Answers to activity questions Check your skills 1.

2.

(a)

Length units such as mm, cm, m, km

(b)

Area units such as

(c)

Volume units such as

(a)

P = 2l + 2w P = 2 × 140 + 2 × 55 P = 390m

(c)

mm 2 , cm 2 , m 2 , ha or km 2 mm3 , cm3 , m3 , km3 (b)

A = π r2 ÷ 2 A = π × 7.52 ÷ 2 A = 88.36m 2

1 V = × s2h 3 1 V = × 1.4 2 × 2.1 3 V = 1.372m3 or 1.372kL or 1372 L

3.

(a)

(b)

The perimeter is basically the circumference of two circles with a diameter of 10cm. If the diameter is 10 cm, the radius is 5 cm. The circumference C = 2 x 3.142 x 5 = 31.42cm The perimeter of the shape is 2 x 31.42 = 62.84 cm The area of the shape is 2 circles with a radius of 5 cm and a square with side length 5 cm. Area of circle = 3.142 x 52 = 78.54 cm2 Area of square = 10 x 10 = 100 cm2 Total area = 2 x 78.54 + 100 = 257.08 cm2

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4.

3

4 V = π r3 3 4 200 = π r 3 3 200 × 3 3 =r 4×π 200 × 3 =r 4×π r = 3.63cm

A sphere with radius 3.63cm will have a volume of 200cm3.

Measurement - length, perimeter and circumference 1. (a)

Draw a sketch diagram for each and calculate the perimeter. A square with a side length of 30 cm. P = 4s

P = 4 × 30 P = 120cm (b)

A rectangle with length of 0.6m and width 20cm (0.6m = 60cm)

P = 2(l + w) P = 2 × (60 + 20) P = 2 × 80 P = 160cm

(c)

An equilateral triangle with side length 27cm

P = 3s P = 3 × 27 P = 81cm

(d)

A circle with a radius of 45mm

C = 2π r C = 2 × π × 45 C = 282.7 mm

(e)

An isosceles triangle with sides 47mm and 75mm (x2)

P = 2s1 + s2 P = 2 × 75 + 47 P = 197mm

(f)

A semicircle with a diameter 1.2m

P = 2π r ÷ 2 + d P = 2 × π × 0.6 ÷ 2 + 1.2 P = 3.085m

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2. (a)

Unknown sides are 9m − 4m = 5m and 4.5m − 2m = 2.5m .

9m 2m 4.5 m

P = 9 + 4.5 + 4 + 2.5 + 5 + 2 P = 27m

4m

(b)

60 × 2π r + 2r 360 1 P = × 2×π × 5 + 2 × 5 6 P = 15.236cm P = 2l + 2s P = 2 × 95 + 2 × 33 P = 256 P=

600 5cm

(c)

33 cm 95 cm (d)

Missing side is 35+22 (radius) = 57

22 cm

3 P = × 2π r + 35 + 22 + 57 4 P = 217.7cm

35 cm

(e)

Radius =2m

5m

1 × 2 π r + 2s 2 P = π × 2 + 2×5 P=

P = 16.28m

4m

(f)

6 cm

6 cm

Perimeter is equivalent to the circumference of a small circle (r=3) plus half the circumference of the big circle (r=6)

1 × 2π R 2 P = 2×π × 3 + π × 6 P = 2π r +

P = 37.7cm

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Area 1.

Draw a sketch diagram for each and calculate the area. (a)

A square with a side length of 30 cm.

A = s×s A = 30 × 30 A = 900cm2

(b)

A rectangle with length of 0.6m and width 20cm

A= l×w A = 60 × 20 A = 1200cm2

(c)

An right angled triangle with side lengths 27 cm, 35 cm and 44.2 cm

The base and height are 27cm and 35cm, the 44.2cm side is unnecessary.

1 A = ×b×h 2 1 A = × 27 × 35 2 A = 472.5cm 2 (d)

A circle with a radius of 45mm

A = π r2 A = π × 452 A = 6361.73mm 2 or 63.6173cm 2

(e)

An isosceles triangle with base 47mm and height 71.2mm

1 A = ×b×h 2 1 A = × 47 × 71.2 2 A = 1673.2mm2 = 16.732cm2

(f)

A semicircle with a diameter 1.2m

1 A = ×π r2 2 1 A = × π × 0.62 2 A = 0.565m2

(if d=1.2m, then r=0.6m)

A = 5650cm2

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2.

Calculate the area of these shapes. (a)

AT = l × w + l × w

9m

AT = 4 × 4.5 + 2 × (9 − 4)

2m

AT = 28m 2

4.5 m

4m

(b)

A=

60

1 6

×π r2

360 1 A = × π × 52 6 A = 13.1cm2 A = b×h

60 0 5cm

(c)

A = 95 × 25

25cm

A = 2375cm 2

95 cm A parallelogram (d)

3 AT = × π r 2 + l × w 4 3 AT = × π × 222 + 57 × 22 4 AT = 2394.4cm 2

22 cm 35 cm

(e)

Ashaded = Atriangle − Asemicircle 15 m 3.

