Practice Maths

Unit 1 Topic 2 Review — Shape and Measurement — Solutions

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This review covers all five lessons: Units and Conversions, Perimeter and Area, Surface Area, Volume, and Similarity and Scale. Allow approximately 75–90 minutes for this review.

Review Questions

  1. Q1 — Unit conversions

    Fluency

    Convert each of the following:

    (a) 3.6 km to m

    (b) 580 mm² to cm²

    (c) 4500 mL to L

    (d) 2.3 t to kg

    (a) 3.6 km to m:

    3.6 × 1000 = 3600 m

    (b) 580 mm² to cm²:

    1 cm = 10 mm, so 1 cm² = 100 mm²

    580 ÷ 100 = 5.8 cm²

    (c) 4500 mL to L:

    4500 ÷ 1000 = 4.5 L

    (d) 2.3 t to kg:

    2.3 × 1000 = 2300 kg

  2. Q2 — Area of shapes

    Fluency

    (a) Trapezium with parallel sides 8 cm and 14 cm, height 5 cm.

    (b) Circle with diameter 18 cm.

    (c) Sector with r = 10 cm and θ = 60°.

    (a) Trapezium:

    A = ½(a + b)h = ½(8 + 14) × 5 = ½ × 22 × 5 = 55 cm²

    (b) Circle: d = 18, r = 9:

    A = πr² = π(81) = 81π ≈ 254.47 cm²

    (c) Sector: r = 10, θ = 60°:

    A = (60/360) × π(100) = (1/6) × 100π ≈ 52.36 cm²

  3. Q3 — Perimeter of composite shape

    Fluency

    A composite shape consists of a rectangle 12 cm × 7 cm with a semicircle removed from one short end (the short end has length 7 cm, so the semicircle has diameter 7 cm and radius 3.5 cm). Find the perimeter of the remaining shape.

    The semicircle is removed from one short end. The outer boundary consists of:

    • Two long sides: 2 × 12 = 24 cm
    • The remaining short end (far end): 7 cm
    • The semicircular indent (replaces the near short end): semicircle perimeter = πr = π × 3.5 = 3.5π ≈ 10.996 cm

    Total perimeter = 24 + 7 + 10.996 = 41.99 cm ≈ 42.0 cm

  4. Q4 — Surface area of a cylinder

    Fluency

    Find the total surface area of a closed cylinder with r = 6 cm and h = 14 cm.

    SA = 2πr² + 2πrh

    = 2π(36) + 2π(6)(14)

    = 72π + 168π

    = 240π ≈ 753.98 cm²

  5. Q5 — Volume of a cone

    Fluency

    Find the volume of a cone with r = 4 cm and h = 9 cm.

    V = ⅓πr²h = ⅓ × π × 16 × 9 = ⅓ × 144π = 48π ≈ 150.80 cm³

  6. Q6 — L-shaped floor: area and tiling cost

    Understanding

    An L-shaped floor has outer dimensions 10 m × 8 m with a 3 m × 3 m square cut from one corner.

    (a) Find the area of the floor.

    (b) Find the cost to tile at $38 per m².

    (a) Area:

    Full rectangle = 10 × 8 = 80 m²

    Removed corner = 3 × 3 = 9 m²

    Floor area = 80 − 9 = 71 m²

    (b) Tiling cost:

    71 × $38 = $2698

  7. Q7 — Surface area and painting cost

    Understanding

    A rectangular prism has dimensions 15 cm × 8 cm × 6 cm. All outer surfaces are to be painted at $0.05 per cm². Find the surface area and total cost.

    SA = 2(lw + lh + wh)

    = 2(15×8 + 15×6 + 8×6)

    = 2(120 + 90 + 48)

    = 2 × 258 = 516 cm²

    Cost = 516 × $0.05 = $25.80

  8. Q8 — Grain silo volume and capacity

    Understanding

    A grain silo consists of a cylinder (r = 2 m, h = 5 m) topped by a cone (r = 2 m, h = 3 m). Find the total volume and total capacity in kL.

    Cylinder volume:

    Vcyl = πr²h = π(4)(5) = 20π ≈ 62.83 m³

    Cone volume:

    Vcone = ⅓π(4)(3) = 4π ≈ 12.57 m³

    Total volume:

    20π + 4π = 24π ≈ 75.40 m³

    Capacity: 1 m³ = 1 kL, so 75.40 kL

  9. Q9 — Similar triangles: scale factor, perimeter, area

    Understanding

    Two similar triangles: the small triangle has sides 6 cm, 8 cm, and 10 cm. The large triangle has a longest side of 25 cm.

    (a) Find the scale factor.

    (b) Find the perimeter of the large triangle.

    (c) Find the ratio of their areas.

    (a) Scale factor:

    k = 25 ÷ 10 = 2.5

    (b) Perimeter of large triangle:

    Small perimeter = 6 + 8 + 10 = 24 cm

    Large perimeter = k × 24 = 2.5 × 24 = 60 cm

    (Or: sides = 15, 20, 25 → 15+20+25=60 ✓)

    (c) Area ratio (small : large):

    k² = 6.25, so area ratio = 1 : 6.25

  10. Q10 — Circular lawn: radius, area, and fertiliser cost

    Understanding

    A circular lawn has a circumference of 62.8 m.

