – 2(100) = 200 ∵ the length of the remaining track the two semicircles is 200 m, then one semicircle or curved part is 200 ÷ 2 = 100 m. ∴, the length of each curved part is 100 m long.

b) Determine the width of the field. ∵ the width of the field is the diameter of the semicircles, we can use the formula c = πd. 200 = πd 200/π = d 63.66 = d ∴, the width of the field is 63.66 m.

#10 continued…

c) Determine the area of the field to the nearest 10 m². ARECTANGLE = lw ARECTANGLE = (100)(63.66)

ARECTANGLE = 6366 m² ASEMICIRCLE = πr² ASEMICIRCLE = π(63.66/2)² ASEMICIRCLE = π(31.83)² ASEMICIRCLE = π(1013.15)

ASEMICIRCLE = 3182.9 m² AFIELD = ARECTANGLE + ASEMICIRCLE

AFIELD = 6366 + 3182.9

AFIELD = 9548.9

AFIELD ≈ 9550 m²

∴, the area of the field is about 9550 m².

d) Determine the number of bags of seed required. Bags = Area / Seed Area Bags = 9550 / 80 Bags ≈ 120 ∴, you need about 120 bags of seed to cover the field.

e) One 1.5-kg bag of grass seed costs $ 12.64. How much does it cost to seed the field? Cost = Bags x Cost Cost = 120 x 12.64 Cost = 1516.80 ∴, it costs $1516.80 to seed the field.

15. Grain elevators are common on the prairies. The front of this grain

elevator is to be painted. One can of paint covers an area of 4 m².

Determine how many cans of paint need to be purchased.

6 m

10 m

2 m

2 m

1 m

3 m

ABIGRECTANGLE = lw ABIGRECTANGLE = (10)(6)

ABIGRECTANGLE = 60 m²

ASMALLRECTANGLE = lw ASMALLRECTANGLE = (3)(2)

ASMALLRECTANGLE = 6 m²

ATRAPEZIOD = (a+b)h/2 ATRAPEZIOD = (6+3)2/2 ATRAPEZIOD = (9)2/2 ATRAPEZIOD = 18/2

ATRAPEZIOD = 9 m²

ATRIANGLE = bh/2 ATRIANGLE = (3)(1)/2 ATRIANGLE = 3/2

ATRIANGLE = 1.5 m²

AELEVATOR = ABIGRECTANGLE + ASMALLRECTANGLE + ATRAPEZIOD + ATRIANGLE

AELEVATOR = 60 + 6 + 9 + 1.5

AELEVATOR = 76.5 m² Cans = Area / Paint Area Cans = 76.5 / 4 Cans = 19.125 Cans ≈ 20 ∴, you need 20 cans to paint the front of the grain elevator.

16. A dog is tied to a 2-m leash at ground level on the side of a building.

The leash is attached 1 m from the corner of the building.

a) Copy this diagram. Sketch the region within the dog’s reach.

b) Determine the area of this region, to the nearest square meter. ∵ the dog’s leash is 2 m long, than the radius of the circle is 2 m.

A = πr² A = π(2)² A = π4 A ≈ 13 ∵ the dog can only reach about half of this area, we have to divide 13 by 2. AREACH = 13/2 AREACH = 6.5 AREACH ≈ 7

∴, the area the dog can reach is about 7 m².

(10.2) Surface Area & Volume of a Prism Pp.454-456 #1, 2, 5-7, 9

1. Calculate the surface area and volume of each prism.

a)

b)

4 cm

Surface Area SA = 6(lw) SA = 6[(4)(4)] SA = 6(16)

SA = 96 cm² Volume V = lwh V = (4)(4)(4) V = 64 cm3

∴, the surface area is 96 cm², and the volume is 64 𝑐𝑚3.

Surface Area SA = 2(wh + lw + lh) SA = 2[(2.5)(3.5) + (12)(2.5) + (12)(3.5)] SA = 2(8.75 + 30 + 42) SA = 2(80.75)

SA = 161.5 cm² 12 cm

2.5 cm

3.5 cm

Volume V = lwh V = (12)(2.5)(3.5) V = 105 cm3

∴, the surface area is 161.5 cm², and the volume is 105 𝑐𝑚3.

#1 continued…

c)

d)

10 cm

3 cm

2 cm

Find Missing Side of Triangle c² = a² + b² c² = 2² + 3² c² = 4 + 6 b² = 10

b² = 10 b² = 3.16 cm

Surface Area SA = 2(bh/2) + (wb) + (wh1) + (wh2) SA = 2(2)(3)/2 + (10)(2) + (10)(3.16) + (10)(3) SA = 6 + 20 + 31.6 + 30

SA = 87.6 cm²

6 cm

10 cm

4 cm

Volume V = 1/2bhl V = ½ (2)(3)(10) V = 30 cm3

∴, the surface area is 87.6 cm², and the volume is 22

𝑐𝑚3.

Surface Area SA = 2(bh/2) + 2(sl) + bl SA = 2(6)(4)/2 + 2(5)(10) + (6)(10) SA = 24 + 100 + 60

SA = 184 cm²

5 cm

Volume V = 1/2bhl V = ½ (6)(4)(10) V = 120 cm3

∴, the surface area is 184 cm², and the volume is 120

𝑐𝑚3.

3.16 cm

2. For each prism, several measurements are given. Calculate each unknown

measurement.

a)

b)

c)

5 cm

12 cm

V = 180 𝑐𝑚3

w

w = v/lh w = 180/(12)(5) w = 180/60 w = 3 cm

∴, the length of the missing side is 3 cm.

3 cm 3 cm

h

h = v/(bh/2) h = 280/[(7)(5)/2] h = 280/17.5 h = 16 cm

A = 138 𝑐𝑚2


V = 280 𝑐𝑚3

7 cm

h

5 cm

h = (a - 2lw)/4/3 h = [138 – 2(3)(3)]/4/3 h = (138 – 18)/4/3 h = 120/4/3 h = 30/3 h = 10 cm


5. Calculate the surface area and volume of each solid.

a)

b)

12.5 cm

25 cm

20 cm

15 cm

x

To find x x ² = 12.5² + 12.5²

x² = 156.25 + 156.25

x² = 312.5

x = 312.5 x = 17.6 cm

Surface Area SA = 2[(25)(12.5)/2] + 2(25)(15) + 2(20)(15) + 2(20)(17.7) + (25)(20) SA = 312.5 + 750 + 600 + 708 + 500

SA = 2870.5 cm² Volume V = (25)(20)(15) + (25)(12.5)/2(20) V = 7500 + 3125 V = 10625 𝑐𝑚3

∴, the surface area is 2870.5 cm², and the volume is

10625 𝑐𝑚3.

