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10

10A Pythagoras’ theorem in two dimensions 10B Pythagoras’ theorem in three dimensions 10C Perimeter and area 10D Total surface area (TSA) 10E Volume 10F Capacity 10G Similar figures 10H Similar triangles 10I Symmetry

Shape and measurement arEaS oF Study

• Mensuration (angle, length, boundary, area, surface area and volume) • Pythagoras’ theorem in two dimensions and simple examples in three dimensions

• Similarity and symmetry in two dimensions and applications to maps, art, tessellations, and plans • Similarity in three dimensions and application to scale models • Tests for similarity and symmetry eBook plus

10a

Pythagoras’ theorem in two dimensions

Digital doc

10 Quick Questions

Pythagoras was a Greek philosopher and mathematician who is usually credited with formulating the following theorem relating the sides of right-angled triangles: In any right-angled triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides. Hypotenuse (Length of hypotenuse)2 = (length of side 1)2 + (length of side 2)2 a c It is conventional to use the letters a and b for labelling the sides and c for labelling the hypotenuse of the right-angled triangle. b Pythagoras’ theorem is then written as c2 = a2 + b2.

WorkEd ExaMPlE 1

Find the value of the pronumeral to 1 decimal place.

A y 25

think

456

B 12

WritE

1

Identify the right-angled triangle which contains the unknown side.

y (or AB) can be found from ∆ABC.

2

In ∆ABC only BC is known; therefore AC should be found first.

Need to find AC.

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

D

17

C

3

Use ∆ADC to find AC using Pythagoras’ theorem. Note that there is no need to evaluate 914 , as we will need to square the value of AC in the next step.

∆ADC: (AC)2 = (AD)2 + (DC)2 = 252 + 172 = 625 + 289 = 914 AC = 914

4

With the added information, find AB (y).

∆ABC: (AC)2 = (AB)2 + (CB)2 ( 914 )2 = (AB)2 + 122 914 = (AB)2 + 144 (AB)2 = 914 − 144 (AB)2 = 770 AB = 770 = 27.748 873 85

5

Replace AB with the pronumeral in question and round to the required number of decimal places.

y ≈ 27.7

rEMEMBEr

Pythagoras’ theorem: In any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Hypotenuse c

a

c2 = a2 + b2 ExErCiSE

10a eBook plus

b

Pythagoras’ theorem in two dimensions 1 For each of the following triangles, find the length of the unknown side correct to 1 decimal place. a

b

Digital docs

a

SkillSHEET 10.1

12

Pythagoras’ theorem

10

Spreadsheet 102

c

4

5 c

b

19

9

Pythagoras’ theorem

d

4.2

6.3 d

e

f

e

16.2

f

14.8 6.7

2 What is the length of a diagonal of: a a square with side lengths of 62 cm (correct to 1 decimal place) b a rectangle with dimensions of 0.8 m by 37 cm? (Give your answer in metres correct to 2 decimal places.) 3 A rectangle is twice as long as it is wide. If its perimeter is 186 cm, then the length of the diagonal is closest to: A 68 cm B 69 cm C 70 cm D 71 cm E 72 cm

Chapter 10

Shape and measurement

457

  4 For a ladder to be stable enough to be climbed, the base needs to be at least 60 cm out from the base of the wall. If the top of a 2.2-m ladder is to reach up to a window 2 m from the ground, would the ladder be classified as stable?   5   WE1  Find the value of the pronumeral to 1 decimal place in each of the following shapes. b c a 32.8 1.8 m

3x

2.5 m 1.2 m

a

1.4 m

16

b

18.2 x 11.2

Home   6 A long-distance runner wants to complete a 30-km run. She has completed 18 km and is at y Park crossing the marked point. If she took the short cut across Mary 4.5 km St x the park, would she reach her goal or should she 2 km 5 km 5 km go along the streets? Justify your answer with Runner Hampton St mathematical evidence.   7 In technology class you have been given the following design brief. Frame Construct a small wooden stool which is to be made in three parts. Part 1 is a frame, part 2 is a platform top, while part 3 comprises supports for Wooden 30 cm supports the corners of the frame. The supports are to divide the top section into 42 cm three even parts, while they are to reach two-thirds of the way down the legs of the frame. A plan of part 1 and part 3 has been drawn for you. a How long do the supports need to be to meet the stated criteria? Give your answer to the nearest millimetre. Platform A platform will sit on top of the frame (as shown at right) so that it has 1 an overhang at each end of 14 of the size of the frame’s top. b How long is the platform? c The timber comes in only one specific width, while the length can be cut. What length of timber (to the nearest centimetre) is needed to y construct this stool?   8 Once you have finished the stool you are then required to build this y shelving unit. It is designed to hold a sound system and CDs. 180 cm The shelves are to be equally spaced within the unit. y a Find the distance indicated by the pronumeral (  y) to the nearest y millimetre. b Calculate the lengths (to the nearest centimetre) of the other three 1m shelves in ascending order. c Overall, what will it cost to construct this unit if the timber required to make the sides and shelves cost $12.00 per metre and the back board (already cut) will cost $55.00?   9 The three different courses (or legs) for the swim, cycle and run stages of a triathlon are set in a triangular format. The swim The swim leg has the competitors heading out due north to buoy 1, turning due east to the second buoy and then heading straight back to the starting position. a Draw a diagram of this situation. b Calculate the total length of the swim if the first two legs are 300 m and 400 m respectively. The cycle The athletes leave the transition area and ride in a southerly direction for 12 km. They then turn due west and ride for 16 km. The last leg has the cyclists heading straight back to the transition area. c Competitor 42’s cycle develops a puncture at the start of the last leg. How far does competitor 42 have to walk to get back to the transition area?

458

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

The run The course designer told the road crew (who will place the markers out for the three legs of the run) that the overall run length is to be 7500 m and that the distances of each leg were those of a Pythagorean triple. [A Pythagorean triple (or triad) is a set of three numbers that satisfy Pythagoras’ theorem.] He gave them a hint that he had calculated the distances by increasing a basic triple by a factor of 250. d What are the distances of the three legs in the run? e Show by calculation that the values you found in part d are a Pythagorean triple. f Overall, how far do the triathletes swim, cycle and run?

10B

Pythagoras’ theorem in three dimensions So far, we have used Pythagoras’ theorem in situations where the right-angled triangle is drawn in 2 dimensions. Three-dimensional objects can have right-angled triangles within them that can be re-drawn in 2 dimensions so that Pythagoras’ theorem can be used to find the missing measurement.

Worked Example 2

Find the height labelled x. Round your answer to 1 decimal place. think 1

2

State the length that needs to be found.

Identify a right-angled triangle that contains AB and redraw it in two dimensions.

E

write B

The unknown length is AB or x.

x A

A

8 12

F

D 17 C

12

x B

C

8

=

a2

+

3

Write Pythagoras’ formula.

c2

b2

4

Identify the values of a, b and c.

c = 12, a = 8, b = x

5

Substitute values of a, b and c into the formula.

6

On the Main screen, tap: •  Action •  Advanced •  solve Complete the entry line as: solve(122 = 82 + x2, x) Then press E.

122 = 82 + x2

Chapter 10  Shape and measurement

459

7

Given that x is the side length of a triangle then its value is positive.

Solving 122 = 82 + x2 for x, gives x = −8.94427 or x = 8.94427. Since x > 0, x = 8.94427.

8

Write the answer, correct to 1 decimal place.

The height labelled x is 8.9 units.

Sometimes the right-angled triangle containing the unknown length also has other dimensions missing. In such cases you will first need to find other right-angled triangles within the figure that will allow you to calculate those missing values using Pythagoras’ theorem.

Worked Example 3

Find the length of the side labelled x. Round your answer to 1 decimal place. think 1

write x

(a) The unknown length x (EH) can be found from triangle EHC. (b) In this triangle, the length of HC is missing as well so it has to be found first. The length of HC is half the length of AC. (c) In turn, AC can be found from triangle ABC. Redraw this triangle in 2 dimensions and include all measurements.

D A C H A

20

B

c2 = a2 + b2 c = AC; a = b = 20 (AC)2 = 202 + 202 = 400 + 400 = 800 AC = 800 = 400 × 2 = 20 2

Use Pythagoras’ theorem to find AC. Leave the answer in surd form.

3

Find the length of HC by halving the length of AC. (There is no need to evaluate 20 2 , as we will have to square this value in the steps that follow.)

HC = 2 AC

Redraw the triangle that contains the unknown length x in 2 dimensions (triangle EHC). Write the measurements, including the length of HC found in the previous step.

E

1

1

= 2 × 20 2 = 10 2

x H

460

20

2

4

E

45

10 2

C

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

20 H B

45 C

5

c2 = a2 + b2 c = 45, a = x, b = 10 2 452 = x2 + (10 2)2 2025 = x2 + 200 x2 = 2025 − 200 = 1825 x = 1825 x ≈ 42.7

Use Pythagoras’ theorem to find x. (Round your answer to 1 decimal place.)

Solving 3-dimensional problems (especially worded ones), is made easier by following the algorithm outlined below. 1. If a diagram of the situation is not supplied, draw one. 2. Label all vertices in the diagram and all known lengths (dimensions). 3. Identify the length that needs to be found. 4. Identify the triangle containing the unknown length. 5. Redraw this triangle in 2 dimensions. If the two other lengths are known, proceed to step 6; if not, work them out first. 6. Use Pythagoras’ theorem to find the unknown length. 7. If the question is presented in words, write an answer sentence, rounding the value appropriately for the situation.

rEMEMBEr

1. Use the seven steps as your guide to solving 3-dimensional problems. 2. Unless stated in the problem, do not round your calculations until the very last value is calculated and then give an answer with the appropriate number of decimal places. 3. If the triangle containing the unknown length has other measurements missing, find these missing values first from other triangles within the figure.

ExErCiSE

10B

Pythagoras’ theorem in three dimensions 1 WE2 In each of the following, find the length of the line labelled x. Round your answers to 1 decimal place. a b E F E

A B

G x 20

F

C

15

B

A

D 10

x

H G C

c

15

D

d

A

B

C 12

16

Digital doc

Spreadsheet 102 Pythagoras’ theorem

24

8 H A 28.2

62.6 x

eBook plus

B

x

C

D

Chapter 10

Shape and measurement

461

  2   WE3  In each of the following, find the length of the line labelled x. Round your answers to 1 decimal place. a

G A 8

x

E

C

H

16

b

c

B x D

12

5

F

20

x

22

10 12

e

d

f

30

x

x

6

9 6

15

x

15

11

12

  3 The diagram at right shows a rectangular metal pencil case. Find: a the length of the longest pen that can be placed flat on the bottom of the pencil case b the length of the longest pen that can be placed in this case.

10 cm 20 cm

15 cm

  4 Rock concerts usually require some form of scaffolding to support screens, lighting, amplifiers and speakers. The lengths E 8m and overall amount need to be pre-ordered for each venue.   Before each concert a manager would be responsible for assessing 12 m G F A the arrangement, re-designing B the structure if required and preordering the lengths.   The structure at right is designed D 2m I 3m to hold a large screen. Some of 4m H C the lengths have been manually measured and others can be determined from these measurements. However, there are a select number that need to be calculated. Your task is to calculate the missing lengths stated below so that the scaffolding can be ordered for the next big night. Round off all dimensions to 1 decimal place as the scaffolding only comes in lengths to the nearest centimetre. a  Find the following lengths. i AB ii CD iii DE iv EF v GH vi GI b The price of the scaffolding is $5.00 per metre. i How many metres of scaffolding are required for this structure? Windows E B ii What will be the total cost of the screen F 2m support?   5 Factories have been designed so that a set of windows can be placed at certain intervals along the factory ceiling allowing natural light to enter. The trusses (roof supports) used to construct this system are called saw tooth trusses.

462

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

D

1st

A

2nd Sections 30 m

C

3rd

6m

10 m

  Rounding values as you go to 1 decimal place, calculate the following dimensions in the factory structure: a AB and AC in the third section    b  DE and DF in the second section.   6 The bottom part of a slide at a water park is constructed upon a series of different sized triangular blocks (prisms). Using the dimensions given in the diagram at right, calculate the length AB of this part of the slide. Round all values as you go to 1 decimal place.

1.8 m

5m

7.5 m

3.8 m 7.4 m

B

8.1 m

Perimeter and area The table below shows the formulas for finding the area and perimeter of some common shapes. Shape

Area A=

Square

Perimeter P = 4L

L2

L

Rectangle

L

A=L×W

P = 2(L + W)

A=b×h where the height measurement must be at a right angle to the base measurement.

P = sum of all sides

W

Parallelogram  h b

Trapezium

a h b x

–

P = sum of all sides

1

P = sum of all sides

1

P = sum of all sides

A=2 x×y

–

Rhombus

1

A = 2 (a + b)h where the height measurement must be at a right angle to the base measurement.

–

y –

10C

A

Triangle  h

A = 2 bh where the height measurement must be at a right angle to the base measurement.

h

b

b h

b (continued)

Chapter 10  Shape and measurement

463

Shape

Area

Triangle

c

b a

Perimeter

A = s(s − a)(s − b)(s − c) 1 where s = 2 (a + b + c) (Use when height measurement is unknown.) A = π r2

Circle r

P=a+b+c

C = π d or C = 2π r

d

Worked Example 4

Calculate a the area and b the perimeter of this shape. think a

1

7m

8m

write

Identify the shape (in this case it is a triangle with no height measurement) and write the appropriate formula for the area.

a A=

s(s − a)(s − b)(s − c)

10 m

1

where s = 2 (a + b + c)

2

Identify the values of the pronumerals.

a = 7, b = 8, c = 10

3

To find s, substitute a, b and c values into the formula and simplify.

s = 2 (7 + 8 + 10)

1

1

= 2 × 25 = 12.5

4

Substitute the values of a, b, c, and s into the formula for the area.

5

Simplify. (a)  Evaluate the brackets first.

1 2

A = 12.5 × 5.5 × 4.5 × 2.5

(b)  Multiply the values together.

= 773.4375

(c)  Take the square root.

= 27.810 744 33

(Round the answer to 1 decimal place and include the units.) b

A = 12.5(12.5 − 7)(12.5 − 8)(12.5 − 10)

Find the perimeter by adding all the side measurements. Write the answer, including the appropriate units.

A = 27.8 m2 b P = 7 + 8 + 10

= 25

P = 25 m

Composite figures The term composite means ‘made up of distinct parts’. Composite figures in geometry are figures comprising a number of distinct shapes. Depending upon the composite figure, to find the overall area or perimeter you may need to add these individual shapes or subtract one from another.

464

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

A1 A2

For example, the composite figure in the diagram at the bottom of page 464 has been formed using a semicircle and a square. The area of this shape can be found as follows: Area of total figure = Area of a semicircle (A1) + Area of a square (A2) When finding the area or perimeter of a composite figure, follow the steps given below. 1. Identify the basic shapes that make up the total figure and number them. 2. Write the expression for the total area/perimeter in terms of the individual shapes. 3. Calculate the area/perimeter of each individual shape. 4. Add or subtract areas or dimensions to find the total area/perimeter of the given shape.

