Printable · GCSE Foundation · ages 14-16
Geometry and measures worksheet — GCSE Foundation
Fifteen questions across the geometry and measures statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Calculator
Answer key: Geometry and measures worksheet — GCSE Foundation
MathsUKwww.geekhero.co.uk
- (d) 42.4 cm² — Sector area = (angle ÷ 360) × π × r². First, 60 ÷ 360 = 1/6. Next, π × r² = 3.14 × 81 = 254.34 cm². So the area = (1/6) × 254.34 = 42.39 cm², which rounds to 42.4 cm². (4.7 cm² comes from forgetting to square the radius: using π × r = 3.14 × 9 = 28.26, then (1/6) × 28.26 = 4.71 cm²; 254.3 cm² comes from finding the area of the whole circle, 254.34 cm², and forgetting the angle fraction; 9.4 cm² comes from using the arc length formula instead: 2 × π × r = 2 × 3.14 × 9 = 56.52 cm, then (1/6) × 56.52 = 9.42 cm.)
- (b) (22, −13) — The vector from the ship to the lighthouse is (12, −4) − (2, 5) = (10, −9). Sailing along this vector twice from the start gives (2, 5) + 2 × (10, −9) = (2 + 20, 5 − 18) = (22, −13). '(12, −4)' stops after the ship reaches the lighthouse and ignores the second identical leg. '(−18, 23)' comes from finding the vector the wrong way round, as (2, 5) − (12, −4) = (−10, 9), and then doubling that. '(2, 5)' comes from adding the vector and then subtracting it again, wrongly cancelling the two legs instead of adding them.
- (d) 60° — Method: the 4 cm side is next to the angle wanted and the 8 cm side is the hypotenuse, so the ratio built from them is cos θ = adjacent ÷ hypotenuse, and the angle comes from the inverse cosine. Working: cos θ = 4 ÷ 8 = 0.5, so θ = cos⁻¹(0.5). Answer: 60°. The distractors: 30° comes from using sin⁻¹(0.5), which treats the 4 cm side as the side opposite the angle when it is the side next to it; 45° comes from assuming the two acute angles of the triangle must be equal; 90° comes from writing down the right angle the question already gives instead of the angle it asks for.
- (c) £7.85 — Arc length = (90 ÷ 360) × 2 × 3.14 × 10 = 0.25 × 62.8 = 15.7 cm. Cost = 15.7 × £0.50 = £7.85. (£31.40 comes from finding the full circumference and forgetting the angle fraction; £15.70 comes from using the diameter, 20 cm, in place of the radius; £157.00 comes from multiplying the arc length by the radius instead of by the cost per centimetre.)
- (d) 38 m² — The two fences meet at a right angle, so the rope sweeps a quarter of a circle: 90 ÷ 360 = 1/4. Grazing area = (90 ÷ 360) × 3.14 × 7² = 0.25 × 153.86 = 38.465 m², which rounds to 38 m². (5 m² comes from forgetting to square the rope length; 154 m² comes from finding the area of a full circle and forgetting the angle fraction; 11 m² comes from using the arc length formula instead of the sector area formula.)
- (b) £26.25 — First multiply the base and height: 2.4 × 1.75 = 4.2. The area of the triangular sail is half of that: half of 4.2 is 2.1 m². Then multiply by the cost per m²: 2.1 × £12.50 = £26.25. £52.50 forgets to halve in the area formula, giving an area of 4.2 m² and doubling the true cost. £30.00 multiplies the base length by the cost per m² (2.4 × £12.50) without ever finding the area. £25.00 rounds the area to 2 m² before multiplying by the cost, losing accuracy.
- (b) 16.747 m³ — Volume of a sphere = (4/3)πr³, so a hemisphere is half that: (2/3)πr³. Substitute r = 2: (2/3) × 3.14 × 2³ = (2/3) × 3.14 × 8 = (2/3) × 25.12 = 16.7467 m³, which rounds to 16.747 m³.