5 cm

25 cm

20 cm

15 cm

Surface Area SA = 2(20)(15) + 2(25)(20) + [2(25)(15) – 2(5)(5)] SA = 600 + 1000 + 700

SA = 2300 cm²

#5 b) continued…

c)

Volume V = (25)(20)(15) - (5)(5)(20) V = 7500 - 500 V = 7000 𝑐𝑚3 ∴, the surface area is 2300 cm², and the volume is 7000

𝑐𝑚3.

5 cm

25 cm

20 cm

15 cm

Surface Area SA = 2(20)(15) + 2(25)(20) + 2(15)(25) – (5)(5) + 5(5)(5) SA = 600 + 1000 + 750 – 25 + 125

SA = 2450 cm² Volume V = (25)(20)(15) + (5)(5)(5) V = 7500 + 125 V = 7625 𝑐𝑚3 ∴, the surface area is 2450 cm², and the volume is 7625

𝑐𝑚3.

#5 continued…

d)

6. The box of a dump truck has dimensions 1m by 2m by 4m. Explain how

this truck was able to carry 9𝑚3 of soil.

1 x 2 x 4 = 8𝑚3.

The truck is able to carry 9𝑚3of soil because they piled the soil over the sides of the box.

7. Appliances such as refrigerators, freezers, and microwave ovens, have volumes

measured in cubit feet. A fridge has a volume of 24 cubic feet. Determine at least

6 sets of possible dimensions for the inside of this fridge.

1 ft x 1 ft x 24 ft, 1 ft x 2 ft x 12 ft, 1 ft x 3 ft x 8 ft, 1 ft x 4 ft x 6 ft, 2 ft x 2 ft x 6 ft, 2 ft x 3 ft x 4 ft 2 ft x 2 ft x 6 ft would be the most likely for a fridge.

25 cm

20 cm

15 cm

5 cm

4 cm

3 cm

Surface Area SA = 2(20)(15) + 2(25)(20) + [2(25)(15) – 2(3)(4)/2] SA = 600 + 1000 + 738

SA = 2338 cm² Volume V = (25)(20)(15) – (3)(4)/2(20) V = 7500 - 120 V = 7380 𝑐𝑚3 ∴, the surface area is 2338 cm², and the volume is 7380

𝑐𝑚3.

2 ft 2 ft

4 ft

9. Determine the minimum amount of packaging needed to completely cover

a Toblerone® bar with these dimensions: length 20.7 cm; triangular face

has edges 3.5 cm and height 3.0 cm. Express the surface area to the

nearest square centimeter.

To find length of missing side

s² = 3.5² + 3² s² = 12.25 + 9 s² = 21.25

s = 21.25 s = 4.6

20.7 cm

3 cm

3.5 cm

Surface Area SA = 2[(3.5)(3)/2] + 2(3.5)(20.7) + (3.5)(20.7) SA = 10.5 + 144.9 + 72.45

SA = 227.85 cm²

SA ≈ 228 cm²

∴, you need at least 228 cm² of packaging to

completely cover a Toblerone bar.

(10.3) Optimal Value of Measurements (2D) Pp.447-450 #1-6, 8, 9

1. A city zoo rents strollers and wagons. Zoo workers are building a rectangular enclosure

to contain the strollers and wagons. They have 30 metal stands, identical to the ones

described in Investigation 2.

a) What are the dimensions for the largest rectangular area of the enclosure?

The dimensions for the largest rectangular area of the enclosure are 21 m (7 stands)

by 24 m (8 stands), or an area of 504 m².

b) Can they make an enclosure with an area of 50 m²? If so, what are its dimensions?

No, because no combination of stands can make an area of 50 m². To make it

50 m², you would have to cut the stands to the appropriate length.

2. Melanie has 36 patio tiles, each 0.6 m square.

a) Suppose she uses the patio tiles to form a path. What are the dimensions of

the narrowest rectangular path she can make?

The dimensions of the narrowest path she can make are 0.6 m (1 tile) by 21.6

m (36 tiles).

b) Suppose she uses the patio tiles to form a square patio. What are the

dimensions of the patio?

The dimensions of the patio would be 3.6 m (6 tiles) by 3.6 m (6 tiles).

3. Campbell has 10 railroad ties, each 1.8 m long. He uses the ties to enclose a rectangular

garden.

a) What are the possible dimensions of the different rectangles he could form?

b) What is the area of each rectangle in part a?

~ Chart in part a) ~

Number of Ties (l)

Number of Ties (w)

Length (cm) Width (cm) Area (cm²)

3 2 5.4 3.6 19.44

4 1 7.2 1.8 12.96

4. Steve wants to fence a rectangular garden. The fencing material comes in 1-m long units

that cannot be cut. Suppose Steve has 20 m of fencing. What are the dimensions of the

largest garden he can make? Explain your answer.

The biggest garden Steve can make is 25 m², or 5 m (5 fences) by 5 m (5 fences).

5. If a spreadsheet program is available, use it to complete this exercise. A lifeguard has

400 m of rope to enclose a rectangular swimming area at a beach. The diagrams show

different ways she can do this.

a) Do you think the area of the enclosed region depends on the way the rope is

arranged? Explain.

Yes, because different lengths of rope on each side will make the area either larger

or smaller.

b) Suppose the side parallel to the beach measures 300m. How long is the other side of

the rectangle? Calculate the area of water enclosed by the rope.

~ In part d) table ~

c) Repeat part b if the side parallel to the beach measures:

i) 250 m ii) 200 m iii) 150 m iv) 100 m

~ In part d) table ~

d) Record your results in a table.

Length of side parallel to beach (m)

Length of side perpendicular to beach (m)

Total length of rope (m)

Area enclosed (m²)

300 50 400 15000

250 75 400 18750

200 100 400 20000

150 125 400 18750

100 150 400 15000

e) What are the dimensions of the largest possible rectangular swimming area? How

are these dimensions related?

The largest possible dimensions are 200 m x 100 m. The length is twice the width.