WorkEd ExaMPlE 5

ulu

An

Circles are said to be concentric if they have the same centre point. The area between the two concentric circles is referred to as an annulus. Area of annulus = area of larger circle − area of smaller circle = π R2 − π r2 = π (R2 − r2) where R = radius of the large circle r = radius of the small circle.

n

area of an annulus s

r R

eBook plus

In one full revolution, the 6-cm-long minute hand of a clock would sweep out a Tutorial larger circle than the 3-cm-long hour hand. What is the difference in the area they int-0904 cover to the nearest square centimetre? Worked example 5 think 1

The area required is the annulus. Write the appropriate formula.

WritE

A = π (R2 − r2) 6 cm

2

Identify the value of R (radius of the larger circle) R = 6, r = 3 and the value of r (radius of the smaller circle).

3

Substitute the values of the pronumerals into the formula.

4

Evaluate. On the Main screen, complete the entry line as: π (62 − 32) Then press E.

3 cm

A = π (62 − 32)

A = 84.823 5

Write an answer sentence with the value rounded to the nearest square centimetre.

The difference in area covered by the two hands is approximately 85 cm2.

Chapter 10

Shape and measurement

465

area of a sector and arc length A sector of a circle can be thought of as a wedgeshaped slice of pie. The area of the sector can be determined by finding the fraction of the whole circle it represents. The angle of the sector removed is divided by the total angle of a circle (360°) and then multiplied by the total area of the circle. θ × π r2 , Area of sector = 360° where θ is the angle of the sector and r is the radius of the circle.

O

Q A

Angle of sector B

Sector of circle WorkEd ExaMPlE 6

A 10-cm-long minute hand moving from the number 12 to the number 4 position sweeps out a sector. What is the area of this sector?

12

1 2

think

WritE

3

1

Write the formula for the area of the sector.

Area of sector =

2

Identify the value of the radius.

r = 10 cm

3

Calculate the angle of the sector: The angle between consecutive numbers on a clock = 360° ÷ 12 = 30°. From 12 to 4 there are four intervals between the numbers. So to find the angle of a sector, multiply 30° by 4.

4 5

θ × πr2 360°

θ = 30° × 4 = 120° 120

Substitute the values of r and θ into the formula and evaluate.

Area of sector = 360 × π × 102

Write an answer sentence with the number rounded appropriately and units given.

The minute hand as it rotates through an angle of 120° sweeps through an area of 104.7 cm2.

= 104.719 755 1

The length of the circumference between the two points (A and B) of the sector shown is known as the arc. To be specific, a minor arc is formed if the angle of the sector is less than 180°; a major arc is formed if the angle of the sector is greater than 180°. Finding the arc length of a circle is a similar procedure to finding the area of a sector. We must determine what fraction of the total circumference the arc length represents. The angle of the sector is divided by the total angle of a circle (360°) and then multiplied by the total circumference of the circle.

466

10 cm

4

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

y 280n 80n A B x x  Minor arc y  Major arc

Arc length = or

θ ×πd 360°

θ × 2π r 360° where θ is the angle of a sector, d is the diameter of a circle r is the radius of a circle. Arc length =

Worked Example 7

What distance did the tip end of the 10-cm minute hand travel when it moved from pointing to the number 12 to pointing to the number 4?

12

1 2

think 1 2

write

Since the radius is known, write the formula for arc length involving the radius. State the value of r.

θ Arc length = × 2π r 360°

3

10 cm

4

r = 10

3

Find the size of the angle of a sector. (a) The angle between each number on a clock = 30°. (b) There are 4 intervals between the numbers; θ = 30° × 4 therefore the angle of the sector can be found = 120° by multiplying 30° by 4.

4

Substitute values of r and θ into the formula and evaluate.

5

Write an answer sentence rounding the value appropriately and writing in the units.

120° × 2 × π × 10 360° = 20.943 951 02 The tip of the minute hand travelled 20.9 cm. Arc length =

REMEMBER

1. The perimeter is the distance around a closed 2-dimensional shape. 2. An area is the space enclosed by the boundaries of a 2-dimensional shape. 3. The formula for the area of a parallelogram, triangle and trapezium require the height to be perpendicular to the base measurement. 4. To find the area/perimeter of composite figures, the area/perimeter of individual shapes that form the figure must be calculated first. 5. An annulus is the area between two concentric circles. Area of annulus: A = π (R2 − r2), where R is the radius of the larger circle and r is the radius of the smaller circle. 6. The area of a sector is given by the area between two radii of the circle. θ Area of a sector = × πr2 360° and the arc length of a sector is given by the length of the circumference between two radii of the circle:

θ × 2π r 360° θ = × πr, 180° where θ is the angle of a sector and r is the radius of a circle. Arc length =

Chapter 10  Shape and measurement

467

10C eBook plus Digital doc

SkillSHEET 10.2

Perimeter and area 1 WE4 Find: i the area and i i the perimeter of the following shapes, to 2 decimal places. a

Conversion of units — length

b

14 mm

140 cm 122 cm –

12 cm

–

5 cm 9 cm

–

20 mm

c

–

ExErCiSE

2m

d

e 16 cm

5.5 m

4 cm 18 cm

1.5 m 2.5 m

2 Calculate i the area and i i the perimeter of the following shapes. Give your answers to 2 decimal places. a

b

10 cm

c

15 cm 25 cm

13 cm

10 m

30 cm eBook plus Digital doc

SkillSHEET 10.3 Area and perimeter of composite shapes

3 MC Examine the diagram at right. a The circles cover an area of approximately: A 402 cm2 B 201 cm2 2 D 805 cm E 603 cm2

6 cm 32 cm

C 804 cm2

b The shaded area is approximately: A 219 cm2 B 421 cm2 C 622 cm2 D 823 cm2 E 220 cm2 c A metal manufacturer is able to cut only four discs from every sheet of metal. What percentage of metal is wasted? A 80% B 21% C 22% D 41% E 61% 4 a A guard dog inside a used car salesyard is tied to a corner post of the fence surrounding the yard. The fence sides meet at a right angle and the dog is on a rope 1.2 m long. i Draw a diagram of this situation showing the area accessible to the dog. ii To 1 decimal place, how much area does the dog have in which to exercise? b One night the dog is moved to the outside corner of a small rectangular building measuring 2 m × 6 m. The dog’s rope has been lengthened to 3 m. i Draw a diagram of this situation showing the area accessible to the dog. ii Calculate the area available for the dog for exercise. Give your answer to the nearest square metre. c If the owner ties the dog back to the fence post and he wants the dog to have as much room to exercise in as it did when it was tied to the building, what length of rope (to the nearest centimetre) does he need to purchase? Geometry is used extensively in design. Questions 5 –10 explore the use of geometrical shapes in the design of country flags, company logos, and some commonly used signs. 5 The flag of Japan is a red circle on a white background. Calculate (to 1 decimal place) the radius of the circle, if the area of the circle must be 20% of the total flag area.

2m 3m

468

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

6 The flag of Israel is blue and white (see figure 1). The star is made from two triangularformations (figure 2) with the dimensions shown. The triangles that form the star overlap in 6 places. The total area of overlap is equivalent to 10% of their combined (blue) area. Without rounding off any values until the final answer (to 1 decimal place), find out what percentage of the flag is blue. 5.5 m 0.6 m 4m

1m 1.5 m

Figure 1

Figure 2

cm

7 The Commonwealth Bank logo is made up of a yellow square with a black trapezium overlaid at one corner. Calculate, to the nearest whole number, what percentage of the overall design is black. 1 cm

2.5

3.0 cm

8 The National Australia Bank logo is a red star with two white strips. The star is made up of 14 equilateral triangles. a What geometrical shape do the white strips represent? b Calculate (to 1 decimal place) the area of the white stripes in the diagram given. c Copy the outline of the star into your workbook. By construction, show how 14 equilateral triangles would make up this shape. d Calculate (to 1 decimal place) the red area of the star. 0.5 cm 1.5 cm

1.5 cm

9 WE5 The collectable plate shown at right is 22 cm in diameter and has a golden 0.5-cm-wide ring. Find (to 1 decimal place) the area of the golden ring if its outer edge is 1 cm from the edge of the plate.

10 The big question is, ‘Should Santa enter this house by the circular chimney or by the front door?’ It is known that: i Santa’s waist measurement is 150 cm. ii The distance from the centre of the chimney to the outer edge is 30 cm. iii The area of the chimney concrete is 1018 cm2. Support your answer with mathematical evidence.

1 cm 0.5 cm 22 cm

1018 cm2 30 cm

Chimney cross-section

Chapter 10

Shape and measurement

469

11 WE6 A family-size pizza is cut into 8 equal slices. If the diameter of the pizza is 33 cm, find (to the nearest square centimetre) the area of the top part of each slice. 12 WE7 The diagram at right shows a surveyor’s compass. The radius of this compass is 2.5 cm. Find the distance, when measured in a clockwise direction, along the edge of the compass from the north marker to the south-western marker.

13 The bicycle chain on Robert’s bicycle is broken. The broken chain measures 198 cm in length. With all the shops shut for Christmas Day and Robert desperate to ride down to the beach, he spies his sister’s bicycle in the back shed. Using the measurements of his sister’s bicycle given in the diagram, is it worth Robert’s effort to take her bike chain? Support your answer with mathematical calculations.

42 cm

10 cm

14 Early-model vehicles had a single windscreen-wiper blade to remove water from the windscreen. (The bus at right has two single blades of this type.) Using the dimensions given in the diagram: a what area (to the nearest whole number) did the blade cover? b what percentage (to 1 decimal place) of the windscreen was cleared? c what distance (to the nearest whole number) does the tip of the blade travel in one full sweep?

6 cm 150n 160n

120 cm 60 cm

140n 45 cm

15 How does the percentage of windscreen cleared on an early-model car compare with a car of today? The dimensions of a typical Holden windscreen are shown. Allow a 5% reduction in the overall area covered due to the overlap of the two wipers. Support your answer with mathematical evidence.

10d

120 cm

110n

110n

65 cm

50 cm 56 cm

total surface area (tSa) The sum of the surfaces of a 3-dimensional object is referred to as the object’s total surface area or TSA. To shorten the process of finding the TSA of some commonly used objects, mathematicians have been able to work out a specific formula for each object. When the nominated surface area of an object does not totally comply with a formula, you need to first draw the net of the object. This should enable you to work out what parts of the formula are needed.

470

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

Calculation of the total surface area of these buildings is a complex task.

Object

Net

TSA TSA =

Cube

6L2

2 1

4

3 6

L

5

L

L

Rectangular prism h h

w

l

TSA = 2(wh + lw + lh)

l h

l

w

w h

Cylinder

r h r

h

TSA = area of 2 circles + curved surface = 2π r2 + 2π rh = 2π r(r + h)

2 Pr r

Sphere

Not shown

TSA = 4π r2

(continued)

Chapter 10  Shape and measurement

471

Object

Net

Cone

TSA TSA = area of base (circle) + area of curved surface 2π r = π r2 + × π S2 2π S = π r2 + π rS = π r(r + S)

S 2P r r

S  Slant height

h r

Note: The slant height of the cone is formed from the radius of the circle, so their values are the same. The arc length of the sector used to form the cone becomes the circumference of the base of the cone, so their values are the same. Curved surface of a cone is formed by removing a sector out of a circle. Arc length Major sector rS

Slant height  radius of sector

Minor sector Circumference of base  arc length of sector

TSA = area of square + area of 4 triangles 1  = b2 + 4 ×  bh

Square-based pyramid h

= b2 + 2bh

2

b

h b

Worked Example 8

Find the total surface area of the object shown at right. 17 cm think

472

write 19 cm

1

Identify the shape.

Shape: rectangular prism

2

Write the formula for the TSA of a rectangular prism.

TSA = 2(wh + lw + lh)

3

Allocate a value to the pronumerals.

w = 9, h = 17, l = 19

4

Substitute the values of the pronumerals into the formula. TSA = 2(9 × 17 + 19 × 9 + 19 × 17)

5

Evaluate (brackets first, then multiply by 2).

6

Write the answer, including units.

= 2(153 + 171 + 323) = 2 × 647 = 1294 TSA = 1294 cm2

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

9 cm

In some situations you may know the total surface area of an object but be missing a dimension. In this case, you can use the total surface formula rearranged to make the unknown dimension the subject of the equation. WorkEd ExaMPlE 9

A tennis ball has a surface area of 154 cm2. Will it fit through a circular hole with a diameter of 6 cm? think

WritE

1

Write the formula for the TSA of a sphere.

TSA = 4π r2

2

Allocate the pronumerals a value.

TSA = 154 r = ?

3

Substitute known values into the formula.

4

On the Main screen, tap: •  Action •  Advanced •  solve Complete the entry line as: solve(154 = 4πr2, r) Then press E.

5

Given that r is the radius of a tennis ball, its value is positive.

Solving 154 = 4 × π × r2 for r gives r = −3.5007 or r = 3.5007. Since r > 0, r = 3.5007.

6

The question requires the diameter of the tennis ball so multiply the radius by 2.

d=2×r = 2 × 3.5007 ≈ 7.0

7

Compare the diameter of the hole with the diameter of the ball and write an answer sentence.

The tennis ball will not fit through the circular hole because its diameter is approximately 7.0 cm, while the hole’s diameter is 6 cm.

154 = 4π r2

As discussed in the previous section on area, a composite figure is a figure comprising a number of distinct shapes. To find the TSA of a 3-dimensional composite figure, we first need to decide what individual shapes make up the total object and then work out the TSA of each of these shapes separately. Remember that the surfaces where the distinct shapes come into contact are not included in the TSA. This is demonstrated in worked example 10.

Many buildings are composite figures made up of prisms and pyramids.

Chapter 10

Shape and measurement

473

WorkEd ExaMPlE 10

eBook plus

The diagram shows the proposed shape for a new container for takeaway Chinese food. The shape will be used if the TSA of the container is less than 750 cm2. If the TSA is greater than or equal to 750 cm2 then production and manufacturing costs are too high and the takeaway shop will have to stay with the old cylindrical container. Next time I order my Chinese takeaway, could it come in the new design? think

10 cm

Tutorial

int-0905 Worked example 10

10 cm

WritE

1

Identify the distinct shapes that make up the total object: these are a square-based pyramid and a cube. The base of the pyramid and one face of the cube are not on the surface and therefore their area should not be included.

TSA = square pyramid (no base) + 5 faces of a cube

2

Calculate the TSA of a square-based pyramid with no base. (a) Alter the general square-based pyramid formula so as not to include the square base. (b) Allocate a value to the pronumerals. (c) Substitute the values into the formula and evaluate.

TSA of square-based pyramid:

3

1

A=4×2 ×b×h = 2bh b = 10, h = 10 A = 2 × 10 × 10 = 200

Calculate the TSA of a cube (5 faces only). (a) Alter the general cube formula to include 5 faces instead of 6. (b) Allocate a value to the length. (c) Substitute the value of the length into the formula and evaluate.

TSA of a cube (5 faces only): A = 5L2

4

Add the individual TSA together to find the TSA of the whole object.

TSA = 200 + 500 = 700 cm2

5

Write an answer sentence.

Next time I order Chinese takeaway there could be a newly designed container with a surface area of only 700 cm2.