- (c) 0.5 — The real distance is 2 × 25000 = 50000 cm. Converting to metres, by dividing by 100, gives 500 m, and converting to kilometres, by dividing by 1000, gives 0.5 km. A candidate who divides the 50000 cm by 1000 in one go, applying the metres-to-kilometres factor straight to the centimetres, gets 50 km. A candidate who divides by 100 twice, treating 100 m as 1 km, gets 5 km. A candidate who slips one extra decimal place when converting 500 m to kilometres gets 0.05 km. The real distance is 0.5 km.
- (a) 3.75 km — To convert metres to kilometres, divide by 1000: 3750 ÷ 1000 = 3.75 km. Dividing by 100 instead of 1000 gives 37.5 km. Dividing by 10 instead of 1000 gives 375 km. Dividing by 10 000 instead of 1000 gives 0.375 km.
- (d) £50 — Area = 1/2 × (3.5 + 6.5) × 4 = 1/2 × 10 × 4 = 20 m². Cost = 20 × £2.50 = £50. (£100 comes from forgetting to halve the trapezium area, giving 40 m² instead of 20 m²; £65 comes from using only the longer parallel side, 6.5 × 4 = 26 m², instead of the trapezium formula; £35 comes from adding all three given lengths, 3.5 + 6.5 + 4, and treating that total as the area in square metres.)
- (c) 27.7 cm — Arc length = (150 ÷ 360) × 2 × 3.14 × 6 = (5 ÷ 12) × 37.68 = 15.7 cm. The perimeter of a sector also includes the two straight radii, so perimeter = 15.7 + 6 + 6 = 27.7 cm. (15.7 cm comes from stopping after the arc length and forgetting the two straight edges; 21.7 cm comes from adding only one radius instead of two; 31.4 cm comes from doubling the arc length instead of adding the two straight edges.)
- (b) 113.04 cm² — Sector area is (angle ÷ 360) × π × radius², and radius means the RADIUS, not the diameter: here the diameter is 24 cm, so the radius is 12 cm. The fraction is 90 ÷ 360 = 1/4, and 12² = 144, so the area is 0.25 × 3.14 × 144 = 113.04 cm². Using the diameter itself as if it were the radius gives 0.25 × 3.14 × 576 = 452.16 cm². Using the arc-length formula, 2 × π × radius, instead of the area formula gives 0.25 × 2 × 3.14 × 12 = 18.84 cm². Using the radius instead of its square gives 0.25 × 3.14 × 12 = 9.42 cm².
- (d) 56.5 cm — Circumference = 2πr. With r = 9 cm and π = 3.14, circumference = 2 × 3.14 × 9 = 56.52 cm, which rounds to 56.5 cm. A student who uses πr instead of 2πr, forgetting to double, gets 3.14 × 9 = 28.26 cm, rounding to 28.3 cm. A student who uses the area formula πr² instead of the circumference formula gets 3.14 × 81 = 254.34 cm², rounding to 254.3. A student who doubles the correct circumference by mistake gets 56.52 × 2 = 113.04 cm, rounding to 113.0 cm.
- (d) sector — The region enclosed by two radii and the arc between them is called a sector. A segment is a different region, enclosed by a CHORD and an arc — it does not touch the centre at all. A chord is a straight line, not a region, so it cannot be the answer to a question asking for a region. A semicircle is only correct when the angle between the two radii is exactly 180°, which is not stated here, so it is too specific to be the general answer.
- (b) 21 — Both legs of the journey are on the same bearing, 070°, so the ship travels in one straight line the whole way and the distances simply add: 12 + 9 = 21 km. A candidate who subtracts instead of adding gets 12 − 9 = 3 km. A candidate who multiplies the two distances gets 12 × 9 = 108. A candidate who assumes the ship changed direction and treats the two legs as the sides of a right-angled triangle works out √(12² + 9²) = √225 = 15 km — but the bearing does not change, so there is no triangle and no hypotenuse to find. Because the ship stays on one straight line, the total distance is 21 km.
Build your own mix at the worksheet builder.