6. A store owner wants to create a rectangular area for a store display. He has 6 m of rope.

What are the dimensions of the largest area he can enclose in each situation? Explain

your answers.

a) The rope encloses the entire area.

The largest area he can enclose would be 1.5 m x 1.5 m, or 2.25 m² of space. This is

the largest area because all the other measurements, such as 1 m x 2 m, or 2

m² of space, are smaller.

b) There is a wall on one side.

The largest area he can enclose would be 1.5 m x 3.0 m, or 4.5 m² of space. This is

the largest area because all the other possible measurements, such as 1 m x

4 m, or 4 m² of space, are smaller.

c) There are walls on 2 sides.

The largest area he can enclose would be 3 m by 3 m, since that is the

maximum amount of rope the store owner has.

8. Refer to Investigation 2. Determine the dimensions of the largest enclosure the

resort workers can make for each situation.

a) Use metal stands for 3 sides of the enclosure, with a wall on the remaining

side.

The dimensions of the largest enclosure in this case would be 36 m (12

stands) by 21 m (7 stands) for an area of 756 m².

b) Use metal stands for 2 sides of the enclosure, with walls on two adjacent

sides.

The dimensions of the largest enclosure in this case would be 39 m (13

stands) by 39 m (13 stands) for an area of 1521 m².

9. The resort workers in Investigation 2 decide to separate the outdoor equipment

from the furniture. They set up two storage enclosures of equal area, with a

common side. The diagram shows one way to do this, using 26 metal stands.

a) Calculate the combined are of the two storage enclosures.

Since each stand is 3 m, and there are 6 stands for the length and 4 stands

for the width, then the dimensions are 18 m by 12 m. This would make the

area of the enclosures 216 m².

b) Find some other ways to arrange the metal stands to make the two storage

areas. Calculate the combined area for each. Record your results in a table.

Number of Stands in the Two Lengths

Number of Stands in the Three Widths

Total Length (m)

Total Width (m)

Area (m²)

10 2 30 6 180

4 6 12 18 216

c) How should the stands be arranged to make the largest possible enclosure?

To make the largest possible enclosure, you would need 7 stands for the two

lengths each, and 4 stands for the widths and separator. This would give you an area

of 252 m².

(10.4) Optimal Value of Measurements (3D) Pp.447-450 #1-4, 10, 11, 12

1. a) List several possible sets of dimensions for a rectangular prism with volume 24 𝑐𝑚3.

Each dimension should be a whole number.

Length (cm) Width (cm) Height (cm) Volume (cm²) Surface Area (𝑐𝑚3)

2 3 4 24 2(2)(3) + 2(2)(4) + 2(3)(4) = 52

1 1 24 24 2(1)(1) + 2(1)(24) + 2(1)(24) = 98

24 1 1 24 2(1)(1) + 2(1)(24) + 2(1)(24) = 98

1 24 1 24 2(1)(1) + 2(1)(24) + 2(1)(24) = 98

4 1 6 24 2(4)(1) + 2(4)(6) + 2(1)(6) = 68

1 3 8 24 2(1)(3) + 2(1)(8) + 2(3)(8) = 70

b) Which prism has the least surface area?

The first on does, with an area of 52 𝑐𝑚3.

2. a) List several possible sets of whole-number dimensions for a rectangular prism with

volume 32 𝑐𝑚3.


1 1 32 32 2(1)(1) + 2(1)(32) + 2(1)(32) = 130

32 1 1 32 2(1)(1) + 2(1)(32) + 2(1)(32) = 130

2 1 16 32 2(2)(1) + 2(2)(16) + 2(1)(16) = 100

4 8 1 32 2(4)(8) + 2(4)(1) + 2(8)(1) = 88

4 2 4 32 2(4)(2) + 2(4)(4) + 2(2)(4) = 64

2 2 8 32 2(2)(2) + 2(2)(8) + 2(2)(8) = 72


The 5th one does, with an surface area of 64 𝑐𝑚3.

3. a) List several possible sets of whole-number dimensions for a rectangular prisms with

volume 100 𝑐𝑚3.


The 7th one does, with an surface area of 130 𝑐𝑚3.


1 1 100 100 2(1)(1) + 2(1)(100) + 2(1)(100) = 402

10 10 1 100 2(10)(10) + 2(10)(1) + 2(10)(1) = 240

1 2 50 100 2(1)(2) + 2(1)(50) + 2(2)(50) = 304

2 5 10 100 2(2)(5) + 2(2)(10) + 2(5)(10) = 160

2 2 25 100 2(2)(2) + 2(2)(25) + 2(2)(25) = 208

4 1 25 100 2(4)(1) + 2(4)(25) + 2(1)(25) = 258

5 5 4 100 2(5)(5) + 2(5)(4) + 2(5)(4) = 130

5 1 20 100 2(5)(1) + 2(5)(20) + 2(1)(20) = 250

4. The Acme Box Company has hired to design boxes to hold 4000 𝑐𝑚3 of popcorn for

movie theatres. The boxes have an open top and a square base.

a) Suppose the base of 10 cm long and 10 cm wide. What is the height for a volume of

4000𝑐𝑚3? Calculate the surface area of this box.

𝑣 = 𝑙𝑤ℎ 𝑆𝐴 = 10 10 + 4(10)(40)

4000 = 10 10 ℎ 𝑆𝐴 = 100 + (1600)

4000 = 100 ℎ 𝑆𝐴 = 1700 𝑐𝑚²

4000

100= ℎ

40 = ℎ

b) Repeat part a) for each base described.

i) 12 cm by 12 cm 𝑆𝐴 = 12 12 + 4(12)(27.78)

𝑣 = 𝑙𝑤ℎ 𝑆𝐴 = 144 + (1333)

4000 = (12) 12 ℎ 𝑆𝐴 = 1477 𝑐𝑚

4000 = 144 ℎ 4000

144= ℎ

27.78 = ℎ

ii) 14 cm by 14 cm 𝑆𝐴 = 14 14 + 4(14)(20.41)

𝑣 = 𝑙𝑤ℎ 𝑆𝐴 = 196 + (1143)

4000 = 14 14 ℎ 𝑆𝐴 = 1339 𝑐𝑚

4000 = 196 ℎ 4000

196= ℎ

20.41 = ℎ

c) Record your results from parts a) and b) in a table.