L = 10 A = 5 × 102 = 500

rEMEMBEr

1. The TSA is the sum of the areas of the outside surfaces of a 3-dimensional object. 2. Formulas for all types of objects are not possible. For those objects without a formula you will need to follow these steps. (i) Draw the net of the object. (ii) Work out the different shapes that make up the net. (iii) Calculate their individual areas. (iv) Add all the individual parts together. 3. Do not include in the TSA the surfaces of contact of the distinct shapes that make up a composite figure.

474

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

Exercise

10D

Total surface area (TSA)   1   WE8  Find the total surface area of the following objects. Round your answer to 1 decimal place. a

b

c

20 cm

6.2 m

12.2 cm

32.5 cm 7.5 cm

10 cm

e

d

f

3 cm

14 mm

10.5 m 16 cm 42 mm

9.0 cm

8.4 m

  2 Match the formulas below with the object. a

b h

d

S

r

w

l

r

c

h

(Base not included) Open-ended cone

e

h

f

r

h b Sphere

g

i

h r

S

h

r Half sphere

Base included



i TSA = 2π r2 + 2π rh = 2π r(r + h) iii TSA = 4 v TSA =

( bh) 1 2

4π r2

(

)

vii TSA = 2 × 1 bh + (3 × bl) 2

l

b



ii TSA = π rS



iv TSA = π r(r + S)



vi TSA = 2(wh + lw + lh) 1

viii TSA = 2 [2π r(r + h)] + 2rh = π r(r + h) + 2rh

4π r 2 2 = 2π r2

ix TSA =

Chapter 10  Shape and measurement

475

  3 For the following diagrams, find the value of S for the formula: TSA (cone) = π r2 + π rS = π r(r + S) b a c 10.5 mm 12 cm 10.4 cm

10 mm

d

60 cm

9m

6 cm 7m

  4 The diagram at right represents a waffle ice-cream cone (no top). Using the dimensions given, calculate the TSA (to the nearest square centimetre) of waffle used.   5 For each of the following objects find the value of the pronumeral, rounding your answer to 1 decimal place when required. a

4m

b

w=? 6m TSA = 148 m2

c

3m h=? TSA = 188.5 m2

d d = 6 cm z=?

b=? TSA of 1 sphere = 55.4 cm2

TSA = 100 cm2

  6   WE9  A cylindrical cork is 6 cm high and has a TSA of 33.38 cm2. Could it be used to close a bottle whose neck is 3 cm in diameter? Justify your answer.   7   MC  A 60-cm-high cone has a base radius of 32 cm. The TSA could be calculated by using the formula: B π 60(32 + 60) A π 32(32 + 60) D π 68(32 + 68) E π 60(32 + 68)

60 cm

C π 32(32 + 68)

32 cm

A spherical candle of TSA = 201 cm2 is to be gift-boxed. The dimensions

  8   MC  in centimetres of several differently shaped boxes are given below (length × height × width). The box which will best fit the candle with the least amount of wasted space is: A 6 × 6 × 6 B 4 × 4 × 6 C 8 × 4 × 4 D 2 × 8 × 4 E 8 × 8 × 8   9   MC  The area covered by the rolling pin shown in one complete turn is: A 2π 5(5 + 30) B 2π 2.5(2.5 + 30) 5 cm C π 2.5(2.5 + 30) D 2π 2.5 × 30 E 2π 5 × 30

30 cm

10 What area of cardboard needs to be purchased to construct a box 1 m × 1 m × 1 m to contain a television set? Allow an extra 5% of the total surface area to cater for overlaps. 40 cm 11 The latest footstool design is in the shape of a cylinder. a Your client would like to know how much fabric to purchase, so you 35 cm as the upholsterer can cover the footstool. (When calculating the area of fabric you always allow an extra 20% of the total surface area to cater for seams and for placing pattern pieces on the fabric.)

476

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

b The fabric your client has chosen comes on a roll 150 cm wide. The length of fabric can be cut to the nearest centimetre. How many metres should be purchased? 12 A church steeple is a square-based pyramid in design. The steeple cap needs new tiles. The vicar has asked the plumber for a costing on its renovation. The plumber stated that the tiles cost $60 per square metre, his labour is $20 per square metre and when he calculates the TSA he always adds 25% of the TSA to 25 m allow for tile overlap.   Prepare an estimate for the vicar which breaks down the plumber’s costs 5m and shows how he came to the amount. Suggested subheadings could be: a TSA of roof only b TSA (including 25% allowance) c Cost of tiles d Cost of labour e Final overall cost. 13 Tennis balls sold as a group of three are normally encased in a vacuum-sealed cylindrical container. A department store manager has complained to the manufacturer that the container is difficult to stack and display. She asks, ‘Would it be possible to investigate a different shaped container?’ The manufacturer agrees to look into the situation. However, the TSA of any new container can be the same as the cylindrical one, or less, but not more. a Calculate to the nearest square centimetre the TSA of the three containers shown below. Based on their TSA, rank them in order of best container to worst container.

7 cm

7 cm

Cylindrical container

12 cm

Rectangular prism

Triangular prism

b The manufacturer changes his mind and says he will consider a different container provided its surface area is no more than 30% above the TSA of the cylindrical container. Do either of the other containers meet this criteria? 14 a  What is the name of the shape (a type of prism) of this chocolate package? b In your mind, unfold the 3-dimensional shape to 3.5 cm 19.5 cm its 2-dimensional net. Draw this net, making no allowance for overlap. c Using the dimensions given, calculate the approximate TSA of a medium sized chocolate package. (Round all values to 1 decimal place.) d What is the size of the smallest rectangle of cardboard from which one packet could be cut? 15 a  What is the area of the label on the curved surface of the can? Allow an extra 5% of the TSA for an overlap. b Theoretically, how many labels can be printed and cut from a sheet of paper 1 m × 1 m? c What are the dimensions of one label? (Remember the 5% overlap will affect the length, not the height, of the label.) d By modelling the situation, how many labels can really be printed and cut out of a sheet 1 m × 1 m? e What area of the 1 m × 1 m sheet is actually used? f What percentage of the area (to the nearest whole number) is wasted?

Chapter 10  Shape and measurement

477

16 Juice Fresh frozen flavoured ice blocks are made and wrapped in a pyramidal container. The triangles that make up the pyramid are equilaterals, with side lengths of 10 cm. a ‘Unfold’ this 3-dimensional diagram to form its 2-dimensional net. b Calculate the TSA of the container. (Round the answer up to the nearest 10 cm whole number.) c Theoretically, how many containers can be cut out from a 1-m × 1-m sheet of specially lined cardboard? d By modelling the situation, how many containers can actually be cut out from this sheet of cardboard? 17 The following questions relate to the hemispherical sugar bowl at right. a If its radius is 4 cm, what is the TSA of the sugar bowl to the nearest square centimetre? b The dish is made from a disc cut out of a sheet of metal. It is heated and modelled to form its finished shape. What is the radius of this disc to one decimal place? c The discs are cut out of a metal sheet as shown at right. What are the dimensions of this sheet? d What percentage of metal is wasted in making 8 bowls? 18 WE10 Certain medicines come in capsules, as shown at right. Find the area of plastic (in square millimetres) needed to produce one such capsule.

4 cm

10 mm 5 mm

19 A commercial bread bin has dimensions as shown in the diagram at right. Find the TSA of the bin. 20 A garbage bin in a designer shop is constructed out of plastic. The dimensions are shown in the diagram. The rubbish opening is equal to 10% of the TSA of the hemispherical top. However, the bin comes only in white and you would like a black one for your bedroom. You decide to paint the container and feel it will need two coats. You have enough black paint at home to cover one square metre. Will you have enough to cover the container or do you have to purchase another tin? Support your final decision with mathematical calculations. eBook plus Digital docs

WorkSHEET 10.1 Investigation Cone heads

478

21 Lampshades for a particular range are made by cutting a cone of fabric in the ratio of 1:2. The top (one-third of the cone’s height) is discarded and the remaining section (two-thirds of the cone’s height), called a frustum, is wired and used as the shade. Five per cent of the final TSA is allowed for an overlap. The original cones are 30 cm high and have a base radius of 12 cm. Find the TSA of the frustum so that fabric can be ordered to cover the lampshade for a particular client. State your answer to the nearest centimetre or square centimetre.

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

90 cm 60 cm

33 cm

28 cm

10E

Volume The volume of an object is the amount of space that the object occupies. Volume of a prism = cross-sectional area × height of the prism V = CSA × H The height is the dimension perpendicular to the cross-sectional area. Shape

Cross-sectional shape

Volume V = area of a circle × height = π r2 × H

Cylinder r H

Area = π r2

r

V = area of a triangle × height 1 = 2 bh × H Note: Lowercase h represents the height of the triangle.

Triangular prism h b h

1

H

Area = 2 bh

b

Rectangular prism

V = area of a rectangle × height =L×W×H

W L

H W

L

Area = L × W V = area of a square × height = L2 × H = L2 × L = L3 (since in a cube, H = L)

Cube

H

L L

Area = L2

Worked Example 11

Find the volume of the shape shown correct to 1 decimal place.

2.3 m

think

2.6 m

write

1

Identify the shape.

Triangular prism

2

Write the appropriate formula for the volume.

V = 2 bh × H

3

Allocate values to the pronumerals keeping in mind that b and h are the base and height of a triangular cross-section or base of the prism, while H is the height of the prism.

b = 2.6, h = 2.3, H = 3.2

4

Substitute and evaluate, rounding the answer to 1 decimal place. Include the units.

V = 2 × 2.6 × 2.3 × 3.2 = 9.568 ≈ 9.6 m3

3.2 m

1

1

Chapter 10  Shape and measurement

479

Odd-shaped prisms The object at right would be classified as a prism, because horizontal cuts would show a uniform crosssection, the same shape as the ends. We could also say that this shape is a prism because the ends are the same and the sides of the object are parallel. To calculate the volume of an odd-shaped prism you would need to be given the value for the area of the cross-sectional shape and the height of the prism.

Height

Cross-section

Worked Example 12

Find the volume of the shape shown at right. think 1

write

V = CSA × height

Write the general formula for the volume of a prism.

8.2 m

CSA = 32

m2,

H = 8.2 m

2

Allocate the pronumerals a value.

3

Substitute the values of the pronumerals into the formula and evaluate.

V = 32 × 8.2 = 262.4

4

Write the answer including units.

V = 262.4 m3

Area  32 m2

Tapered objects A tapered object is one that has a flat base at one end and tapers to a point at the other. A cone and the family of pyramids (square-based, rectangular-based and triangular-based) are examples of tapered objects.

Cone

Square pyramid

Rectangular pyramid

Triangular pyramid

A tapered object does not have a uniform cross-section. The cross-sectional area becomes smaller as it nears the apex (point). The internal capacity or volume of a tapered object is 1 a fraction of the volume of a prism. Mathematicians found this fraction to be one-third (3). They defined the base of a tapered object to be the flat end opposite the apex. To calculate the volume of a tapered solid we find the area of the flat end, multiply this by the height of the 1 pyramid (which must be perpendicular to the base) and then multiply by 3 (or divide by 3). 1

Volume of a tapered object = 3 × area of base × height of object 1

=3×A×H

480

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

The following table shows the formulas for the volume of some common tapered objects. Shape

Flat end (base) shape

Volume

Cone

1

V = 3 × area of a circle × height

r

1

V = 3π r2 × H

H r

Square pyramid

1

V = 3 × area of a square × height 1

V = 3L2 × H

H L L

Rectangular pyramid

1

V = 3 × area of a rectangle × height

W

1

L

= 3L × W × H

H W

L

Triangular pyramid

1

V = 3 × area of a triangle × height

1  V = 3  2 bh × H Note: Lowercase h represents the height of the triangle. 1

h H b

b

h

Spheres The volume of a sphere is given by the following formula: 4

Volume of a sphere = 3 π r3

r

where r is the radius of the sphere. A hemisphere is half of a sphere. Its volume, therefore, is half of the volume of a sphere. 1

Volume of hemisphere = 2 (volume of sphere) 1

r

4

= 2 × 3 π r3 2

= 3 π r3

Composite figures The principles used to calculate the TSA of 3-dimensional composite figures can also be applied when calculating the volume of 3-dimensional composite objects. The volumes of the individual objects need to be found before they are added or subtracted to find the total volume.

Chapter 10  Shape and measurement

481

WorkEd ExaMPlE 13

eBook plus

Find the volume (in m3) of the toy shown, correct to 1 decimal place. think

int-0906 Worked example 13

1

Identify the components of the shape.

2

Determine the volume of the hemisphere.

Total volume = volume of a cone + volume of a hemisphere Volume of hemisphere:

5 6

80 cm

2

= 0.134 041 3 m3 Volume of cone:

Find the volume of the cone.

1

V = 3π r2 × H r = 0.4 m, H = 1.4 − r = 1.4 − 0.4 =1m 1 2 V = 3π (0.4) × 1

(a) Write the formula. (b) Allocate values to the pronumerals. (Height = 1.4 − radius of circle) 4

1.4 m

V = 3πr 3 r=d÷2 = 80 ÷ 2 = 40 cm = 0.4 m 2 V = 3 π (0.4)3

(a) Write the formula. (b) Allocate values to the pronumerals. Note that some units are in cm, others in m. As the answer requires cubic metres, change centimetres to metres. (c) Substitute the value of r into the formula and evaluate. 3

Tutorial

WritE

Substitute the values of H and r into the formula and evaluate.

= 0.167 551 6 m3 Total volume = 0.134 041 3 + 0.167 551 6 = 0.301 592 9 V = 0.3 m3

Add the individual volumes together to find the total volume of the given shape. Round your answer to 1 decimal place and include units.

rEMEMBEr

1. Volume of a prism = CSA × perpendicular height. 1 2. The volume of a tapered object = 3 × area of base × perpendicular height. 3. Volume formulas: Trapezoidal prism Triangular prism Triangular pyramid 1 1 1 1 V = 2 bh × H V =  2 (a + b) × h  × H V = 3 2 bbhh × H  

( )

a h

H h

H

b

H

h

b b

h = height of triangle 4. Volume of a sphere: V =

482

h = height of triangle 4 3

πr3

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

h = height of trapezium

ExErCiSE

10E

Volume 1 Match the volume formula with the appropriate 3-dimensional object. a

b

c

H

L

H r

H

H

A cm2

d

f

e

a

h

b h H

b

r

1

1

iv V = eBook plus

iii V =  1 (a + b)h  × H

ii V = 3π r2 × H

i V = 2 bh × H

π r2

H

H

×H

v V=

1 2 L 3

2

×H



vi V = A × H

2 WE11 Find the volume of each of the following shapes correct to 1 decimal place. a

Digital doc

8 cm

b

6 cm

c 7.4 cm

SkillSHEET 10.4 Volume

10.2 cm

d

10 cm

22.4 cm

4 cm

12 cm

e

g

f

12.6 cm

h

30 cm

16 cm

18.5 cm

10.2 m

6.5 m 32 cm

12.4 cm

40 cm

3 WE12 Find the volume of each of the following shapes. b

a

Shaded area = 42 cm2 eBook plus Digital doc

SkillSHEET 10.5 Conversion of units — volume

8 cm

Shaded area  23 cm2

c

CSA  116 mm2

d

3.5 cm 10.5 mm

0.5 m 55 cm2

4 Alexander is ordering a concrete base (in the shape of the trapezoidal prism shown) for his favourite garden sculpture. How much will he have to pay if concrete costs $50 per cubic metre and the cost of labour is $45?

