Printable · GCSE Higher · ages 14-16
Geometry and measures worksheet — GCSE Higher
Fifteen questions across the geometry and measures statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Geometry and measures worksheet — GCSE Higher
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- 1.A circular decoration has a radius of 10 cm. A ribbon costs £0.50 per centimetre and is fixed along the curved edge (the arc) of a sector-shaped section with an angle of 90°. Using π = 3.14, work out the cost of the ribbon for this section.
- 2.Triangle JKL and triangle MNP are similar, with JK ÷ MN = JL ÷ MP = 2. A student says triangle JKL must be congruent to triangle MNP. Write down why the student is wrong.
- 3.A market stall sells apples at £2.40 per kilogram. Which of these expressions gives the cost, in pounds, of buying m kilograms of apples?
- 4.A garden flower bed is in the shape of a trapezium. Its two parallel edges are 3.5 m and 6.5 m long, and the perpendicular distance between them is 4 m. Bedding plants cost £2.50 per square metre to cover the bed. Work out the total cost of the bedding plants.
- 5.A sector of a circle has radius 6 cm and angle 150°. Using π = 3.14, work out the perimeter of the sector.
- 6.A shop sells rice in packs of 250 g. Aisha buys 4 packs. Work out the total mass of rice she buys, in kilograms.
- 7.In triangle ABC, angle ABC = 47°, AC = 12.5 cm and BC = 9.2 cm. Work out the size of angle BAC. Give your answer to 1 decimal place.
- 8.Two similar ponds have perimeters in the ratio 4 : 11. The perimeter of the larger pond is 88 m. Work out the perimeter of the smaller pond.
- 9.Two similar company logos are printed at different sizes. The area of the larger logo is 2.25 times the area of the smaller logo. Work out the length scale factor from the smaller logo to the larger logo.
- 10.A sector of a circle has an angle of 90° and an arc length of 15.7 cm. Using π = 3.14, work out the radius of the circle.
- 11.In triangle ABC, AB = 9 cm, AC = 6 cm, and angle BAC = 65°. Work out the length of BC. Give your answer to 1 decimal place.
- 12.A flagpole is 12 m tall. Freya stands 9 m from the base of the flagpole on level ground. Work out the angle of elevation of the top of the flagpole from where Freya stands. Give your answer correct to 1 decimal place.
- 13.A tent has the cross-section of a right-angled triangle. The sloping side of the cross-section is 10 m long, and it makes an angle of 60° with the horizontal ground. Using the exact value of cos 60°, work out the horizontal distance covered by the sloping side.
- 14.A water tank is a cube with edges of length 2 m. Work out how many cubic centimetres the tank holds when it is full.
- 15.A robotic arm's tip starts at the point (12.5, −4.25) on a grid measured in centimetres. It moves by the vector to pick up a component, then by the vector to place it. Work out the coordinates of the tip after both moves.
Answer key
- (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.)
- (c) The scale factor is 2, not 1, so the sides are not equal — Congruent shapes must be exactly the same size as well as the same shape, which means a scale factor of 1. Here the scale factor between the triangles is 2, so the sides are different lengths and the triangles cannot be congruent, even though they are similar. 'Similar triangles are never congruent' is too strong — a scale factor of exactly 1 would make them both similar and congruent. 'The angles are not necessarily equal' is wrong, since similar triangles always have equal matching angles. 'Congruent triangles must have a right angle' is an unrelated, false fact about congruence.
- (a) 2.40m — Method: a cost found from a rate is the rate multiplied by the amount bought. Working: the rate is £2.40 per kilogram and the amount is m kilograms, so the cost is 2.40 × m. Answer: 2.40m. A candidate who divides the amount by the rate instead of multiplying writes m/2.40. A candidate who adds the rate to the amount instead of multiplying writes 2.40 + m. A candidate who subtracts the rate from the amount instead of multiplying writes m − 2.40.
- (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.)
- (d) 1 kg — First find the total mass in grams: 4 × 250 = 1000 g. Then convert to kilograms by dividing by 1000: 1000 ÷ 1000 = 1 kg. Finding the correct total in grams but forgetting to divide by 1000 gives 1000 kg. Adding the number of packs to the pack mass instead of multiplying, 4 + 250 = 254 g, gives 0.254 kg. Dividing the pack mass by the number of packs instead of multiplying, 250 ÷ 4 = 62.5 g, gives 0.0625 kg.
- (c) 32.6° — Method: an angle is wanted from two sides and the angle facing one of them, so use the sine rule in the form sin A/a = sin B/b. Working: BC = 9.2 cm faces angle BAC and AC = 12.5 cm faces the 47° angle, so sin BAC = 9.2 × sin 47° ÷ 12.5 = 6.7285 ÷ 12.5 = 0.5383, and the inverse sine of 0.5383 is 32.566°. Because BC is shorter than AC, angle BAC must be smaller than 47°, so the acute value is the only one that fits. Answer: angle BAC = 32.6° to 1 decimal place. The distractors: 147.4° comes from taking 180° − 32.566°, the obtuse angle with the same sine, without checking it — 147.4° and 47° already add to more than 180°, so no such triangle exists; 83.6° comes from putting the sides the wrong way up, 12.5 × sin 47° ÷ 9.2, which gives a sine of 0.9937; 57.4° comes from pressing the inverse cosine key on 0.5383 instead of the inverse sine key.