10 10 40 4000 10 10 + 4 10 40 = 1700

12 12 27.78 4000 12 12 + 4 12 27.78 = 1477

14 14 20.41 4000 14 14 + 4 14 20.41 = 1339

d) Determine the dimensions of the box that requires the minimum amount of

cardboard. Describe the box. Do you think it would be a good idea for movie

theatres to use this shape? Explain.

(20)(20) + 4(10)(20) = 1200 𝑐𝑚3

The dimensions that require the least amount of cardboard are 20 cm x 20 cm x 10

cm.

This would not be a good design, because the popcorn would fall out of the box

easily.

10. Suppose you have an 8-cm cube.

The 8-cm cube can be divided into 4-cm cubes.

These 4-cm cubes can be divided further into 2-cm cubes.

a) Calculate the total surface area of the cubes in each step above.

Step 1 Cube

SA = 6(lw)

SA = 6(8)(8)

SA = 384 cm²

8 cm

Step 2 Cube

SA = 8[6(lw)]

SA = 8[6(4)(4)]

SA = 8(96)

SA = 768 cm²

Step 3 Cube

SA = 64[6(lw)]

SA = 64[6(2)(2)]

SA = 64(24)

SA = 1536 cm²

b) SA = 512[6(lw)]

SA = 512[6(1)(1)]

SA = 512(6)

SA = 3072 cm²

c) You can multiply the previous surface area by two to get the surface area of the

cubes in the next step.

d) 3072*2 = 6144*2 = 12288*2 = 24576*2 = 49152*2 = 98304*2 = 196608 cm²

11. Sugar cubes come in boxes of 144 cubes. There are 2 layers of cubes. Each layer forms a

12 by 6 rectangle. The company wants to design a box that uses less cardboard and still

holds 144 sugar cubes.

a) Calculate the surface area of the box, in square units, that would enclose the cubes

shown.

SA = 2(lw) + 2(lh) + 2(wh)

SA = 2(12)(6) + 2(12)(2) + 2(6)(2)

SA = 144 + 48 + 24

SA = 216 square units

b) Determine three other ways to arrange 144 sugar cubes in a box. Calculate the

surface area of each box.

Box 1 (1 x 1 x 144) SA = 2(lw) + 2(lh) + 2(wh)

SA = 2(12)(1) + 2(12)(1) + 2(1)(1)

SA = 24 + 24 + 2


Box 2 (2 x 8 x 9)

SA = 2(lw) + 2(lh) + 2(wh)

SA = 2(9)(8) + 2(9)(2) + 2(8)(2)

SA = 144 + 48 + 24


Box 3 (2 x 3 x24)

SA = 2(lw) + 2(lh) + 2(wh)

SA = 2(24)(3) + 2(24)(2) + 2(3)(2)

SA = 144 + 96 + 24


c) How would you arrange 144 sugar cubes to use the least amount of cardboard?

SA = 2(lw) + 2(lh) + 2(wh)

SA = 2(6)(6) + 2(6)(4) + 2(6)(4)

SA = 72 + 48 + 48


I would use the dimensions 6 x 6 x 4, giving it a surface area of 168 units squared.

d) Yes, because using 6 x 6 x 4 would make the box more compact and easy to carry.

12. Mary bout some caramels at the bulk food store. She wants to pack them in a box. The

caramels are 2 cm by 2 cm by 1 cm.

a) Mary found a box measuring 8 cm by 5 cm by 4 cm. What is the maximum number

of caramels she can pack in this box? Explain your answers.

𝑉𝐵𝑂𝑋 = lwh 𝑉𝐶𝐴𝑅𝐴𝑀𝐸𝐿 = lwh

𝑉𝐵𝑂𝑋 = (8)(5)(4) 𝑉𝐶𝐴𝑅𝐴𝑀𝐸𝐿 = (2)(2)(1)

𝑉𝐵𝑂𝑋 = 160 cm² 𝑉𝐶𝐴𝑅𝐴𝑀𝐸𝐿 = 4 cm²

𝑉𝐵𝑂𝑋

𝑉𝐶𝐴𝑅𝐴𝑀𝐸𝐿 = Number of Caramels ∴, the number of caramels

𝑉𝐵𝑂𝑋

𝑉𝐶𝐴𝑅𝐴𝑀𝐸𝐿 =

160

4 she can hold is 40.

𝑉𝐵𝑂𝑋

𝑉𝐶𝐴𝑅𝐴𝑀𝐸𝐿 = 40

b) Mary found another box whose dimensions are double the dimensions of the box in

part a. What is the maximum number of caramels she can pack in it?

Since the dimensions are doubled, that means the volume is 8x bigger.

40 x 8 = 320 caramels.

∴, she can hold 320 caramels in the box.

(10.5) Surface Area of a Cylinder Pp.473-476 #1-7, 9, 10, 12

1. Estimate the area of each circle. Calculate the area of each circle. Give the answers to 1

decimal place.

a) Estimate = 314 A = πr²

A = π(10)² A = π100 A = 314.2

b) Estimate = 42 A = πr²

A = π(4)² A = π16 A = 50.3

c) Estimate = 150 A = πr²

A = π(6.7)² A = π44.89 A = 141

2. A net for a cylinder is shown.

a) Calculate the circumference of each circle.

C = 2πr C = 2π(1) C = 2(3.14) C = 6.28 cm

10 cm

4 cm

6.7 cm

2 cm

6.28 cm

3 cm

b) Divide the length of the rectangle by 2π. 6.28 / 2π = 1

c) How does the length of the rectangle compare with the circumference of each circle?

They are the same.

d) Determine the total surface area of the cylinder.