60 cm

50 cm 50 cm

1.2 m

Chapter 10

Shape and measurement

483

5 A cosmetic eye-mask is 8 mm thick and has a cross-sectional area of 120 cm2. Find the volume of the special liquid that fills the mask. 6 WE13 Find the volume of each of the following shapes correct to 1 decimal place. ExaM tiP Rounding off should only be done at the final answer stage in a question, not part way through the calculation. The continued use of a rounded answer will often compound the rounding error. [Assessment report 2007]

60 cm

b

1.4 m

113 cm

a

2m

1.8 m

64 cm

c

22 cm

10 cm 25 cm

15 cm

7 cm

7 The diagram at right shows 3 tennis balls packed in a cylindrical container. Find: a the volume of each ball b the volume of the cylinder c the volume of space that remains free.

eBook plus Digital doc

SkillSHEET 10.6 Finding unknown lengths

8 A chocolate company wants to make chocolate Christmas ball decorations containing 50 small, candy-coated chocolates to hang on Christmas trees. If each candy-coated chocolate has a volume of approximately 0.8 cm3, what is the diameter of the Christmas ball required to contain them? 9 The teapot shown at right is made up of a number of geometrical shapes. The hemispherical bowl can be filled to the brim with tea. a How many cylindrical cups (with radius of 3 cm and height of 9 cm) could be filled to a mark 2 cm from the top from one full pot of tea? b If you wanted to design a teapot that when filled would enable six of the above cylindrical cups to be filled to a mark 1 cm from the top, what would be the radius of the hemispherical bowl?

r = 7 cm

10 A large apple takes up approximately 512 cm3 of space. I have 160 apples to pack into one of the following containers. With the aim of having the minimal amount of wasted space, which container would be best for this purpose and why? 40 cm

a

60 cm

484

38 cm

b

38 cm

70 cm

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

40 cm

c

38 cm 42 cm 34.2 cm

40 cm

11 You have just whipped up the most delicious chocolate mud cake mixture ready for baking. It fills a hemispherical mixing bowl to the brim. In the cupboard you have 4 different types of cake tins used for baking. The decision is which one to use. You know that the cake will rise 3 cm as it cooks and if it rises more than 1 cm over the brim of the container it will sink in the middle, ooze down the sides of the container and be ruined. a For each of the containers shown below, calculate how far up the sides of the container the unbaked mixture could go. Round all values to 1 decimal place as you go. b Using these results, justify which container(s) can and can not be used. i

7 cm

ii

iii 7 cm

15 cm Square-based tin

18 cm Round tin

7 cm iv

16.4 cm

10.5 cm

9 cm 23 cm Rectangular tin

7 cm 20 cm Ring tin

12 The flower vases shown below are from a designer shop. They are all examples of frustums. A frustum is the part of a solid shape cut by two parallel planes. c a b 18 cm 21 cm 12 cm 24 cm

Vase 2 Vase 1 Vase 3 They are all made from a different geometrical shape; however, their overall construction was based on the following constraints. i The frustum of the tapered objects was used to make the vases. The frustum height is 2 of the height of the original tapered object. 3 1

ii The remaining 3 was discarded. iii All vases are 24 cm high. Note: If one dimension of a cone is reduced/enlarged by a scale factor then the other dimensions of the cone are also affected by this factor. For example, if the height of a 36-cm cone is reduced by a scale factor of 6, then the radius value would also be reduced by this scale factor. a Calculate the total volume of each vase, rounding all calculations to 1 decimal place as you go. b What volume of water would be held in the vase with the overall greatest volume, if the water is poured into the vase to a vertical height of 21 cm? c What percentage of the base’s total capacity is this volume?

Chapter 10

r=6 h = 36 r=1 h=6 eBook plus Digital docs

Investigation Cone volume

Spreadsheet 010 Cone volume

Shape and measurement

485

10F

Capacity The capacity of a container refers to the amount that it can hold. The capacity or volume of a container is usually measured in cubic units; however, when the volume of a liquid is being discussed it can be referred to in terms of millilitres, litres and kilolitres. Recall the following facts: 1000 millilitres (mL) = 1 litre (L) 1000 litres = 1 kilolitre (kL) Cubic units are connected to the fluid capacity units as follows: 1 cm3 = 1 mL 1000 cm3 = 1 L 1 m3 = 1000 L = 1 kL

Worked Example 14

Convert: a   400 cm3 to mL b   1200 cm3 to mL and to L c   2 kL to m3. think

write

a

Since 1 cm3 is equivalent to 1 mL, then 400 cm3 is equivalent to 400 mL.

b

1

Each 1 cm3 will hold 1 mL of liquid. Therefore, 1200 cm3 will hold 1200 mL of liquid.

2

To change mL to L, divide by 1000 (since there are 1000 mL in 1 L).

c

One kL is equivalent to 1 m3. Therefore, 2 kL is equivalent to 2 m3.

a

400 cm3 = 400 mL

b 1200 cm3 = 1200 mL

= 1.2 L 2 kL = 2 m3

c

To find the capacity of a container in litres, find its volume in cubic units first and then convert. Worked Example 15

Find the capacity in mL of a rectangular container measuring 10 cm × 12 cm × 14 cm. think

486

write

V=l×w×h = 10 × 12 × 14 = 1680 cm3

1

Find the volume of the container in cm3. (Since the container is rectangular, use the formula for the volume of a rectangular prism.)

2

Change cubic centimetres to millilitres.

1680 cm3 = 1680 mL

3

Write the answer in words.

The capacity of the given container is 1680 mL.

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

rEMEMBEr

To find the capacity of a container in litres, find its volume in cubic units first and then convert, using the following conversions: 1 cm3 = 1 mL 1000 cm3 = 1 L 1 m3 = 1000 L = 1 kL ExErCiSE

10F

Capacity 1 WE14

Convert the following units as indicated. = _____ mL

a

750 cm3

c

2500 cm3

= _____ mL

b

eBook plus

800 cm3

= _____ L

d 40 000 cm3 = _____ L

Digital doc

Spreadsheet 006 Capacity

e 6 m3 = _____ cm3 = _____ mL = _____ L

f 12 m3 = _____ L

g 4.2 m3 = _____ kL

h 7.5 m3 = _____ kL = _____ L

i 5.2 mL = _____ cm3

j 6 L = _____ cm3

k 20 L = _____ mL = _____ cm3

l 5.3 KL = _____ m3

2 WE15 How many millilitres will a rectangular drink container hold if its dimensions are 11 cm × 4 cm × 15 cm? 30 cm

3 A hemispherical bowl with a diameter of 30 cm will be used to hold a pre-mixed fruit drink for a party. If you want to fill it to the brim, how many litres of pre-mixed drink can you pour in? 4 One litre of orange and mango juice is packed in a rectangular container of height 172 mm and width 93 mm. Find the length of the container to the nearest mm. 5 A plastic bottle contains 1.25 L of soft drink. a How many cylindrical cups 10 cm high and 7 cm in diameter can be filled to capacity from this bottle? b What is the volume (to the nearest mL) of the drink that remains in the bottle?

172 mm l

93 mm

6 A rectangular swimming pool measures 4 m by 3.5 m by 2.5 m. a What is the capacity of the pool in kL? b If the pool is being filled at a rate of 14 L per second, how long will it take for it to be: i 70% full? ii filled to capacity?

7 A tub in the shape of a trapezoidal prism has dimensions as shown. If the capacity of the tub is 943.5 L, how deep is it?

0.85 m

2.2 m

d 1.5 m

Chapter 10

Shape and measurement

487

10G

Similar figures Imagine! You have just seen the best photograph in a magazine. It was .  .  . [insert fantasy of your choice!]: 1. the car you will buy when you get your licence 2. the dress you have dreamed about for the end-of-year formal dance 3. the place you are going to go to when all your studies are over. You place the photograph on the photocopier screen, set the enlargement to 145% and press start. Out comes a copy of the photograph exactly the same shape as the original but larger. This copy can go on the wall near your desk for inspiration. Now you need a copy for the inside of your school locker door. You re-set the size to 66% and out comes another copy of the photograph, exactly the same shape but smaller.

Enlarged 145%

Original

Reduced 66%

You have produced two similar photographs or similar figures by dilating (enlarging) or reducing about a fixed point. To achieve a similar enlargement or reduction of an image by hand you would need to: 1. fix a point on the page 2. draw lines to specific places on the image 3. measure the length of these lines 4. multiply or divide these measurements by a given amount 5. follow along the lines already drawn and either extend or reduce these lines to the length of the new measurement 6. redraw the object by connecting the ends of new length lines. The photocopier accomplishes all these tasks at the press of a button! By drawing from a fixed point, the angles in the original photograph and the enlarged/ reduced copy stay the same. So the figures produced are the same shape, but not the same size. Using a photocopier on its normal setting produces a copy that is identical in size and shape.   Figures that are the same in shape but not in size are called similar figures.   Figures that are identical in both size and shape are called congruent figures. The trapezium labelled ABCD at the top of page 489 has been enlarged by a scale factor of 2 from a fixed point to produce a new trapezium labelled A′B′C′D′. As both trapeziums are the same shape but a different size they are said to be similar figures. Mathematicians label a similar figure with the same vertex labels, but with a ‘prime’ (the symbol ′) to indicate it is an image of the original. Mathematicians would write ‘A′B′C′D′ is similar to ABCD’ as: A′B′C′D′ ∼ ABCD (∼ means ‘is similar to’) 488

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

2.8 cm

A' 123n A A 1.7 cm D

1.4 cm

123n

120n

57n

3.4 cm

B 1.35 cm 60n

3.25 cm

C

B' 120n B 2.7 cm

O D

C

D' 57n

60n C'

6.5 cm

A′B′C′D′ ∼ ABCD, because if you measure: 1. the angles in each trapezium, the corresponding angles will be equal (for example, ∠DAB = 123° so ∠D′A′B′ = 123°) 2. the sides in each trapezium, the length of each pair of corresponding sides will be in the same ratio, equal to the scale factor. 6.5 Image length C' D' : = =2 3.25 Original length CD C' B' 2.7 = =2 CB 1.35 A' B' 2.8 = =2 AB 1.4 A' D' 3.4 = =2 AD 1.7 The following example shows how a figure can be enlarged from a fixed point. Worked Example 16

Enlarge this shape by a scale factor of 2. think 1

A write/Draw

Select a point inside the shape, say, point O and draw lines from this fixed point to the shape’s vertices (that is points A, B, C and D).

A

B

D

2

Measure the lengths of OA, OB, OC and OD. Note: The measurements are in cm.

3

Multiply each of the measurements by a scale factor of 2.

4

Extend the lines to the new measurements; that is, continue line OB until it measures 2.1 cm, line OC until it measures 3.5 cm and so on. Join the ends of the extended lines to form the similar figure. Label the vertices appropriately.

C

O D

5

B

C

OA = 0.95, OB = 1.05, OC = 1.75, OD = 1.75 O′A′ = 0.95 × 2 O′C′ = 1.75 × 2 = 1.9 = 3.50 O′B′ = 1.05 × 2 O′D′ = 1.75 × 2 = 2.1 = 3.50 A'

B' A

B O

\ D

C

D'

C'

The following example shows how to reduce a figure from a fixed point.

Chapter 10  Shape and measurement

489

WorkEd ExaMPlE 17

eBook plus

Reduce the pentagon labelled ABCDE by a scale factor of 2.

A

Tutorial

int-0907

B

E

D think 1

Worked example 17

C

WritE/draW

Select a point outside the shape and label it O. Draw lines from this fixed point O to the shape’s vertices and measure their lengths. Note: The measurements are in cm.

A B

E O D

C

OA = 6.8, OB = 7.6, OC = 7.1, OD = 5.8, OE = 5.6 2

Divide these measurements by the scale factor of 2.

6.8

OA′ = 2 = 3.4 7.6 OB′ = 2 = 3.8

3

4

Locate the vertices of a reduced pentagon by measuring new distances (that is OA′, OB′ and so on) from point O along the existing lines. Join the vertices A′, B′, C′, D′ and E′ to form a reduced figure.

7.1

5.6

OC′ = 2 = 3.55 5.8 OD′ = 2

OE′ = 2 = 2.8

= 2.9 A B

E

A' B'

E' O D'

C'

D

C

Scale drawing Obviously, the principles of enlarging and reducing lengths in similar figures can be applied to scale drawing problems. In scale drawing, there are three basic situations: 1. Given the distance between two points on a scale drawing, and the scale factor, you have been asked to find the distance between the points in real life. 2. You have measured the distance between two points in real life and now need to draw this measurement for a plan using a specific scale factor. 3. You know the distance in real life and the distance you would like to achieve on the plan. What scale factor should you use? Reduced scale drawing

Real-life measurement

Scale factor

490

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

Enlarged scale drawing

Scale factor

Scale factor The scale factor for scale drawings is usually represented as a ratio (1:200, 6:113). The ratio symbol (:) means ‘to’, so 1:100 is read as ‘one to one hundred’. This means that one unit on the drawing represents 100 units in real life. In the scale factor ratio, the first number represents the measurement on the scale drawing, while the second number represents the real-life measurement. When expressing scale factors as a ratio, the order of the figures in a ratio is important, as 2:5 is not the same as 5:2. The ratio 2:5 is read as, ‘2 units on the scale drawing is equivalent to 5 units in real life’. The scale drawing (for example, a house plan) is a reduction of the real-life object. The ratio 5:2 is read as, ‘5 units on the scale drawing is equivalent to only 2 units in real life’. The scale drawing (for example, an enlarged drawing of an ant) is actually an enlargement of the real-life object. Sometimes the relationship between the scale plans and the real-life object can be written as ‘1 cm to 1 m’, or ‘5 cm to 2 mm’. These are not ratio values. To change scale values into ratio values, the units for the scale drawing and the real-life object must be the same. Worked Example 18

A plan uses the scale 2 cm to 1 m. Write this scale as a ratio in simplest form. think

write

1

2 cm to 1 m on a plan means that 2 cm on the scale drawing represents 1 m in real life. To write this as a ratio, the units must be the same.

2 cm to 1 m

2

As 1 m = 100 cm, replace 1 m with 100 cm.

2 cm to 100 cm

3

Simplify by dividing both numbers by 2.

1 cm to 50 cm

4

Omit the units and write in ratio form.

1:50

A ratio equation that will allow you to find other values in the scale drawing or similar figure can be formed from a ratio. For example, to determine a rule for the scale drawing situation where the ratio is 1:50, the ratio will need to be first changed into a fraction. Usually, this would 1 50 be 50 , but 1 is also correct. 1 scale measurement , 50 real-life measurement or = = 50 real-life measurement 1 scale measurement By rearranging either of these equations we end up with the following rule/equation: Real-life measurement = scale measurement × 50 Therefore, a scale measurement of 10 cm would represent a length of the object in real life of 10 × 50 = 500 cm. If the object in real life is 10 m high, then the object’s height on the scale plan can be calculated as follows: 10 = scale measurement × 50 10

Scale measurement = 50 = 0.2 m = 20 cm The rule formed is based on a ratio and because of this we can enter values for the scale length or real-life length in any unit. However, if the value for scale length is entered in

Chapter 10  Shape and measurement

491

millimetres then the real-life length will be calculated in millimetres. Likewise, if metres are entered for the real-life length then the scale length will be calculated in metres.