- (a) 32 m — The scale factor from the larger pond to the smaller pond is 4 ÷ 11, so the smaller perimeter is 88 × 4 ÷ 11 = 32 m. The distractor 242 m comes from using the ratio the wrong way round, 88 × 11 ÷ 4 = 242. The distractor 84 m comes from subtracting the smaller ratio number, 88 − 4 = 84, instead of scaling. The distractor 121 m comes from multiplying 11 × 11 = 121, ignoring the given perimeter altogether.
- (c) 1.5 — The area scale factor is the length scale factor squared, so if n is the length factor, n² = 2.25. Taking the positive square root gives n = 1.5 (check: 1.5² = 2.25). 2.25 is just the area factor restated, with no root taken. 1.125 comes from halving 2.25 instead of taking its square root. 5.0625 comes from squaring 2.25 instead of rooting it.
- (b) 10 cm — Arc length = (angle ÷ 360) × 2 × π × r, so 15.7 = (90 ÷ 360) × 2 × 3.14 × r = 1.57r, giving r = 15.7 ÷ 1.57 = 10 cm. (2.5 cm comes from forgetting the angle fraction and dividing by 2π alone; 20 cm comes from leaving out the factor of 2 in the arc length formula before dividing; 5 cm comes from dividing the arc length by π alone, ignoring both the angle fraction and the factor of 2.)
- (b) 8.4 cm — Two sides and the angle between them are known, so use the cosine rule: BC² = AB² + AC² − 2 × AB × AC × cos(BAC). Substituting, BC² = 81 + 36 − 45.64 = 71.36. Taking the square root: BC = √71.36 = 8.4 cm (1 d.p.). Leaving out the factor of 2 in the formula gives BC² = 81 + 36 − 22.82 = 94.18, so BC = 9.7 cm. Adding the cosine term instead of subtracting it gives BC² = 81 + 36 + 45.64 = 162.64, so BC = 12.8 cm. Using sin65° in place of cos65° gives BC² = 81 + 36 − 97.88 = 19.12, so BC = 4.4 cm. Keep the factor of 2, subtract the cosine term, and the correct length is 8.4 cm.
- (b) 53.1° — The angle of elevation is opposite the height of the flagpole, 12 m, and adjacent to the distance from its base, 9 m, so tan θ = 12/9 = 1.333..., giving θ = tan⁻¹(1.333...) = 53.13...° ≈ 53.1°. "36.9°" finds the OTHER acute angle of the triangle, 90° − 53.1°, the angle at the top of the flagpole rather than the angle of elevation at Freya's position. "48.6°" comes from wrongly treating 9/12 as a sine ratio and finding sin⁻¹(0.75) = 48.6°, when neither side here is the hypotenuse. "41.4°" comes from wrongly treating 9/12 as a cosine ratio and finding cos⁻¹(0.75) = 41.4°, again without a hypotenuse in the ratio at all.
- (c) 5 m — The horizontal distance is the side adjacent to the 60° angle, and the sloping side is the hypotenuse, so horizontal distance = hypotenuse × cos 60°. The exact value of cos 60° is 1/2, so horizontal distance = 10 × 1/2 = 5 m. Using the sloping side itself as the horizontal distance, without using any trigonometry at all, gives 10 m. Using sin 60° = √3/2 instead of cos 60° finds the vertical height of the tent rather than the horizontal distance: 10 × √3/2 = 5√3 = 8.7 m (1 d.p.). Dividing the sloping side by cos 60° instead of multiplying by it, 10 ÷ 0.5 = 20 m, treats the sloping side as though it were the adjacent side rather than the hypotenuse.
- (a) 8,000,000 cm³ — Method: change the edge length into centimetres first and then cube it, because 1 m = 100 cm and a volume needs that conversion applied to all three dimensions. Working: 2 m = 2 × 100 = 200 cm, so the volume is 200 × 200 × 200. 200 × 200 = 40,000 and 40,000 × 200 = 8,000,000. Answer: 8,000,000 cm³. The distractors: 8,000 cm³ comes from converting 2 m to 20 cm and cubing that; 80,000 cm³ comes from cubing in metres to get 8 m³ and then multiplying by 10,000, the conversion factor for an area rather than the 1,000,000 a volume needs; 8 cm³ comes from cubing the 2 without converting at all and simply writing cm³ because the question asked for that unit.
- (d) (11, 0.75) — Add the moves to the starting point one component at a time. x: 12.5 + (−3.75) + 2.25 = 11; y: −4.25 + 6.5 + (−1.5) = 0.75, giving (11, 0.75). (8.75, 2.25) stops after the first move only and never applies the second vector. (6.5, 3.75) comes from subtracting the second vector instead of adding it. (0.75, 11) comes from swapping the final x-coordinate and y-coordinate.
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