SACIRCLE = πr² SACIRCLE = π(1)² SACIRCLE = π(1)

SACIRCLE = 3.14 cm² 2(3.14) + (6.28)(3) 6.28 + 18.84

25.12 cm²

3. Estimate the total surface area of each cylinder. Calculate the total surface area. Give the answers to 1 decimal place. a) Estimate: 20 cm²


SACIRCLE = 3.14 cm² C = 2πr C = 2π(1) C = 2(3.14) C = 6.28 cm 2(3.14) + (6.28)(2) 6.28 + 12.56

18.84 cm²

1 cm

2 cm

b) Estimate: 1880 cm²


SACIRCLE = 314 cm² C = 2πr C = 2π(10) C = 2(31.4) C = 62.8 cm 2(314) + (62.8)(20) 628 + 1256

1884 cm²

c) Estimate: 20 cm²


SACIRCLE = 50.27 cm² C = 2πr C = 2π(4) C = 2(12.57) C = 25.14 cm 2(50.27) + (25.14)(3) 100.64 + 75.42 176.6 cm²

10 cm

20 cm

4 cm

3 cm

d) Estimate: 20 cm²

SACIRCLE = πr² SACIRCLE = π(2.2)² SACIRCLE = π(4.84)

SACIRCLE = 15.2 cm² C = 2πr C = 2π(2.2) C = 2(6.91) C = 13.82 cm 2(15.2) + (13.82)(12.4) 30.4 + 171.37

201.8 cm²

4. A cylinder has a radius of 10 cm and a height of 40 cm.

a) Calculate the surface area in terms of π.

SACIRCLE = πr² SACIRCLE = π(10)² SACIRCLE = 100π

C = 2πr C = 2π(10) C = 20π 2(100π) + (20π)(40) 200π + 800π 1000π

b) 1000π

3141.59 cm²

5. One cylinder has a radius of 3 cm and a height of 4 cm. Another cylinder has a

base radius of 4 cm and a height of 3 cm. Do you think their total surface areas

are equal? If not, which one do you think has the greater total surface area?

Explain.

No, I don’t think they will be the same. The second one will be bigger because it has a

greater radius, which means the circle is larger, making the overall cylinder larger.

2.2 cm

12.4 cm

10 cm

40 cm

6. a) Check your answer to exercise 5 by calculating the total surface area of each cylinder.

SACIRCLE1 = πr² SACIRCLE1 = π(3)² SACIRCLE1 = π9

SACIRCLE1 = 28.27 cm² C = 2πr C = 2π(3) C = 2(9.42) C = 18.84 cm 2(28.27) + (18.84)(4) 56.54 + 75.36

131.9 cm²

________________________________________________________________________

SACIRCLE2 = πr² SACIRCLE2 = π(4)² SACIRCLE2 = π16

SACIRCLE2 = 50.27 cm² C = 2πr C = 2π(4) C = 2(12.57) C = 25.14 cm 2(50.27) + (25.14)(3) 100.54 + 75.42

176 cm²

b) What is the ratio of the total surface areas?

The ratio is about 3:4

3 cm

4 cm

4 cm

3 cm

7. Estimate, and then calculate the total surface area of each cylinder.

a) Estimate: 600 cm²



595 cm²

b) Estimate: 1000 cm²



1098 cm²

c) Estimate: 600 cm²


SACIRCLE = 11.34 cm² C = 2πr C = 2π(1.9) C = 2(5.97) C = 11.94 cm 2(11.34) + (11.94)(45) 22.68 + 537.3

560 cm²

9. A square piece of cardboard measuring 20 cm on a side is used to form the curved surface of a cylinder. Circle cut from another piece of cardboard are used for the top and the bottom.

a) Calculate the radius of the cylinder to 2 decimal places.

C = 2πr

20 = 2πr 20/2 = πr 10 = πr 10/π = r 3.18 = r

b) SACIRCLE = πr² SACIRCLE = π(3.18)² SACIRCLE = π(10.11)

SACIRCLE = 31.76 cm² C = 2πr C = 2π(3.18) C = 2(10) C = 20 cm 2(31.76) + (20)(20) 63.52 + 400

464 cm²

20 cm

10. A juice can has a cardboard curved surface and two metal ends.

a) What area of cardboard is needed to make the can?

C = 2πr C = 2π(3.3) C = 2(10.37) C = 20.74 cm A = lw A = (20.74)(11.6)

A = 240.58 cm²

The area needed to make the cardboard surface 240.58 cm².

b) What is the area of each metal end? SACIRCLE = πr² SACIRCLE = π(3.3)² SACIRCLE = π(10.89)

SACIRCLE = 34.21 cm² The area of each metal end is 32.21 cm².

c) What is the total surface area of the juice can?

2(34.21) + 240.58 68.42 + 240.58

309 cm²

12. Newsprint is one of Canada’s major exports. It is shipped in cylindrical rolls. The rolls shown have a diameter of 102 cm and a length of 137 cm. What is the area of the outer wrapping of the roll?


SACIRCLE = 8171 cm² C = 2πr C = 2π(51) C = 2(160.22) C = 320.44 cm 2(8171) + (320.44)(137) 16342 + 43900.28

60242 cm²

The area of the outer wrapping is approximately 60242 cm².

(10.6) Volume of a Cylinder Pp.479-482 #1-5, 8a, 15-17

1. Estimate the area of each circle. Give answers to 1 decimal place.

a) a = πr²

a = π(12)²

a = π144

a = 452.4 cm²

b) a = πr²

a = π(0.4)²

a = π(0.16)

a = 0.5 cm²

c) a = πr²

a = π(4.5)²

a = π(20.25)

a = 63.6. cm²

d) a = πr²

a = π(6.5)²

a = π(42.25)

a = 132.7 cm²

2. Estimate, and then calculate the volume of each cylinder. Give the answers to 1

decimal place.

a) a = πr²

a = π(1)²

a = π(1)

a = 3.1 cm²

v = bh

v = (3.1)(2)

v = 6.2 𝑐𝑚3

12 cm

0.4 cm

9 cm

6.5 cm

1 cm

2 cm

b) a = πr²

a = π(10)²

a = π(100)

a = 314.2 cm²

v = bh

v = (314.2)(20)

v = 6284 𝑐𝑚3

c) a = πr²

a = π(4)²

a = π(16)

a = 50.27 cm²

v = bh

v = (50.27)(3)

v = 150.8 𝑐𝑚3

d) a = πr²

a = π(2.2)²

a = π(4.84)

a = 15.2 cm²

v = bh

v = (15.2)(12.4)

v = 188.5 𝑐𝑚3

3. One litre is often represented by a cube measuring 10 cm along each edge.

a) a = πr²

a = π(5)²

a = π(25)

a = 78.54 cm²

v = bh

v = (78.54)(10)

v = 785.4 𝑐𝑚3

2.2 cm

12.4 cm

4 cm

3 cm

10 cm

20 cm

b) What percent of the space in the cube is occupied by the cylinder? Give the

answers to 1 decimal place.