Worked Example 19

A pool company produces a brochure containing diagrams of their differently shaped pools. The dimensions are given for the maximum length, width and depth. The brochure states that the diagrams were drawn to the scale of 2 cm to 2 m. The small plunge pool you are interested in is missing the length and depth measurements. Determine these values. Overhead view

Ground

Depth  ? m

Width  2.4 m

Scale depth  2 cm

Length  ? m Scale length  4.4 cm think 1

We need to determine a rule joining scale and object measurements with the scale factor. Write the given scale.

2 cm to 2 m

2

To change the scale into a ratio we need to have both numbers in the same units, so convert metres to centimetres (2 m = 200 cm).

2 cm to 200 cm

3

Write the scale as a ratio and simplify it.

4

To form a rule, change the ratio into a fraction and cross-multiply to simplify the expression.

2:200 1:100 1 scale measurement = 100 real-life measurement Real-life measurement = 100 × scale measurement

5

To calculate the length of the real-life pool: (a) determine the scale measurement of the length (b)  substitute it into the rule and simplify. (c) Since the measurement we substituted was in centimetres, the resultant measurement is also in centimetres. Change the answer to more appropriate units (in this case to metres).

Length of pool:

To find the depth of the pool: (a) determine the scale measurement of the depth (b)  substitute it into the rule and simplify (c)  change the answer to metres.

Depth of pool: Scale measurement = 2 cm

Write an answer sentence.

The pool is 4.4 m long and 2.0 m deep.

6

7

492

write

Scale measurement = 4.4 cm Real-life measurement = 100 × 4.4 = 440 cm = 4.4 m

Real-life measurement = 100 × 2 = 200 cm = 2.0 m

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

WorkEd ExaMPlE 20

eBook plus

A hobbyist has built a scale model of a 25-m-long aeroplane. If the model is 25 cm Tutorial in length, calculate: int-0908 a the scale factor he used to reduce the aeroplane’s dimensions Worked example 20 b what lengths he will need to make the scale model’s wings if the aeroplane’s wingspan is 30 metres. think a

b

WritE a

Scale factor =

model length aeroplane length

1

Write the formula for scale factor.

2

State the values of the aeroplane length and the length of the model. Since the units are different, change one of them, say, centimetres to metres.

3

Substitute the values into the equation and simplify.

4

Write a brief comment interpreting what the fraction represents.

1

Substitute the value of the scale factor into the formula and cross-multiply to form a rule.

2

Substitute 30 metres instead of the aeroplane length into the rule and hence find the corresponding length of the model. Note: Since the aeroplane’s wingspan was entered in metres then the model’s wingspan will be also calculated in metres.

If an aeroplane’s wingspan is 30 m, 100 × model length = 30 30 model length = 100 = 0.3

3

Write an answer sentence.

The wingspan of the model aeroplane is 0.3 m or 30 cm.

Aeroplane length = 25 m Model length = 25 cm = 0.25 m 0 . 25 25 1 = 100 That is, one unit on the model represents 100 units on the aeroplane. Scale factor =

b

1 model length = 100 aeroplane length 100 × model length = aeroplane length

rEMEMBEr

1. Similar figures are the same shape but not the same size. The corresponding angles in each figure are the same; however, the corresponding side lengths differ by a scale factor. 2. The scale factor can be written in several ways, for example: 1

2 cm to 10 m = 2 cm to 1000 cm = 2:1000 = 1:500 = 500 3. When the scale factor is written as a ratio, the first number represents the measurement on a scale drawing, while the second number represents the real-life measurement. 4. When forming a rule, the following format is usually used. scale measurement Scale factor = real-life measurement 5. When finding measurements with the rule, the measurement that we substitute and the resulting measurement (that is, the answer) are in the same units.

Chapter 10

Shape and measurement

493

Exercise

10G

Similar figures   1 Write the following sentences into your workbook, deleting the inappropriate words. a Multiplying an object’s measurements by a scale factor < 1 will produce (an enlarged/ a reduced) image. b Multiplying an object’s measurements by a scale factor > 1 will produce (an enlarged/ a reduced) image.   2   WE16  Enlarge each of the following shapes by the given scale factor. a Scale factor = 3 b Scale factor = 1.6 c Scale factor = 2.2 A B A B O

C

O

D

C

F E

O A

D

  3   WE17  Reduce each of the following shapes by the given scale factor. a Scale factor = 1.3 b Scale factor = 2

O



Hint: Select points along the circumference to use as reference points (like vertices). c Scale factor = 1.25 O

Questions 4 to 8 refer to the following scales given on a series of maps. a 1 mm to 1 m

b 2 cm to 10 m

c 3 cm to 25 m

d 40 cm to 1 m

e 20 cm to 10 cm

f 175 mm to 1 m

  4   WE 18 

494

Write each of these scales as a ratio in simplest form.

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

scale length . real-life length   6 Determine the rule for each map situation. Have the real-life length as the subject of the rule, that is: Real-life length =   5

Change each ratio from question 4 into a fraction of the form:

  7   WE 19  Find the real-life length in millimetres, centimetres and metres for each map situation if the length on the map is 3 cm.   8 Find the length on the map in millimetres, centimetres and metres for each situation if the reallife length is 12.5 m.   9   MC  The construction plan of a television cupboard is drawn to the scale of 3 cm to 20 cm. If the height on the scale plan is 14 cm, the real-life height to the nearest centimetre would be: A 102 B 93 C 89 D 98 E 94 10 In a furniture catalogue, the real-life dimensions of a bookcase you want were given as 74 cm × 22 cm × 174 cm. The catalogue did not state the height, width or depth. It did say to look at the scale drawings (which follow). However, there were three different scale drawings. Determine which diagram is the scale drawing of the bookcase you are interested in purchasing. a

c

b 1.85 cm

5.4 cm

2 cm

4.35 cm

2.2 cm

0.55 cm

3.87 cm

0.93 cm

0.7 cm

11   WE20  On a photograph, a 30-m-high tree measures 5 cm. Calculate: a the scale factor b the diameter of the tree at its base, as shown on the photograph, if the real diameter is 1.2 m. 12 When designing kitchens, the draftsperson can use 1-cm-square graph paper to scale the drawings. a What is the scale factor of this kitchen design? b In scale length, how long is x? c In real-life, how long is x? d What are the dimensions of the cook top? e What area (to 1 decimal place) on the cook top is taken up by the hot plates?

250 cm Cupboard space

Fridge

x cm Cook top

13 You are visiting the wonderful Funland, in the country of Amazina. The distance from Fun Hotel to the main entrance is 700 yards on foot. In Amazina, the imperial system of measurement is used. To change this to metric-system measurements we need to use the conversion factors of 36 inches to 1 yard and 39.37 inches to 1 metre.

Chapter 10  Shape and measurement

495

a 700 yards converts to _________ inches. b From part a, _________ inches converts to _________ metres (to the nearest whole number). c If the scale-plan distance from the hotel to the main entrance is 6.4 cm, write the scale length and the real-life distance from the hotel to the entrance as: i a scale _________ cm to _________ m ii a simplified ratio _________:_________. d Write a general rule to calculate the real-life distance if you know the scale factor and scale length; that is, real-life distance = _________. Questions 14 to 19 will help you to investigate the area and volume ratios of similar figures. 14 Which of the following figures are similar? a All rectangles b All squares c All pentagons d All circles e All equilateral triangles f All isosceles triangles g All right-angled triangles h All isosceles right-angled triangles Mathematically it can be shown that if the side lengths of two similar figures are in the ratio a:b then the corresponding areas of the similar figures are in the ratio a2:b2.   Squares are always similar figures because of they always have four equal angles (90°). For the two squares shown: if the side length ratio = 2:6 then the area ratio = 22:62 = 4:36 2 6 = 1:9. Therefore, if the area of the smaller square is 4 cm2, then the area of the larger square is: large square area = scale factor (ratio) small square area large square area 9 = 1 4 large square area = 36 cm2 15 Copy the following similar rectangles into your workbook. a Copy and complete the following statements and calculations.

1 4

Smaller:Larger Side length ratio Area ratio

2

2

5

10

Smaller:Larger

2:     ?     

or

     ?     :10

22:     ?     2

or

     ?     2:102

4:     ?     

or

     ?     :100

(Cancel if possible.)      ?     :25 It does not matter which pair of dimensions you choose; the area ratio will finally work out to be the same. b If the area of the larger rectangle is 50 cm2, use the area ratio to calculate the area of the smaller rectangle: area of small rectangle = scale factor area of large rectangle area of small rectangle 4 = 25 50

496

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

area of small rectangle =      ?      cm2 Check: Area of rectangle = L × W L = 4, W = 2 = 4 × 2 =      ?      cm2 16 a   Show that ∆ABC ∼ ∆DEF. b What is the ratio of the two triangles’ lengths in simplest form? c What is the ratio of their areas? d If the area of ∆ABC is 6 cm2, use the area ratio to show that the area of ∆DEF is 13.5 cm2.

D A 4 B

5 3

7.5

6 C E

4.5

F

17   MC  An estate has an area of 25000 m2. If the area of the scale drawing of this estate is 10 cm2, then 1 cm on the scale drawing represents an actual length of: A 20 m B 25 m C 50 m D 250 m E 500 m Exam tip   In this question, students are given the actual area of the estate. They are also given its area on a scale drawing. From this information a scale factor for area (k2 = 2500) can be determined. The majority of students apparently obtained this area scale factor but then incorrectly applied it directly to scaling the given length rather than first converting it into the corresponding linear scale factor, k. [Assessment report 2000]

18 A map’s scale is 1:4000. a If an area on the map is 5 cm2, what does this represent in real-life terms? Convert your answer to square metres. b If this area is a square and is to be used as a fun-park with rides and games, what are the dimensions of the park, rounded to 1 decimal place? It follows that if the area (a 2-dimensional property) ratio of similar figures is the side length ratio squared, then the volume (a 3-dimensional property) ratio of similar figures is the side length ratio cubed. In summary: side length ratio a:b area ratio a2:b2 volume ratio a3:b3 a

19 Draw the following cylinders into your workbook.

b

10 cm 6 cm

r = 5 cm r = 3 cm

a Copy and complete the following statements and calculations. Using the height dimensions:

Using the radius dimensions:

Side length ratio

     ?     :10

or

3:     ?     

Volume ratio

     ?     :103

or

33:     ?     

(Evaluate)

     ?     :1000

or

27:     ?     

(Cancel)

     ?     :125

Chapter 10  Shape and measurement

497

b If the volume of the larger cylinder is approximately 785.4 cm3, use the volume ratio to calculate the volume of the smaller cylinder to 1 decimal place: Volume of small cylinder = scale factor Volume of large cylinder 27 ? = Volume of large cylinder 125 eBook plus Digital docs

WorkSHEET 10.2 Investigation Making an aeroplane

10h eBook plus Interactivity

int-0811

Volume of small cylinder = ? cm3 c Check this value by using the formula: volume = area of cross-section × height. 20 a What is the volume ratio (in simplest form) between the cones shown at right? b If the volume of the larger cone (to 1 decimal place) is 56.5 cm3, use the volume ratio to calculate the volume of the smaller cone to 1 decimal place. c Check your answer using the formula for the volume of a tapered object.

3 cm 2 cm 6 cm

4 cm

Similar triangles We have seen the principles of similarity applied to scale drawings and 3-dimensional objects. In this section we shall concentrate on similarity between triangles and how this can be used to find dimensions in situations in which a direct measurement is impossible or inconvenient. There are three formal test rules that can be applied to triangles to see if they are similar.

Similar triangles

A

4.5 80n 3 40n 60n E F 4

80n 9

B

6

40n

60n 8

G

D

C

80n 9

H

6

40n

60n 8

I

Test 1 Triangles are said to be similar if all the corresponding angles are equal (abbreviated to AAA). That is, the three angles in one of the triangles are equal to the three angles in the other triangle. ∆ABC ∼ ∆DEF, because ∠A = ∠D ∆DEF ∼ ∆GHI, because ∠D = ∠G ∠B = ∠E ∠E = ∠H ∠C = ∠F ∠F = ∠I Test 2 Triangles are said to be similar if the ratios between the corresponding side lengths are equal (abbreviated to SSS). ∆ABC ∼ ∆DEF, because the ratio of corresponding side lengths is 2: AB 9 AC 6 BC 8 = = 2, = = 2, and = =2 DE 4.5 DF 3 EF 4 1 ∆DEF ∼ ∆GHI, because the ratio of corresponding side lenghs is : 2 EF 4 1 DE 4.5 1 DF 3 1 = = = = , = = , and HI 8 2 GH 9 2 GI 6 2 The exception ∆ABC and ∆GHI are not classified as being similar, but belong to a sub-group of similar triangles called congruent triangles.

498

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

Congruent triangles are identical in shape and size. Their corresponding angles are equal and the corresponding sides are in the same ratio, equal to 1; that is, the corresponding side lengths are exactly the same. The sign used to denote congruency is ≅, so ∆ABC ≅ ∆GHI. Test 3 Triangles are said to be similar if two of their D corresponding sides are in the same ratio (ratio value does not equate to 1) and the angle 18 between these two sides (the included angle) is A the same in both triangles (abbreviated to SAS). 6 ∆ABC ∼ ∆DEF, because the corresponding 60n 60n B 8 C E sides have the same ratio value that does not F 24 equate to one and the included angle for both Included angle triangles is the same: EF 24 ED 18 = = 3 and = = 3; ∠B = ∠E BC 8 AB 6 Sometimes it is hard to decide whether the two triangles are similar or not, because they are not orientated the same way. In such cases it is helpful to re-draw the triangles so that the sides and angles that we think might be corresponding are in the same order. Worked Example 21

Compare each of the following triangles with ∆ABC and state whether they are similar, congruent or there is not enough information given for a decision to be made. Justify your answers. a   ∆ DEF b   ∆GHI c   ∆ JKL d   ∆ MNO

N E 30n

1

B

20

110n

40n

C 40n

10

30n

J

10

40n 110n H 4.5 I

110n D

110n

20 7.5

F

6

G

M

6.6 110n K 40n O

L

think a

6.6

A

write/draw

Re-draw ∆DEF so that its angles correspond to those of ∆ABC.

a

B 2

Compare the corresponding angles and write what you observe.

3

Compare the side measurements given to test for congruency.

D

110n A 6.6 6

C 30n 10 40n

110n E

30n

20

40n

F

∠A = ∠D ∠B = ∠E ∠C = ∠F All corresponding angles are equal. FE 20 = =2 CB 10 2 ≠ 1, so the triangles are not congruent.

4

State whether the triangles are similar and specify the test upon which your conclusion is based.

∆ABC ∼ ∆DEF because all corresponding angles are equal (AAA).

Chapter 10  Shape and measurement

499

b

1

Re-draw ∆ABC and ∆GHI so that angles/ sides correspond.

b

A 6.6 B

2

30n

110n

H

6

10

40n

110n 4.5

C G

7.5 40n

I

AC 6 = HI 4.5

Compare the side measurements.