    (a) Find the radius (use π ≈ 3.14).

    (b) Find the area.

    (c) Find the cost to fertilise at $3.50 per m².

    (a) Radius:

    C = 2πr ⇒ r = C ÷ (2π) = 62.8 ÷ (2 × 3.14) = 62.8 ÷ 6.28 = 10 m

    (b) Area:

    A = πr² = 3.14 × 100 = 314 m²

    (c) Cost:

    314 × $3.50 = $1099

  11. Q11 — Fish tank volume and capacity

    Understanding

    A rectangular fish tank measures 90 cm × 40 cm × 45 cm.

    (a) Find the volume in cm³.

    (b) Find the capacity in litres.

    (c) If filled to 80%, how many litres does it contain?

    (a) Volume:

    V = 90 × 40 × 45 = 162 000 cm³

    (b) Capacity in litres:

    162 000 cm³ ÷ 1000 = 162 L

    (c) At 80% capacity:

    0.80 × 162 = 129.6 L

  12. Q12 — Cone: find height, slant height, and TSA

    Problem Solving

    A cone has volume 452.4 cm³ and radius 6 cm.

    (a) Find the perpendicular height.

    (b) Find the slant height.

    (c) Find the total surface area.

    (a) Perpendicular height:

    V = ⅓πr²h ⇒ h = 3V ÷ (πr²)

    h = 3 × 452.4 ÷ (π × 36) = 1357.2 ÷ 113.097 ≈ 12 cm

    (b) Slant height:

    l = √(r² + h²) = √(36 + 144) = √180 = 6√5 ≈ 13.42 cm

    (c) TSA:

    TSA = πr² + πrl = π(36) + π(6)(13.42) = 36π + 80.52π = 116.52π ≈ 366.0 cm²

  13. Q13 — Painting a sphere

    Problem Solving

    A sphere has diameter 2 m. Paint covers 6 m² per litre and is sold in 500 mL tins for $18 each.

    (a) Find the surface area.

    (b) Find the litres of paint needed.

    (c) Find the number of 500 mL tins needed (round up).

    (d) Find the total cost.

    d = 2 m, r = 1 m

    (a) Surface area:

    SA = 4πr² = 4π(1) = 4π ≈ 12.566 m²

    (b) Litres needed:

    12.566 ÷ 6 = 2.094 L

    (c) Number of 500 mL tins:

    2.094 L = 2094 mL

    Tins = 2094 ÷ 500 = 4.188 → round up to 5 tins

    (d) Total cost:

    5 × $18 = $90

  14. Q14 — Similar cylinders: scale factor and volume

    Problem Solving

    Two similar cylinders: the small has r = 3 cm, h = 4 cm. The large has r = 7.5 cm.

    (a) Find the scale factor (small to large).

    (b) Find the height of the large cylinder.

    (c) Find the ratio of their volumes.

    (d) If the small cylinder has volume 113.1 cm³, find the volume of the large cylinder.

    (a) Scale factor:

    k = 7.5 ÷ 3 = 2.5

    (b) Height of large cylinder:

    hlarge = k × hsmall = 2.5 × 4 = 10 cm

    (c) Volume ratio:

    k³ = 2.5³ = 15.625

    Volume ratio = 1 : 15.625

    (d) Volume of large cylinder:

    Vlarge = 15.625 × 113.1 = 1767.19 cm³

    Check: V = π(7.5)²(10) = 562.5π ≈ 1767.15 cm³ ✓

  15. Q15 — Variable-depth swimming pool

    Problem Solving

    A swimming pool is 20 m × 10 m. The depth varies uniformly from 1 m at the shallow end to 3 m at the deep end (the cross-section, viewed from the side, forms a trapezium).

    (a) Find the cross-sectional area (trapezium: parallel sides 1 m and 3 m, width 20 m).

    (b) Find the total volume in m³.

    (c) Find the capacity in kL.

    (d) If water is pumped in at 150 L per minute, how long (in hours) does it take to fill?

    (a) Cross-sectional area (trapezium):

    The cross-section runs along the 20 m length; the two parallel sides are the depths: 1 m and 3 m. The “height” of the trapezium is the horizontal length = 20 m.

    A = ½(1 + 3) × 20 = ½ × 4 × 20 = 40 m²

    (b) Volume:

    V = A × width = 40 × 10 = 400 m³

    (Using V = prism formula: cross-section area × depth into page)

    (c) Capacity in kL:

    1 m³ = 1 kL, so capacity = 400 kL

    (d) Time to fill at 150 L/min:

    Total volume = 400 kL = 400 000 L

    Time = 400 000 ÷ 150 = 2666.7 minutes

    Convert to hours: 2666.7 ÷ 60 ≈ 44.4 hours

    (44 hours and 27 minutes)