VCUBE = lwh

VCUBE = (10)(10)(10)

VCUBE = 1000 𝑐𝑚3

Therefore, it takes up 79% of the cube

= VCYLINDER__

VCUBE

=785.4

1000

= 79%

4. A can of paint is marked 978 mL. It has a base diameter of 10.4 cm and a height

of 12.5 cm. Calculate the volume of the can, in cubic centimetres. Does the result

confirm that the can’s capacity is 978 mL? Explain.

a = πr²

a = π(5.2)²

a = π(27.04)

a = 84.95 cm²

v = bh

v = (84.95)(12.5)

v = 1062 𝑐𝑚3

The result does not confirm that the can’s capacity is 978 mL, but there might be

empty space in the can because they won’t fill the can to the rim.

5.2 cm

12.5 cm

5. Calculate the volume of the cylinder formed if rectangle ABCD is rotated about:

a) Side AB

a = πr²

a = π(5)²

a = π(25)

a = 78.5 cm²

v = bh

v = (78.5)(2)

v = 157 𝑐𝑚3

b) Side BC

a = πr²

a = π(2)²

a = π(4)

a = 12.6 cm²

v = bh

v = (12.6)(5)

v = 63 𝑐𝑚3

8. a) Calculate the volume of the newsprint described in exercise 12 on page 475.

a = πr²

a = π(51)²

a = π(2601)

a = 8171 cm²

v = bh

v = (8171)(137)

v = 1119427 𝑐𝑚3

v ≈ 1.12 𝑚3

5.0 cm

2.0 cm

15. Determine the volume of each cylinder with the given dimensions. Give each answer to the nearest cubic centimetre. a) radius 5 cm; height 3.18 cm

a = πr²

a = π(5)²

a = π(25)

a = 78.54 cm²

v = bh

v = (78.54)(3.18)

v = 250 𝑐𝑚3

b) radius 7.5 cm; height 1.42 cm

a = πr²

a = π(7.5)²

a = π(56.25)

a = 176.71 cm²

v = bh

v = (176.71)(1.42)

v = 250 𝑐𝑚3

c) radius 3.99 cm; height 5 cm

a = πr²

a = π(3.99)²

a = π(15.92)

a = 50 cm²

v = bh

v = (50)(5)

v = 250 𝑐𝑚3

d) radius 3.26 cm; height 7.5 cm

a = πr²

a = π(3.26)²

a = π(10.62)

a = 33.36 cm²

v = bh

v = (33.36)(7.5)

v = 250 𝑐𝑚3

The volumes are all about 250𝑐𝑚3.

16. Calculate the volume of each object. Give the answers to 1 decimal place.

a) Volume of Box

V = lwh

V = (3)(6)(5)

V = 90 𝑐𝑚3

Volume of Cylinder

a = πr²

a = π(1)²

a = π(1)

a = 3.14 cm²

v = bh

v = (3.14)(3)

v = 9.4 𝑐𝑚3

Total Volume

90 + 9.4 = 99.4 𝑐𝑚3

b) Volume of Box

V = lwh

V = (12)(7)(2)

V = 168 𝑐𝑚3

Volume of Cylinder

a = πr²

a = π(0.5)²

a = π(0.25)

a = 0.79 cm²

v = bh

v = (0.79)(2)

v = 1.58 𝑐𝑚3

Total Volume

168 + 1.58 = 166.4 𝑐𝑚3

17. A machine bales hay in rolls 1.8 m in diameter and 1.2 m long.

a) What is the radius of each roll?

b) The radius of each roll is 0.9 m, or 90 cm.

a = πr²

a = π(0.9)²

a = π(0.81)

a = 2.54 cm²

v = bh

v = (2.54)(1.2)

v = 3.04 𝑚3

The volume of each hay roll is 3.04 𝑚3

(10.7) Surface Areas of a Pyramid & a Cone Pp.495-497 #4-12, 14, 15, 17

4. One cone has a base radius of 4 cm and a height of 3 cm. Another cone has a

base radius of 3 cm and a height of 4 cm. Do you think their total surface areas

are the same? If not, which one do you think has the greater total surface area?

Give a reason for your answer.

No, I do not think that the cones have an equal surface area. The cone that has a

greater radius than the height, i.e. the first cone, will have the greater surface

area. This is because the height of a cone is not included in the formula for

surface area, only the slant counts.

5. a) Check your answer to exercise 4 by calculating the total surface area of each

cone.

Cone 1

SA = πrs + πr² s² = (b/2)² + (h)²

SA = π(4)(3.6)+ π(4)² s² = (4/2)² + (3)²

SA = π(14.4) + π16 s² = (2)² + (3)²

SA = 45.2 + 50.3 s² = 4 + 9

SA = 95.5 cm² s² = 13

S = 3.6

Cone 2

SA = πrs + πr² s² = (b/2)² + (h)²

SA = π(3)(4.27)+ π(3)² s² = (3/2)² + (4)²

SA = π(12.81) + π9 s² = (1.5)² + 16

SA = 40.24 + 28.27 s² = 2.25 + 16

SA = 68.51 cm² s² = 18.25

S = 4.27

b) What is the ratio of the total surface areas?

About 2:3

6. Which net folds to make a pyramid? Explain how you know.

a) , because the top of b) is flat, and wouldn’t be able to close up entirely, and

would have a big square hole at the top.

7. Determine the surface area of each pyramid. Give the answer to 1 decimal place.

a) SA = 2bs + b² b) SA = 2bs + b²

SA = 2(4)(3) + (4)² SA = 2(5)(6) + (5)²

SA = 24 + 16 SA = 60 + 25

SA = 40 cm² SA = 85 cm²

c) SA = 2bs + b²

SA = 2(10)(8) + (10)²

SA = 160 + 100

SA = 260 cm²

8. Determine the area of the curved surface of each cone. Give the answer to 1

decimal place where necessary.

a) SA = πrs + πr² a² + b² = c²

SA = π(5)(13) + π(5)² 5² + 12² = c²

SA = 204 + 78.5 25 + 144 = c²

SA = 282.5 m² 169 = c²

13 = c

b) SA = πrs + πr² a² + b² = c²

SA = π(4.8)(2)+ π(4.8)² (4.8)² + (4.8)² = c²

SA = 30 + 72.4 23 + 23 = c²

SA = 102.4 m² 46 = c²

6.7 = c

c) SA = πrs + πr² a² + b² = c²

SA = π(20)(54)+ π(20)² 20² + 50² = c²

SA = 3392 + 1256 400 + 2500 = c²

SA = 1256.6 2900 = c²

54 = c²

9. Determine the surface area of each tetrahedron to 1 decimal place.

a) Bh/2

(3)(2.6)/2

3.9

3.9 x 4

15.6 cm²

b) Bh/2

(4.5)(3.9)/2

8.8

8.8 x 4

35.2 cm²

10. Determine the total surface area of each cone. Give the answers to 1 decimal

place.