= 1.333 and BC 10 = GI 7.5 = 1.333

c

3

Compare the angles formed by the sides which have measurements in the same ratio (that is the included angles).

∠C = ∠I = 40°

4

State and justify your conclusion.

∆ABC and ∆GHI are similar because two pairs of corresponding sides are in the same ratio that doesn’t equate to 1 and the included angles are equal (SAS).

1

Re-draw ∆ABC and ∆ JKL so the angles/ sides correspond.

c

B

d

2

Check if the corresponding angles are equal.

3

As side measurements are given, check for congruency.

4

State your conclusion and justify.

1

Re-draw ∆ABC and ∆MNO so the angles/ sides correspond.

40n

6.6 C

J

110n 10

40n

L

∠J° = 180° − 150° = 30° ∠J = ∠B, ∠A = ∠K, ∠C = ∠L BC 10 AB 6.6 = = 1 and = =1 JL 10 KJ 6.6 Initially, the two triangles appeared to be similar because of AAA; however, as the side ratio equates to 1, these triangles are congruent. d

M

B 30n

None of the three tests (AAA, SSS or SAS) can be performed since we know the measurements of only one side and one angle in the ∆MNO. State this in writing.

10

30n

6.6

2

K

A 6.6 110n 6

A 110n 6 CN 10 40n

110n 20

O

Unable to determine whether the triangles are similar, as ∆MNO does not provide enough information to test AAA, SSS or SAS.

Being able to determine the corresponding angles in a triangle is vital to all three tests on similarity. In some cases the actual value of one of the angles is unknown, but by mathematical deduction its size can be found. This requires that you remember your angle properties. Some of these properties are shown in the following table.

500

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

Angle description 1. When two lines intersect, they form vertically opposite angles, which are equal.

Diagram an bn

bn an

2. When a transversal line cuts parallel lines, a number of angles are formed.

Transversal line Parallel lines

3. Alternate angles are on opposite (alternate) sides of the transversal and are always equal. bn

4. Corresponding angles lie on the same side of the transversal and both are either below or above the parallel lines. They are always equal.

an bn an

cn dn an bn cn dn an bn

5. Co-interior angles lie on the same side of the transversal and within the parallel lines. They always add up to 180°.

dn

cn an bn

an bn 180n cn dn 180n

WorkEd ExaMPlE 22

Show that ∆ABC ∼ ∆ADE, because: a the corresponding angles are equal b the corresponding side ratios are equal (to a value other than 1).

eBook plus

7.5 A

1

C

10

think a

3.75 D B 4.8

Tutorial

7.2 5

Draw the triangles separately with angles marked and side measurements shown.

a

7.5 bn

B an 10

2

State the corresponding pairs of angles which are equal in size. Specify the reason.

3

Write your conclusion. Evaluate the ratio between the corresponding sides.

D

4.8 cn C 11.25 A

1

E

WritE

A

b

int-0909 Worked example 22

bn

an 15

7.2 cn E

∠A = ∠A (shared) ∠B = ∠D (corresponding) ∠C = ∠E (corresponding) ∴ ∆ABC ∼ ∆ADE using AAA test. b AD = 11.25 AB 7.5 = 1.5 AE 15 = AC 10 = 1.5 DE 7.2 = BC 4.8 = 1.5

Chapter 10

Shape and measurement

501

2

Write your conclusion.

The corresponding side lengths of the triangles are in the same ratio. (The ratio is not equal to 1.) ∴ ∆ABC ∼ ∆ADE (SSS test).

If, in a pair of similar triangles, the lengths of at least one pair of corresponding sides are given, the ratio (scale factor) can be established. It can then be used to find missing lengths in one of the triangles, provided that the corresponding lengths in the other triangle are known. Worked Example 23

Find the value of the pronumeral in the figure at right. think

write

1

Establish whether the triangles ABC and ADE are similar by applying the AAA test.

2

Since the triangles are similar, their corresponding sides are in the same ratio. Find this ratio using the corresponding pair of sides whose lengths are known. The missing length AD in ∆ADE corresponds to the side AB in ∆ABC. Use this information to form the ratio equation. Identify the values of AD and AB. Substitute the values of AD and AB into the equation and solve for x.

3

4 5

x 7.5

D B C

10

∠A = ∠A (shared) A ∠B = ∠D (corresponding) ∠C = ∠E (corresponding) ∴ ∆ABC ∼ ∆ADE (AAA test) AE 15 = AC 10 = 1.5

15

E

AD = 1.5 AB AD = x, AB = 7.5 x = 1.5 7.5 x = 1.5 × 7.5 = 11.25

Note: When forming an equation, place the unknown value in the numerator to make calculation steps easier or use a CAS calculator to solve the equation. REMEMBER

1. There are three formal tests to identify similar triangles. Test 1:  Test that all corresponding angles are equal (abbreviated to AAA). Test 2: Test the ratio between corresponding sides. This ratio needs to be equal for all pairs and cannot equate to 1. It is abbreviated to SSS. Test 3: This test involves two pairs of corresponding sides and the angle formed by these sides (the included angle). For the condition SAS to hold true the ratios of the sides must be equal (though not 1) and the included angle should also be equal. 2. Congruent triangles are a sub-group of similar triangles. They have all angles equal and the side values are exactly the same. 3. When comparing triangles it is best to have them drawn with the same orientation. 4. When forming an equation to solve for a missing dimension, place the unknown value in the numerator to make the calculation steps easier or use a CAS calculator to solve the equations.

502

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

Exercise

10H

Similar triangles   1   WE21  For each of the following, compare ∆ABC and ∆EFD and state whether they are similar, congruent, or there is not enough information given for a decision to be made. Justify your answers. a

c

b

A

D 120n 5.5 6 4 E 20n 8.25 40n C B 120n 20n

B 3.375 A 7.875 2.5 5.625 E

F

B

6 cm 25n 40n 3 cm A

d D 3.5

1.5

F E

110n

12 6

6 cm

3 cm

40n 6

D A

F

D 115n 40n E C

C 8

F

30n

B

C

  2 Calculate the size of the missing angles. Justify your answer. a 24n

d

78n an

cn

an bn

b 68n

e

cn dn

bn

en

xn

an

f

99n

cn

bn

yn 63n

en an

•

dn en

cn

32n

bn

an

dn

bn

c

cn

32n

dn

58n

  3   MC  The magnitude of angle a° is: A 30° B 180° − 30° D 180° − 60° E 70°

C 60°

  4   MC  The magnitude of angle a° is: A 64° B 128° D 154° E 104°

C 52°

30n

an

64n

an

  5   WE22  Show that ∆ABC is similar to ∆DBE because: a the corresponding angles are equal b the corresponding side ratios are equal (to a value other than 1).

B 6 cm D 4.2 cm A

4 cm 6.8 cm

5 cm E 3.5 cm C

Chapter 10  Shape and measurement

503

  6 In each of the following diagrams, find and re-draw two triangles that are similar. Give reasons for your answer. c a A b A C A

B

B D

E

d

C

C

E

D

e

E

D

B

E

D

A D

A C

B

B

C

  7   WE23  Find the values of the pronumerals (to 1 decimal place). c A a b 2.4 37 cm y

5

9.6 cm

x

B

18

* 6.2 cm 15

6

x

B

e

A

C 5 D y

B

4 E

1.7 m

C

x

D

4m

f

3.4 m

d A

C

x

y

x

*

y

12.4

9.6

31 cm

12

E

A 30n

2 cm

E

D

xn

yn

B 3 cm

C

z

D 9 cm E

g

22n37' A

h

i

15

r=8

x y z 36

9

9

24

x

y r=4

C

x yn D 15

B

  8   MC  The side lengths of triangle ABC are 18 mm, 24 mm and 30 mm. Triangle DEF is similar to triangle ABC and its shortest side is 12 mm. The perimeter of the triangle DEF is: A 48 mm B 52 mm C 64 mm D 72 mm E 108 mm   9 It just happens that you always carry a wooden rod 1.5 m in length and a tape measure. For each of the following situations, draw a diagram showing the two similar triangles and calculate the height of the vertical object. a The clearance sign is missing from a low bridge over a road. The sun has created an image of the bridge and its opening on the ground. The image of the opening is 1.44 m in length. The semi-trailer you are driving is 2.5 m high. You step out of the truck and place your trusty rod on the ground, producing a 90-cm shadow. Will you need to find an alternative route or can you proceed? b The measuring tape for the pole vault height is missing. How high did Emma George pole vault if the pole vault creates a shadow of 2.4 m on the ground, while your wooden rod creates a shadow of 80 cm?

504

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

c Does the fire brigade need to bring an extension ladder longer than 15 m to rescue people from the top of a high-rise building, if the shadow created by the building is 24 m and the shadow created by your pole is 2 m? d Slam-dunking a basketball ring can be dangerous; people have been injured by the ring and back-board collapsing. You have always wanted to know how high the basketball players had to jump to touch the top of the ring. Your wooden rod held next to the basketball pole created a shadow half the length of the basketball pole’s 1-m shadow.   If the ring is 45 cm from the top of the board, how far up from the ground do the players have to reach to touch the ring?

10 A triangle with sides 12 cm, 24 cm and 32 cm is similar to a smaller triangle that has a longest side measurement of 8 cm. Draw a diagram to represent this situation and then calculate the perimeter of the smaller triangle. 11 A curved section of a fun-ride is to be supported by similar triangles. a If the dimensions of the first and last triangle are as shown, calculate the dimensions of the triangle in between. b The construction people decide to make the track even steeper by extending the track with one more similar triangle. What are the dimensions of the next triangle?

13.75

2.2

1.5 x y

1

12 At your 18th birthday party you want to show slides of you and your friends growing up. The diagram at right shows the set-up of the projector lens, slide and projector screen. If:  (a)  a slide is 5 cm × 5 cm     (b)  the distance from A to B is 10 cm      (c)  the projector screen is 1.5 m × 1.5 m how far horizontally from the screen do you need to place the projector’s lens so the image just covers the entire screen?

z

9.375 6.25

A

Slide

B Lens

Projector screen

Chapter 10  Shape and measurement

505

10i

Symmetry A 2-dimensional shape may be described as having either line symmetry and/or rotational symmetry.

line symmetry If it is possible to cut a 2-dimensional shape in such a way that it is divided into two mirror images, then it possesses line symmetry. The line that allows this to occur is called the axis of symmetry. Shapes can have more than one line of symmetry:

One line of symmetry

Two lines of symmetry

Infinite number of lines of symmetry

Three lines of symmetry

An easy way to check whether the line is an axis of symmetry of a certain shape is to fold the shape along that line. If the two parts coincide (that is, the shape folds onto itself), the line is the axis of symmetry. WorkEd ExaMPlE 24

eBook plus

Which of the dotted lines in this figure is an axis of symmetry? think 1

E

A

Tutorial

int-0910

WritE D

C

A

Visualise the triangle as being folded along each dotted line. If the triangle folds onto itself the line is an axis of symmetry.

Worked example 24

F B B

2

AB is an axis of symmetry. CD and EF are not.

E C

D F

AB is an axis of symmetry.

rotational symmetry If a shape can be rotated about a point so that it appears as it did in its original position at least once in less than one complete revolution of 360°, then it is said to have rotational symmetry. The number of times a shape appears as the original in one rotation is called its order of symmetry.

506

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

For example, a square possesses rotational symmetry. By marking one corner and rotating the square in a clockwise direction around point × (its centre of rotation), it can be seen that in a turn of 90°, the square appears as it did originally. r

=

Original

Rotated on itself 90n

Rotated on itself 180n

Rotated on itself 270n

Rotated on itself 360n

With each subsequent turn of 90° the square appears as it originally did. Within the 360° revolution there were four times the square appeared as its original, so a square’s order of rotational symmetry is said to be 4. An equilateral triangle has an order of rotational symmetry of 3. 120n

120n  120n

The principles of line and rotational symmetry can also be applied to 3-dimensional objects.

Planes of symmetry Planes of symmetry for 3-dimensional figures are like lines of symmetry for 2-dimensional figures. Planes of symmetry slice through a 3-dimensional object so that each half is a mirror image of the other. Some 3-dimensional objects may have one plane of symmetry, while other objects can have more than one.

This scalene triangular prism has only one plane of symmetry.

This equilateral triangular prism has four planes of symmetry.

Axes of symmetry An axis of symmetry is a line about which an object can rotate so that it assumes positions identical with those of the original. Like rotational symmetry in 2-dimensional shapes, a 3-dimensional shape can be described by its order of rotational symmetry. This is the number of times the shape assumes an appearance identical with that in the original position, within one revolution of 360° about its axis of symmetry. A square-based pyramid has an order of rotational symmetry of 4. In a square-based pyramid there is only one axis of symmetry; however, other 3-dimensional objects can have none or many.

Chapter 10  Shape and measurement

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An object’s order of rotational symmetry can be affected by the axis chosen. For example, if a cube is rotated around the vertical line, passing through the centre of the top face, its order of rotational symmetry is equal to 4.

Order of rotational symmetry  4 

Order of rotational symmetry  2

If, however, the same cube is rotated around its diagonal, the order of rotational symmetry is then 2.



tessellations

eBook plus Digital doc

Extension — Tessellations

There are many examples in design, art and architecture where shapes are used to cover a surface with patterns which leave no gaps. These patterns are referred to as tessellations. Tessellations can be divided into three main categories: regular, homogeneous (or semiregular) and non-homogeneous (irregular). 1. In regular tessellations, the pattern is made by only one type of regular polygon. 2. In homogeneous tessellations, the pattern is made up of different types of regular polygon. 3. In non-homogeneous tessellations, any type of shape can be used. The subject of ‘tessellations’ is also presented in chapter 6 ‘Geometry in two and three dimensions’ (pages 287–89).

non-homogeneous (irregular) tessellations This category is not restricted to using regular polygons, so any shapes that will fit together to cover a plane surface can be used. The shapes still need to form a definite pattern that is repeated to cover the surface.

rEMEMBEr

1. The line that divides a 2-dimensional shape into two parts that are mirror images of each other is called the axis of symmetry. 2. The plane that divides a 3-dimensional shape into two halves that are mirror images of each other is called a plane of symmetry. 3. The number of lines/planes of symmetry can be 0, 1, more than 1 (for example, 2, 3, …) or infinite. 4. If a 2-dimensional shape can be rotated about a fixed point, or a 3-dimensional shape can be rotated about a line (the axis of symmetry) so that it assumes positions identical with those of the original at least once in less than one complete turn, the shape is said to have a rotational symmetry. 5. The number of times the shape assumes an appearance identical with that in the original position within one revolution is called an order of rotational symmetry. 6. In 3-dimensional shapes, an order of rotational symmetry can be affected by the axis chosen.

508

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

Exercise

10I

Symmetry   1   WE24  Which of the dotted lines in each of the figures shown is an axis of symmetry? a

A

E

G

b

A E

C

D

H

c

E

A

C

d

D

A

B

H F

E G

C

F

B

C

H

G

H

G

F

B

D

B

D

F

  2 Copy the following figures into your workbook and draw in all the possible axes of symmetry. a

b

c

d

e

f

  3 a  Draw a regular pentagon with sides 1.5 cm in length and internal angles all accurately measuring 108°. b Draw in all axes of symmetry. The point where these lines intersect in a regular polygon is known as the centre of rotation. Use this point to rotate the pentagon. c Through what angle must the pentagon be rotated before it superimposes itself for the first time? (It is important to measure the angle very carefully to the nearest degree.) d How many times does the pentagon superimpose itself when turned 360°? e What is its order of rotation?   4 Copy and complete the following table by answering questions 3a to e for each of the regular polygons listed. (A regular polygon has all sides of equal length and all angles of equal size.) Add two extra rows at the foot of the table for the answer to question 5b .