(f) 6 cm

6 cm

1 1 Ashaded = bh − π r 2 2 2 1 1 Ashaded = ×15 ×15 − × π ×1.752 2 2 2 A = 107.7m The area of the small semicircle is the same as the area of the semicircle missing from the large semicircle, only the area of the large semicircle is required.

1 A = π r2 2 1 A = × π × 62 2 A = 56.55cm 2

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The shaded area is one quarter of the difference is area between the area of the square and the circle.

(g) 8.5m

1 2 (s −π r2 ) 4 1 A = ( 8.52 − π × 4.252 ) 4 A = 3.876m 2 A=

8.5m

8.5m

8.5m The shaded area is the area of the trapezium in the centre add the area of the large semi-circle subtract the area of the small semicircle. The trapezium has parallel side lengths 6m and 9m separated by a length 8m.

(h)

9m

6m 8m

1 1 1 ( a + b ) h + π R2 − π r 2 2 2 2 1 1 1 A = ( 6 + 9 ) × 8 + π × 4.52 − π × 32 2 2 2 A = 60 + 31.8 − 14.14 A=

A = 77.66m 2

Volume 1. (a)

Draw a sketch diagram for each and calculate the volume. A cube with a side length of 20 cm. V = s×s×s

V = 20 × 20 × 20 V = 8000cm3 (b)

(c)

A rectangle prism with length of 0.6m, width of 20cm and height 500mm. (0.6m = 60cm, 500mm = 50cm)

A prism with a base area of 28 m2 and a height of 2.7m.

V = l × w× h V = 60 × 20 × 50 V = 60 000cm3 V = Ah V = 28 × 2.7 V = 75.6m3

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(d)

A cylinder with a diameter of 50 cm and a height of 60 cm. Give your answer in mL and L.

V = Ah V = π r 2h V = π × 252 × 60 V = 117810cm3 V = 117810mL = 117.81L

(e)

2.

A prism with a semicircular base ( radius = 51 cm) and a height of 63cm

V = Ah 1 V = π r 2h 2 1 V = π × 512 × 63 2 V = 257 395.4cm3

Calculate the volume of these solids. Express the answer in (i) mm3 (ii) cm3 (iii) m3

65 mm

(a)

V = Ah 1 V = bhtriangle × hprism 2 1 V = × 6.5× 5.5×115 2 V = 2 055.625cm3

5m 1 . 1

V = 2 055.625×1000 = 2 055 625mm3

55 mm

(b)

V = 2 055.625 ÷1000 000 = 0.002 055 625m3 This is a water tank. Express your

2.5 m

1.83 m

answer in L.

(r = 2.5 ÷ 2 = 1.25m)

V = Ah V = π r 2h V = π ×1.252 ×1.83 V = 8.983m3 V = 8.983kL = 8 983L

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(c)

Volume of large cone (h=12m, d=3m) subtract the volume of small cone (h=8m, d=2m).

2m

4m

3m

1 1 V = π R2 H − π r 2h 3 3 1 1 V = π × 1.52 × 12 − π × 12 × 8 3 3 V = 28.27 − 8.38 V = 19.89m 3

(e)

A farmer has a large dam covering 3 ha of land. The average depth when full is 3m. Calculate the volume of water in dam in kL and ML. (3 ha = 30 000m2)

(f)

V = Ah V = 30 000m 2 × 3m V = 90 000m3 V = 90 000kL = 90 ML

500 mm

A farmer has built these troughs to hold water for livestock. How many litres of water will each trough hold?

1.2 m

1 V = π r 2h 2 1 V = π × 0.252 × 1.2 2 V = 0.118m3 V = 0.118kL = 118L

(g)

30 mm

This is an ice-cream cone. Calculate the volume of soft serve ice cream that would fill the cone and a semi-spherical amount on top of the cone. Using cm,

VT = Vcone + Vhemisphere 90 mm

1 1 42 3 VT = π r 2 h + × πr 3 2 3 1 2 VT = π × 32 × 9 + π × 33 3 3 VT = 84.8 + 56.5 VT = 141.3cm3

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(h)

A garden hose has an internal diameter of 15 mm and a length of 5 m. What volume of water fills the hose?

The hose is a cylinder r=7.5mm=0.75cm, h=500cm

V = π r 2h V = π × 0.752 × 500 V = 883.6cm 3 V = 883.6cm 3 V = 883.6mL

(i)

A can contains 375 mL of soft drink and must have a height of 120 mm. What will the diameter need to be? (375 mL=375cm3) (120mm=12cm)

V = π r 2h V = r2 πh V =r πh r=

V πh

375 π × 12 r = 3.154 cm r=

→ d = 6.31cm

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