a) SA = πrs + πr² a² + b² = c²

SA = π(2.5)(4.3)+ π(2.5)² 2.5² + 3.5² = c²

SA = 33.8 + 19.6 6.25 + 12.25 = c²

SA = 53.4 18.5 = c²

4.3 = c

b) SA = πrs + πr² a² + b² = c²

SA = π(15)(57)+ π(15)² 15² + 55² = c²

SA = 3392.9 + 706.8 225 + 3025 = c²

SA = 4100 3250 = c²

57 = c

c) SA = πrs + πr² a² + b² = c²

SA = π(54)(90)+ π(54)² 54² + 72² = c²

SA = 15268 + 9161 2916 + 5184 = c²

SA = 24428 8100 = c²

90 = c

11. Determine the surface area of each pyramid.

a) b² + 2bh

(18)² + 2(18)(12)

324 + 432

756 cm²

b) (l)(w) + (bh1) + (bh2)

(25)(18) + (25)(33) + (18)(34)

450 + 825 +612

1887

c) b² + 2bh

(2.1)² + 2(2.1)(3.4)

4.41 + 14.28

18.7 cm²

12. Draw a net for each described pyramid. Determine its surface area to 1 decimal

place.

a) A square base with side length 5.5 cm, triangular faces with height 8.4 cm

b² + 2bh

(5.5)² + 2(5.5)(8.4)

30.25 + 92.4

122.7 cm²

b) a rectangular base 12 cm by 7.5 cm, the triangular face on the longer side has

height 14.25 cm, the triangular face on the shorter side has height 15 cm

(l)(w) + (bh1) + (bh2)

(12)(7.5) + (12)(14.25) + (7.5)(15)

90 + 171 +112.5

373.5

c) Each face is an equilateral triangle with side lengths 4.1 cm

a² + b² = c²

2.05² + b² = 4.1²

4.2 + b² = 16.8

12.6 = b²

3.5 = c

b² + 2bh

(4.1)² + 2(4.1)(3.5)

16.8 + 28.7

45.5 m²

14. The Louvre is a famous art gallery in Paris, France. It is housed in a historic

palace and was opened in 1793. In 1989, it was expanded. One addition is a large

glass square pyramid that covers this main entrance. Each wall of this pyramid...

4(bh/2) 4(35.4)(27.9)/2

1975.32 m²

15. A carpenter is building two playhouses for a daycare center. The roof of each playhouse is a pyramid. One roof has a square base... a) How much wood is needed for each playhouse?

House 1 4(3.8 x 1) 15.2 2bh 2(3.8)(4.2) 32 32 + 15 = 47 m²

House 2 2(1.9)(1) + 2(2.6)(1) 9 bh (1.9)(2.4) 4.56 Bh (2.6)(2.2) 5.72 9 + 4.56 + 5.72 = 19.3 m²

b) The first playhouse needs more wood.

c) House 1 47 x 15.25 $716.25 House 2 5.72 x 15.25 $87.23

17. For Halloween, a clown’s hat made by stapling together the straight edges of a quarter of a circle with radius 30.0 cm. a) What is the radius of the base of the hat?

7.5 cm

b) How high is the hat? 29 ~ 30 cm, because the radius of the paper that made the hat is 30 cm

c) (πr²/2)/4

[π(30)²/2]/4 1413.7/4

353.4 cm²

(10.8) Volumes of a Pyramid & a Cone Pp.503-505 #1-9, 12, 14 – 18

1. How does the formula for the volume of a rectangular pyramid compare

with the formula for the volume of a cone?

The formula for the volume of a rectangular pyramid and the formula for the volume of a cone are basically the same, except the base areas are different.

2. The diagrams show three views of the same rectangular prism. How do the volumes of the three pyramids compare? Explain your answer.

All the volumes will be the same, because they are all in the same rectangular prism, making all the dimensions they use the same.

3. Determine the volume of each pyramid. a) V = 1/3lwh

V = 1/3(4)(4)(4.5) V = 24 𝑐𝑚3

b) V = 1/3lwh

V = 1/3(7)(6)(9) V = 126 𝑐𝑚3

c) V = 1/3lwh V = 1/3(9.5)(9.5)(12.2) V = 367 𝑐𝑚3

4. A cylinder has volume 96𝑐𝑚3. What is the volume of a cone that fits inside the cylinder? Since 3 cones fit into one cylinder, then if we divide the volume by three, we will

get the volume of 1 cone.

32𝑐𝑚3

5. A cone has volume 54𝑐𝑚3. What is the volume of a cylinder that just holds the

cone?

Since 3 cones fit into one cylinder, then if we divide the multiply by three, we will

get the volume of 1 cone.

162 𝑐𝑚3

6. One litre is often represented by a cube measuring 10 cm along each edge.

a) Calculate the volumes of a cylinder and a cone that fits inside the cube.

Cylinder

V = 1/3lwh

V = 1/3(10)(10)(10)

V = 333.3𝑐𝑚3

Cone

V = 1/3πr²h

V = 1/3π(5)²(10)

V = 261.8𝑐𝑚3

b) Cylinder = 33.3%

Cone = 26.2%

7. A cone and a cylinder have the same base. Supposed they also have the same

volume. How are their heights related? Explain.

The cone’s height will be 3 times the height of the cylinder, because their

volumes have a 3 times difference.