Chapter 10  Shape and measurement

509

Polygon (regular)

Number of sides

Number of lines of symmetry

Angle of first superimposition

Order of rotation

Triangle (angle size 60°) Square Pentagon (angle size 108°) Hexagon (angle size 120°) Octagon (angle size 135°)   5 a Write a brief paragraph describing in general how you can determine the number of lines of symmetry, the angle of first superimposition and the order of rotation of a regular polygon with any (n) number of sides. b Using the concepts you wrote about in part a , add to the table in question 4 a regular nonagon and a regular decagon and fill in their details.   6 For each of the following shapes, state its order of rotational symmetry. a

b

c

d

e

f

  7 Which of the following cuts are planes of symmetry? a

b

c

d

e

f

  8 The diagram at right shows a regular octahedron. a How many planes of symmetry does it have? b How many axes of symmetry does it have and what is its order of rotational symmetry?

510

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

E A

B D F

C

  9 Make a square pyramid by copying onto paper or thin cardboard, the net for the square pyramid shown.

Cut around the edge of the net, fold along the dotted lines and then stick the sides together with adhesive tape. a How many planes of symmetry and axes of symmetry does a square pyramid have? b What is its order of rotational symmetry? 10 Create a table with the following headings. Vertical symmetry

Horizontal symmetry

Rotational symmetry

None

Determine to which column or columns each of the capital letters of the alphabet (shown below) belongs.

ABCDEFGHIJKLMNOPQRSTUVWXYZ Use the table of information from question 10 to answer the following questions.

11 Some words possess rotational symmetry, in that if they are rotated 180° they still match the original word appearance.

180n

Other words can possess a line of symmetry. a Copy the following words into your workbook and determine if they have a line of symmetry.

If so, draw in the line or lines of symmetry. b Use your table of alphabet symmetry information to try to find other example words of three letters, four letters and five letters for each kind of symmetry (horizontal, vertical and rotational). c When placed in front of a mirror, letters with a vertical line of symmetry appear __________, while letters with a horizontal line of symmetry look __________. d Try to write a sentence that would look the same in a mirror.

Chapter 10  Shape and measurement

511

Summary Pythagoras’ theorem in two dimensions

• In any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides: a (length of hypotenuse)2 = (length of side 1)2 + (length of side 2)2 Side 1 or c2 = a2 + b2

Hypotenuse c b Side 2

Pythagoras’ theorem in three dimensions

• Use the following steps when solving 3-dimensional problems. Step 1.  If not supplied, draw a diagram of the situation. Step 2.  Label all vertices in the diagram. Step 3.  Identify the length that needs to be found. Step 4.  Identify the triangle that contains the unknown length. Step 5.  Re-draw this triangle in two dimensions. Step 6.  Use Pythagoras’ theorem to find the unknown length. Step 7. If the question is presented in words, write an answer sentence, rounding the value appropriately for the situation. • If the triangle containing the unknown length has other measurements missing, find these missing values first from other triangles within the figure. Perimeter and area

• Generally: 1. Unless specified, do not round units until the end of the solution. 2. When solving worded questions:   (a)  draw a diagram   (b)  write a comment line on your plan of action   (c)  write the formula   (d)  allocate a value to the pronumeral(s)   (e)  substitute the value(s) into the formula and evaluate   (f)  write an answer sentence. 3. If the measurements in question include units, include them in your final answer. • The perimeter is the distance around a closed 2-dimensional shape. • Area refers to the space enclosed by the boundaries of a 2-dimensional shape. Shape

Area

Square

Perimeter

A = L2

P = 4L

A=L×W

P = 2(L + W)

L

Rectangle

L W

512

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

Shape

Area

Perimeter

A=b×h where the height measurement must be at a right angle to the base measurement.

Parallelogram  h

P = sum of all sides

b

Trapezium

a

1

P = sum of all sides

1

P = sum of all sides

1

P = sum of all sides

A = 2 (a + b)h where the height measurement must be at a right angle to the base measurement.

h b

A=2 x×y

–

Rhombus

x

–

–

–

y

Triangle  h

A = 2 bh where the height measurement must be at a right angle to the base measurement.

h

b

b h

b

Triangle

c

b a

A = s(s − a)(s − b)(s − c) 1 where s = 2 (a + b + c) (Use when height measurement is unknown.) A = π r2

Circle r

P=a+b+c

C = π d or C = 2π r

d

• The formula for the area of a parallelogram, triangle and trapezium require the height to be perpendicular to the base measurement. • An annulus is the area between two concentric circles. Area of annulus: A = π (R2 − r2), where R is the radius of the larger circle and r is the radius of the smaller circle. • A sector is formed between two radii of the circle and a length on the circumference. θ Area of a sector A = πr2 360° θ Arc length of a sector = × 2π r 360° θ = πr , 180° where θ is the angle of a sector and r is the radius of the circle. • To find the area/perimeter of composite shapes, calculate the area/perimeter of the individual shapes first.

Chapter 10  Shape and measurement

513

Total surface area (TSA)

• Total surface area is the sum of the surfaces of a 3-dimensional object. Object

Net

TSA TSA = 6L2

Cube 2 1

4

3 6

L

5

L

L

Rectangular prism h h

w

l

TSA = 2(wh + lw + lh)

l h

l

w

w h

Cylinder

TSA = area of 2 circles + curved surface = 2π r2 + 2π rh = 2π r(r + h)

r h

h

r

2 Pr r

TSA = 4π r2

Not shown

Sphere

Cone

S S  Slant height

h

2P r r

r

514

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

TSA = area of base (circle) + area of curved surface 2π r = π r2 + × π S2 2π S = π r2 + π rS = π r(r + S)

Object

Net

TSA

Note: The slant height of the cone is formed from the radius of the circle, so their values are the same. The arc length of the sector used to form the cone becomes the circumference of the base of the cone, so their values are the same. Curved surface of a cone is formed by removing a sector out of a circle. Arc length

Slant height  radius of sector

Major sector rS Minor sector

Circumference of base  arc length of sector

TSA = area of square + area of 4 triangles 1  = b2 + 4 ×  bh

Square-based pyramid h

= b2 + 2bh

h

2

b b

• Formulas for all types of objects are not possible. For those objects without a formula: 1. draw the net of the object 2. work out the different shapes that make up the net 3. calculate their individual areas 4. add all the individual parts together. • When finding the TSA of a composite figure, do not include the areas of the inside surfaces (that is, the surfaces of contact of the individual shapes that make up the figure). Volume

• The volume of an object is the amount of space that the object occupies. It is measured in cubic units. • Prisms are 3-dimensional objects with uniform cross-sections and parallel sides. The cross-section is represented by the ends of the prism. The height is the dimension perpendicular (at right angles) to the cross-sectional area. Volume of a prism = cross-sectional area × height of the prism Shape

Cross-sectional shape

Cylinder r

Volume V = area of a circle × height = π r2 × H

H

Area = π r2

r

Triangular prism h h b

H

b 1

Area = 2 bh

V = area of a triangle × height 1 = 2 bh × H Note: Lowercase h represents the height of the triangle. (continued)

Chapter 10  Shape and measurement

515

Shape

Cross-sectional shape

Rectangular prism

Volume V = area of a rectangle × height =L×W×H

W L

H

Area = L × W

W

L

V = area of a square × height = L2 × H = L2 × L = L3 (since in a square, H = L)

Cube

H

L

Area = L2

L

• Tapered objects have one flat end (known as the base) and one pointed end (called the apex). The height dimension in the formula for a tapered object must be at right angles to the base (flat end). 1

Volume of a tapered object = 3 × area of base × perpendicular height Shape

Flat end (base) shape

Cone

Volume 1

V = 3 × area of a circle × height

r

1

V = 3π r2 × H

H r

Square pyramid

1

V = 3 × area of a square × height 1

V = 3L2 × H

H L L

Rectangular pyramid

W

1

V = 3 × area of a rectangle × height 1

L

= 3L × W × H

H W

L

Triangular pyramid

1

V = 3 × area of a triangle × height

h H b

h

b

1  V = 3  2 bh × H Note: Lowercase h represents the height of the triangle. 1

• Do not confuse the lowercase h in the formula for a triangular prism/pyramid or trapezoidal prism with the uppercase H. Uppercase H = total height of the prism or pyramid, while lowercase h is the height of the triangle, parallelogram of a trapezium.

516

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

4

• For a sphere: Volume = 3 π r3 Capacity

• The capacity of a container is the amount that it can hold. • To find the fluid capacity of a container, find its volume in cubic units first and then convert, using the following rules: 1 cm3 = 1 mL, 1 m3 = 1 kL = 1000 L Similar figures

• Similar figures have the same shape but different sizes. The corresponding angles in each similar figure are the same; however, the corresponding side lengths differ by a scale factor. • The scale factor can be written as: 1. 2 cm to 10 m 2. a ratio, provided the units are the same 2 cm to 10 m 2 cm to 1000 cm 2:1000 1:500 (The first number in a ratio represents the measurement on a scale drawing; the second represents the reallife measurement.) 3. a fraction

1 500

  or  

500 1

• When forming a rule the following format is usually used: scale measurement Scale factor = real-life measurement where both scale and real-life measurements are in the same units. Similar triangles

• Triangles are similar if: 1. all corresponding angles are equal (AAA), or 2. all corresponding sides are in the same ratio (SSS), or 3. two sides are in the same ratio and the included angles are equal (SAS). • Congruent triangles are a sub-group of similar triangles. They have all angles equal and the side values are exactly the same. • When comparing triangles it is best to have them drawn with the same orientation. • When forming an equation to solve for a missing dimension, place the unknown value in the numerator to make the calculation steps easier. Symmetry

• The line that divides a 2-dimensional shape into two parts that are mirror images of each other is called the axis of symmetry. • A 2-dimensional shape can be classified by the number of lines of symmetry it possesses and the number of times the shape superimposes itself when rotated around a fixed point, called the order of rotational symmetry. • The plane that divides a 3-dimensional shape into two parts that are mirror images of each other is called the plane of symmetry. • A 3-dimensional shape can be classified by the number of planes of symmetry it possesses and the number of times the shape superimposes itself when rotated around the chosen axis of symmetry (its order of rotational symmetry). • The axes chosen can affect the order of rotational symmetry in a 3-dimensional shape.

Chapter 10  Shape and measurement

517

chapter review Multiple choice

  1 Examine the diagram. The value of the pronumeral to 1 decimal place is: A 6.2 C 5.8 E 5.4   2

12 5

x

B 6.0 D 5.6

9

r 50 cm w 80 cm

d cm depth of water in the pipe

The cross-section of a water pipe is circular with a radius, r, of 50 cm, as shown above. The surface of the water has a width, w, of 80 cm. The depth of water in the pipe, d, could be: A 20 cm B 25 cm C 30 cm D 40 cm E 50 cm Exam tip   The question required students to determine the depth of water in a circular pipe, given its surface width. Only 22% of students were successful in choosing option A. Option C, which corresponded to the distance of the surface from the centre of the pipe, was given by 39% of students. A further 22% of students wrongly assumed that the cross-section of the surface was semicircular and chose option D. [Assessment report 2006]

[©VCAA 2006]

  3 The longest stick that can be placed inside a cube with side length of 2 m has the length (to 2 decimal places) of: A 3.46 m B 3.44 m C 3.50 m D 3.48 m E 3.52 m   4 The perimeter of a rectangle with an area of 56 cm2 and one side of 8 cm is: A 30 cm B 7 cm C 15 cm D 28 cm E 26 cm

518

  5 The area of a one-metre wide gravel path around a lawn 25 m × 15 m is: B 80 cm2 A 84 cm2 C 78 cm2 D 76 cm2 2 E 82 cm 8 cm   6 Examine the diagram. The height of the shape shown is: A 23.5 cm B 19.5 cm 16 cm C 25.5 cm D 20.5 cm E 15.5 cm 7 mm   7 The area of the shaded section in this diagram is: A 12.2 cm2 B 141.4 cm2 C 1110 mm2 D 11.0 cm2 2 cm E 1414 mm2 2   8 The TSA of a rectangular prism is 126 cm . If the base of the prism is 6 cm × 5 cm, what is its height? A 2.5 cm B 3 cm C 3.4 cm D 3.5 cm E 4 cm   9 Examine the diagram at right. The TSA of the solid shown is given by: A 3π m2

B

 m C 5π    2

2

 m E 2π  2 

2

m

3

D 4 π m2

10 A solid cylinder has a height of 30 cm and a diameter of 40 cm. A hemisphere is cut out of the top of the cylinder as shown at right. In square centimetres, the total surface area of the remaining solid (including its base) is closest to:

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

3π m 2 2

30 cm

40 cm

A 1260 C 6280 E 10050

B 2510 D 7540

Exam tip   To ensure that they include all of the surfaces involved, students might find it helpful to begin solution to questions such as this by writing down a statement like: total surface area = surface area of the hemispherical bowl + surface area of the side of the cylinder + surface area of the base of the [Assessment report 2007] cylinder.

[©VCAA 2007]

11 Examine the diagram at right. The volume of the shape shown is given by: 1

4

A 3π m2n + 3 π m2

m

n

2

B 3 π m2 + π m2n C π m2n 1

D 3 π n(m2 + 2) 1

E 3 π m2(n + 2m) 12 A cylindrical container of radius r and height h is being filled with sand using a conical container, also of radius r and height h. How many cones of sand are needed to fill the cylinder to capacity? A π  B 3π  C 3 E

D

1 3

π 3

13 The rectangular box shown in the diagram is closed at the top and at the bottom. It has a volume of 6 m3. The base dimensions are 1.5 m × 2 m. The total 2m surface area of this box is: 1.5 m A 10 m2 B 13 m2 C 13.5 m2 D 20 m2 E 27 m2 [©VCAA 2006] 14 A rectangular swimming pool has a capacity of 1000 kL. If the pool is 50 m long and 2 m deep, how wide is it? A 10 m B 20 m C 25 m D 50 m E 100 m

15 If the volume of a container is 1000 mm3, what is its capacity in mL? A 0.1 mL B 1 mL C 10 mL D 100 mL E 1000 mL 16 The volumes of two similar-shaped cylinders are 1000 cm3 and 3375 cm3. In simplest form, what is the ratio of their surface areas? A 10:15 B 4:9 C 4:6 D 2:3 E 9:4 17 On the architect’s plan, a house block is 7.5 cm wide, while the actual (real-life) width of the block is 60 m. What is the ratio scale of the plan? A 1:80 B 8:1 C 1:8 D 1:800 E 80:1 18 The length of x is: A 4 cm B 16 cm 8 6 C 10 cm D 8 cm 12 E 6 cm x cm 19 The lengths of x and y, 12 24 10 respectively, are: x A 12 m and 14 m 15 y B 13 m and 16 m C 18 m and 16 m D 16 m and 18 m E 18 m and 14 m 20 A tree casts a shadow of 10 m, while your 0.45-m wooden stick casts a shadow of 100 cm. What is the height of the tree? A 45 m B 4 m C 4.5 m D 5 m E 5.5 m 21 The number of planes of symmetry of the figure shown is: A 1 B 2 C 3 D 4 E 5 22 The number of axes of symmetry of the figure shown is: A 2 B 5 C 7 D 3 E 9

Chapter 10  Shape and measurement

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short answer

1 To allow people access to the gym, Ramp the manager has decided to place a 16.5 cm ramp over a set of 27 cm stairs. What length of timber (to the nearest centimetre) does he need to purchase? 2 A 2-m surfboard is to be placed in a locker. Will it fit down the side shown by the line AB or will it have to go diagonally across as shown by the line CB? 3 Find (to 1 decimal place): i the perimeter ii the area of the following shapes. b a

A

C

2.4 cm r

All walls, floor and top are 0.25 m thick

3.5 m

a What is the internal radius, r, of the tank? Exam tip   An incorrect value of 1.75 was a common response. [Assessment report 2006]

1.8 m

B

0.62 m

D 0.62 m

b Determine the maximum amount of water this tank can hold. Write your answer correct to the nearest cubic metre. Exam tip   Many students did not find the correct internal height (1.9 m) of the tank, which was necessary for a method mark. [Assessment report 2006]

220n35'

8.2 cm

c

2m

6 cm 15.67 cm

4 To secure the 2.4-m-high poles of a beach volleyball net you need to attach a guy rope to a ring seven-eighths of the way up each pole. The guy ropes are 2.5 m in length. a Draw a diagram of the situation and include all dimensions given. b How far out (to 1 decimal place) from the base of the poles will the guy ropes go? 5 A designer vase has the shape of a truncated, square-based pyramid. The base of the vase is a square with a side length of 15 cm. The area of the square opening is 70.56 cm2. Each of the 4 sides is a trapezium with slant sides 9 cm long. Find (to the nearest square centimetre) the total surface area of the vase. 6 A closed cylindrical water tank has external diameter 3.5 metres. The external height of the tank is 2.4 metres. The walls, floor and top of the tank are made of concrete 0.25 m thick.

520

[©VCAA 2006]

7 If the volume of a cone of height 10 cm is 261.8 cm3, show that this volume is increased by a factor of 8 if the dimensions of the cone are doubled. 8 A cylindrical can of drink holds 375 mL of liquid. If the area of the base is 31.25 cm2, find the height of the can. 9 The shapes below have either been enlarged or reduced by a scale factor from a fixed point. In both cases, calculate by what scale factor the original will need to be multiplied to produce the image. b a 10 cm 1.5 cm 1 cm 0.4 cm

6 cm

5 cm

8 cm

Note: The image is the shaded area. 10 Two rectangles are similar. The larger rectangle’s dimensions are 24 cm × 12 cm. If the longest side on the smaller rectangle is 16 cm, calculate the other dimension. 11 A circle with a radius of 21 cm has an area of 1385.4 cm2 and another circle has an area of 153.9 cm2. Using the principles of ratios, what is the radius of this circle to the nearest centimetre?

Maths Quest 11 Standard General Mathematics for the Casio ClassPad

12 Find the length of x in the diagram shown.

A B 12 cm

x

D

16 cm

C

13 a  Copy the chessboard at right into your workbook and draw in the lines of symmetry. b Rotate the board around its centre of rotation (the point where the lines of symmetry cross). What is its order of rotation?

Extended response

1 Lena and Alex are renovating their bathroom. Inspired by a creative exhibit from the recent Home Show, they decided to use various geometric shapes in their design. The ‘window wall’ is the first section to be renovated. The old window is to be replaced with two new windows as shown below and then the wall is to be tiled.

1.2 m 2.5 m

The window on the right is in the shape of an equilateral triangle, while the one on the left is in the shape of a trapezium. The shorter parallel side of the trapezium is the same length as the sides of the triangular window and the longer parallel side is equal to 2.5 m. Both windows are to be 1.2 m high. Calculate: a the perimeter of each window to the nearest centimetre b the cost of wood (to the nearest dollar) needed to frame the windows, if the wood is sold at $13.50 per metre c the area of each window d the total cost of the glass (to the nearest dollar), priced at $45 per m2. The ‘window wall’ is to be tiled with glass tiles in the shape of a rhombus as shown at right. 20 cm e Will the tiles tessellate? Give reasons for your answer and draw a small segment of the pattern to illustrate it. 30 cm f Small spaces between the tiles are to be filled with a special grout. Find the length (to the nearest centimetre) around each tile that needs to be filled with grout. g Find the area of each tile. h If the ‘window wall’ is 4.6 m × 2.7 m, find the total area that needs to be tiled. i The tiles are sold in boxes of 12. Use the results from parts g and h to find the number of boxes that Lena and Alex need to purchase. Add an extra 10% for breakages to the number of tiles needed. 2 After successfully coping with the ‘window wall’ (see question 1), Alex and Lena decide to improve other parts of their bathroom. First they want to decorate other walls with mosaic features as shown below. 60º 40 cm 25 cm 40 cm

40 cm

a Find the area of each feature to the nearest cm2. b If Lena wants 6 of each type of feature, find the total area of the mosaic needed; then add 10% extra for breakages and cutting.

Chapter 10  Shape and measurement

521

c Mosaic can be purchased in square sheets 30 cm × 30 cm at $22 each. (Each sheet can then be cut as needed, but only Mirror cm whole sheets can be purchased.) Find the total cost of the 50 Shelves mosaic features. 50 cm Next Alex wants to replace the old mirror with a new one as shown at right. The mirror is in the shape of a major segment, that is, a circle with a section of it cut off straight. (The ‘cut-off’ section will be complemented by small glass shelves.) d If both the radius and the length of the ‘cut-off’ edge are 50 cm each, find the total area of the mirror (to the nearest cm2). e Alex wants to have the edges of the mirror bevelled. This can be easily done at the local glass store at a cost of $20 per metre. Find the cost of the bevelling job. The most daring feature of the new bathroom is the extremely modern looking shower screen. It can be thought of as a section of the curved surface of a cylinder 1.4 m long. The curved edge is in the shape of an arc, Curved subtended by a 120° angle from the centre of edge r = —23 Curved 2 edge 120º a circle with a radius of m. 3 1.4 m f Find the length of the curved edge. g Find the total area of the glass in the shower screen. Bela the builder advises Lena and Alex that since they want a frameless screen, it has to be made from a special toughened 10-mm-thick glass. h What fraction of a cubic metre is the volume of glass in this unusual shower screen? 3 A chocolate manufacturer has brought out a new selection of chocolate shapes. They are filled with a delicious honey/chocolate liquid. 1.5 60n 3

2 Cone

1.5

1.5 Sphere

Two triangular prisms joined together 1.5

1.5 Cube

2.5 Rectangular prism

1.5

1.5 1.5 1.5 3 Trapeziodal square-based prism

a For each chocolate shape, calculate: i the TSA of chocolate required ii the volume of honey/chocolate liquid required (to 1 decimal place). b If a box contains four of each type of chocolate shape, what is the total area of chocolate needed to make one box? c Chocolate comes in square sheets of 50 cm × 50 cm. How many boxes can be produced per sheet of chocolate?

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Maths Quest 11 Standard General Mathematics for the Casio ClassPad

d How many millilitres of honey/chocolate liquid are required per box? e If the liquid comes in one-litre containers, how many boxes of chocolate can be produced per litre? 4 Tessa is a student in a woodwork class. The class will construct geometrical solids from a block of wood. Tessa has a piece of wood in the shape of a rectangular prism. This prism, ABCDQRST, shown in Figure 1, has base length 24 cm, base width 28 cm and height 32 cm.

T Q

S R

32 cm

D A 24 cm

C B

28 cm

Figure 1 On the front face of Figure 1, ABRQ, Tessa marks point W halfway between Q and R as shown in Figure 2. She then draws line segments AW and BW as shown.

Q

W

R 32 cm

A

B 24 cm

Figure 2 a Determine the length, in cm, of QW.

Exam tip   A number of students inappropriately tried to apply trigonometric ratios or Pythagoras’ theorem to this length, which is half of AB, since W is a midpoint. [Assessment report 2007]

b What fraction of the area of the rectangle ABRQ does the area of the triangle AWB represent? Tessa carves a triangular prism from her block of wood. Using point V V, halfway between T and S on the back face DCST of Figure 1, she W constructs the triangular prism shown in Figure 3. D B

A 24 cm

32 cm C

28 cm

Figure 3 c Show that, correct to the nearest centimetre, length AW is 34 cm. d Using length AW as 34 cm, find the Exam tip   The solid in this question is a triangular prism where the total surface area, in cm2, of triangle does not have a right angle. A significant number of students the triangular prism ABCDWV in used an incorrect formula that had obviously been copied from a text and Figure 3. 2 2 failed to score any marks. The formula TSA = bh + bl + hl + l b + h applies only to a right triangular prism. Further, the question directed that a length AW = 34 cm had to be used. If this length was not used at all, then no marks were available.

[Assessment report 2007]

Chapter 10  Shape and measurement

523

Tessa’s next task is to carve the right rectangular pyramid ABCDY shown in Figure 4 at right. She marks a new point Y halfway between points W and V in Figure 3. She uses point Y to construct this pyramid.

Y 32 cm D

C 28 cm

A

B

24 cm

Figure 4 e Calculate the volume, in cm3, of the pyramid ABCDY in Figure 4. f Show that, correct to the nearest cm, length AY is 37 cm. g Using AY as 37 cm, demonstrate the Exam tip   This question required the correct use of Heron’s formula. use of Heron’s formula to calculate One mark was allowed for the correct and identified value of s. the area, in cm2, of the triangular Ineffective calculator usage again seems to have led to incorrect values face YAB. for s, often due to not bracketing the numerator of the fraction before dividing by 2. Another quite common error involved using 32 cm for one of the sides of triangle YAB rather than recognising that it was an isosceles triangle. [Assessment report 2007]

Tessa’s final task involves removing the top 24 cm of the height of her pyramid (Figure 4). The shape remaining is shown in Figure 5. The top surface JKLM is parallel to the base ABCD.

L

M D J

C

8 cm

K

A

28 cm

B 24 cm

Figure 5 h What fraction of the height of the pyramid in Figure 4 has Tessa removed to produce Figure 5? i What fraction of the volume of the pyramid in Figure 4 remains in Figure 5? Exam tip   Many students made a mistake by assuming incorrectly that removing the top 3 of the height of a pyramid leaves 1 of the volume 4 4 remaining. [Assessment report 2007]

[©VCAA 2007]

5 The scale drawing at right is of a garden shed. a If the real-life width of the shed is 2 m, what is the scale factor i in ______ cm to ______ m? ii as a ratio ______:______? iii as a fraction? b Use this scale factor to calculate all the real dimensions indicated by the pronumerals. c If the shed is to be painted, and this includes every side except the roof (shown at right): i calculate the TSA to be painted ii how many litres of paint are required if one litre covers 4 m2?

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Maths Quest 11 Standard General Mathematics for the Casio ClassPad

C 3 cm (e)

A

3 cm Width (a)

(f) 3.75 cm 2.4 cm 3 cm (c) (d) 3.75 cm (b)

d Find the total amount of space available inside the shed. Give the answer in cubic metres to 1 decimal place. e The longest stick that can fit inside the shed will go from the corner on the floor, labelled A, to the opposite corner on top of the roof, labelled C. Find the length of such a stick. 6 The diagram below shows a triangular coffee table.

Roof section (not painted)

O

a Enlarge the object by a scale factor of 2. 10 cm The original design of the tabletop is drawn as shown. b If the longest side of the actual table is 1.6 m, what is the scale ratio? c Find the actual lengths of the other two sides of the tabletop. 8 cm The tabletop is made from marble with a glass insert in the middle. The glass insert is also in the shape of the triangle with side lengths 60 cm × 48 cm × 32 cm. d Are the two triangles (that is, the marble top and the glass insert) similar? Justify your answer. e Does the tabletop have: i line symmetry? ii rotational symmetry? f Design a triangular tabletop with three lines of symmetry and an order of rotational symmetry of 3. Specify the lengths of the sides of your tabletop. g Draw the plan of your design from part f using a scale ratio of 1:20.

eBook plus

Digital doc

Test Yourself Chapter 10

Chapter 10

Shape and measurement

525

eBook plus

aCtiVitiES

Chapter opener Digital doc

•  10 Quick Questions: Warm up with ten quick questions on shape and measurement. (page 456) 10A

Pythagoras’ theorem in two dimensions

Digital docs

•  SkillSHEET 10.1: Practise Pythagoras’ theorem. (page 457) •  Spreadsheet 102: Investigate Pythagoras’ theorem. (page 457) 10B

Pythagoras’ theorem in three dimensions

Digital doc

•  Spreadsheet 102: Investigate Pythagoras’ theorem. (page 461) 10 C

Perimeter and area

Tutorial

•  WE5 int-0904: Watch how to find the difference in the area covered by two hands of a clock in one full rotation. (page 465) Digital docs

•  SkillSHEET 10.2: Practise the conversion of units — length. (page 468) •  SkillSHEET 10.3: Practise calculating the area and perimeter of composite shapes. (page 468) 10D

Total surface area (TSA)

Tutorial

•  WE10 int-0905: Watch how to find the total surface area of a composite object. (page 474) Digital docs

•  Investigation: Investigate cone heads. (page 478) •  WorkSHEET 10.1: Calculate measurements of two and three dimensional objects. (page 478) 10E

Volume

Tutorial

•  WE13 int-0906: Watch how to calculate volume correct to one decimal place. (page 482) Digital docs

•  SkillSHEET 10.4: Practise calculating volume. (page 483) •  SkillSHEET 10.5: Practise the conversion of units — volume. (page 483) •  SkillSHEET 10.6: Practise finding unknown lengths. (page 484) •  Investigation: Cone volume. (page 485) •  Spreadsheet 010: Investigate cone volume. (page 485) 10 F

10G

Similar figures

Tutorials

•  WE17 int-0907: Watch how to reduce the size of a pentagon by a factor of two. (page 490) •  WE20 int-0908: Watch how to determine the scale factor to be used to build a model aeroplane and use it to calculate dimensions of the model. (page 493) Digital docs

•  WorkSHEET 10.2: Calculate surface area and volume of objects. (page 498) •  Investigation: Investigate making an aeroplane. (page 498) 10H

Similar triangles

Interactivity int-0811:

•  Similar triangle: Consolidate your understanding of similar triangles. (page 498) Tutorial

•  WE22 int-0909: Watch how to show that two triangles are similar. (page 501) 10I

Symmetry

Tutorial

•  WE24 int-0910: Watch how to determine the axis of symmetry. (page 506) Digital doc

•  Extension: Investigate tessellations. (page 508) Chapter review Digital doc

•  Test Yourself: Take the end-of-chapter test to test your progress. (page 525) To access eBookPLUS activities, log on to www.jacplus.com.au

Capacity

Digital doc

•  Spreadsheet 006: Investigate capacity. (page 487)

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Maths Quest 11 Standard General Mathematics for the Casio ClassPad