8. Determine the volume of each cone.

a) V = 1/3πr²h

V = 1/3π(5)²(9)

V = 234.6𝑐𝑚3

b) V = 1/3πr²h

V = 1/3π(4.3)²(9.7)

V = 187.8𝑐𝑚3

c) V = 1/3πr²h

V = 1/3π(1.3)²(1.0)

V = 1.77𝑐𝑚3

d) V = 1/3πr²h

V = 1/3π(0.7)²(3.5)

V = 1.78𝑐𝑚3

9. Determine the volume of each pyramid.

a) V = 1/3lwh

V = 1/3(12)(12)(4)

V = 192𝑐𝑚3

b) V = 1/3lwh

V = 1/3(24)(12)(4)

V = 384𝑐𝑚3

c) V = 1/3lwh

V = 1/3(12)(12)(8)

V = 384𝑐𝑚3

d) V = 1/3lwh

V = 1/3(12)(12)(16)

V = 768𝑐𝑚3

12. The Louvre is a famous art gallery in Paris, France. See exercise 14, page 496 for

information about the Louvre. Calculate the volume of the pyramid over the

main entrance to the Louvre.

V = 1/3lwh

V = 1/3(35.4)(35.4)(27.9)

V = 11654.388𝑚3

14. Determine the volume of each cone, to 2 decimal places.

a) V = 1/3πr²h

V = 1/3π(1.6)²(3.5)

V = 9.4𝑐𝑚3

b) V = 1/3πr²h

V = 1/3π(0.35)²(1.8)

V = 0.23𝑚3

c) V = 1/3πr²h

V = 1/3π(8)²(27)

V = 1809.6𝑐𝑚3

d) V = 1/3πr²h

V = 1/3π(42)²(57)

V = 105293.6𝑐𝑚3

15. A cone shaped funnel has radius 5.7 cm and height 4.3 cm. How much can the

funnel hold? Express your answer to 1 decimal place.

V = 1/3πr²h

V = 1/3π(5.7)²(4.3)

V = 146.3𝑐𝑚3

16. A farmer stores feed in a cone-shaped storage unit. The storage unit has base

diameter 14.3 m and height 27.4 m. How much feed can this unit store?

V = 1/3πr²h

V = 1/3π(7.15)²(27.4)

V = 1466.87𝑚3

17. An engineer is designing a cone shaped storage unit to hold 5000𝑚3of sand. The

unit has a base radius 15 m. What is its height, to 1 decimal place?

V = 1/3πr²h

5000 = 1/3π(15)²h

5000 = 15.7h

5000/15 = h

h = 333.3

18. Cone A has base radius 25 cm and height 10 m. Cone B has height 25 cm and

base radius 10 m. Which cone has the greater volume?

Cone A will have the greater volume, because the height comes more into play

when calculating the volume.

(10.9) Surface Area & Volume of a Sphere Pp.510-511 #1-3, 6, 7, 10 – 13

1. Calculate the surface area and volume of each sphere. Give the volumes to the

nearest whole unit. The radius is given. a) SA = 4πr²

SA = 4π(10)² SA = 1256.6 cm² V = 4/3π𝑟3 V = 4/3π(10)3 V = 4188𝑐𝑚3

b) SA = 4πr²

SA = 4π(5)² SA = 314.2 cm² V = 4/3π𝑟3 V = 4/3π(5)3 V = 523.6𝑐𝑚3

c) SA = 4πr² SA = 4π(15.2)² SA = 2903.3cm² V = 4/3π𝑟3 V = 4/3π(15.2)3 V = 14710.2𝑐𝑚3

2. Calculate the surface area and the volume of each ball. a) Baseball

SA = 4πr² SA = 4π(3.7)² SA = 172 cm² V = 4/3π𝑟3 V = 4/3π(3.7)3 V = 212.17 𝑐𝑚3

b) Golf SA = 4πr² SA = 4π(2.15)² SA = 58 cm² V = 4/3π𝑟3 V = 4/3π(2.15)3 V = 41.6 𝑐𝑚3

c) Table tennis SA = 4πr² SA = 4π(1.85)² SA = 43 cm² V = 4/3π𝑟3 V = 4/3π(1.85)3 V = 26.5 𝑐𝑚3

d) Volleyball SA = 4πr² SA = 4π(10.45)² SA = 1372.3 cm² V = 4/3π𝑟3 V = 4/3π(10.45)3 V = 4780.1𝑐𝑚3

3. In this self-portrait, the Dutch artist, M.C. Escher, is holding a reflecting sphere. a) Estimate the diameter of the sphere in this picture.

About 10 cm

b) Use your estimate. Calculate the surface area and the volume of the sphere. SA = 4πr² SA = 4π(5)² SA = 314.2 cm² V = 4/3π𝑟3 V = 4/3π(5)3 V = 523.6𝑐𝑚3

6. The mean radius of the moon is approximately 1740 km. Determine the surface area and volume of the moon.

SA = 4πr² SA = 4π(1740)² SA = 38045943.67 cm² V = 4/3π𝑟3 V = 4/3π(1740)3 V = 12681981.22𝑐𝑚3

7. A spherical balloon is blown p from a diameter of 20 cm to one of 60 cm. By how

many times has its: a) Surface area increased?

SA = 4πr² SA = 4π(10)² SA = 1256.6 cm² V = 4/3π𝑟3 V = 4/3π(10)3 V = 4188𝑐𝑚3 SA = 4πr² SA = 4π(30)² SA = 11309 cm² V = 4/3π𝑟3 V = 4/3π(30)3 V = 113097.3𝑐𝑚3 It has increased by 9 times.

b) Volume increased? It has increased 27 times.

10. Calculate the radius of the sphere with each volume. a) 45.7 𝑚3

45.7 = 4/3π𝑟3

r = 2.2

b) 23.8 𝑚3 23.8 = 4/3π𝑟3

r = 1.8

c) 1356 𝑚3 1356 = 4/3π𝑟3

r = 6.9

11. A sphere just fits inside a cube with edges of length 10 cm. Calculate the surface area and the volume of the sphere.

SA = 4πr² SA = 4π(5)² SA = 314.2 cm² V = 4/3π𝑟3 V = 4/3π(5)3 V = 523.6𝑐𝑚3

12. Repeat exercise 11 for a cube with edges of length x cm.

SA = 4πr² SA = 4π(x/2)² SA = πx²

V = 4/3π𝑟3 V = 4/3π(𝑥/2)3 V = (2πx²)/6

13. A basketball has a circumference of 75 cm. Determine: a) Its radius:

C = 2πr 75 = 2πr r = 12 cm

b) It’s surface area SA = 4πr² SA = 4π(12)² SA = 1809.6 cm²

c) V = 4/3π𝑟3 V = 4/3π(12)3 V = 7238.2𝑐𝑚3

gr.9 pre-ib unit 10 measurement

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