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.In triangle ABC, angle BAC = 42°, angle ABC = 63° and BC = 8.4 cm. Work out the length of AC. Give your answer to 1 decimal place.
- 2.Two coastguard stations A and B are 18 km apart, with B due east of A. A boat at C is on a bearing of 062° from A and on a bearing of 315° from B. Work out the distance of the boat from station B. Give your answer to 1 decimal place.
- 3.A regular pentagon has centre O. Each vertex is 8 cm from O, and each of the 5 triangles formed by joining O to the vertices is isosceles with an angle of 72° at O. Work out the area of the pentagon. Give your answer to 1 decimal place.
- 4.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.
- 5.In triangle ABC, AB = 12 cm, AC = 9 cm and angle BAC = 55°. Work out the size of angle ABC. Give your answer to 1 decimal place.
- 6.A shop stacks identical storage crates in a display. The diagram shows the plan of the floor positions used, with one corner position marked. Every floor position is filled with crates stacked 3 high, except the marked corner position, which is left completely empty. Work out how many crates are used in total.
- 7.A scale drawing shows a room that is 5 cm wide. The real room is 15 m wide. Write the scale of the drawing in the form 1 : n.
- 8.A game designer places a coin at (5, 3) on a grid. The game rotates the coin 90° clockwise about the point (2, 3) each time the player presses a button, and this is applied twice in a row. Work out the coordinates of the coin after the button is pressed twice.
- 9.A flower bed in the shape of a parallelogram has a base of 3.5 m and a perpendicular height of 2 m. Compost costs £4.80 per m². Work out the total cost of covering the flower bed with compost.
- 10.A circular tabletop has a radius of 15 cm. Work out the area of the tabletop. Use π = 3.14. Give your answer to 1 decimal place.
- 11.Two similar hexagonal tiles have lengths in the ratio 4 : 7. A side of the smaller tile is 8.4 cm long. Work out the length of the corresponding side of the larger tile.
- 12.A fish tank is a cuboid measuring 40 cm long, 25 cm wide and 30 cm high. Water is poured in to a depth of 20 cm. Work out the volume of water in the tank, in litres.
- 13.In triangle ABC, AB = 15 cm, angle BAC = 38° and the area of the triangle is 61 cm². Work out the length of BC. Give your answer to 1 decimal place.
- 14.Two bronze statues are geometrically similar and made from the same solid bronze. Their heights are 30 cm and 45 cm. The smaller statue has a mass of 12 kg. Work out the mass of the larger statue, in kg.
- 15.A shape is translated by the vector (5, −3), and the image is then translated by the vector (−5, 3). Which single transformation has the same overall effect as this combination, for every point?
Answer key
- (c) 11.2 cm — Method: two angles and a side are given, so use the sine rule, a/sin A = b/sin B = c/sin C, taking care that each side is paired with the angle it faces. Working: BC faces angle BAC = 42°, and AC faces angle ABC = 63°, so AC/sin 63° = 8.4/sin 42°. Multiplying up, AC = 8.4 × sin 63° ÷ sin 42° = 7.4845 ÷ 0.6691 = 11.185. Answer: AC = 11.2 cm to 1 decimal place. The distractors: 6.3 cm comes from writing the ratio upside down, 8.4 × sin 42° ÷ sin 63°, which pairs each side with the angle beside it rather than the angle opposite it; 12.1 cm comes from using the third angle, 180° − 42° − 63° = 75°, in the numerator, which gives the length of AB instead of AC; 12.6 cm comes from assuming the sides are in the same ratio as the angles and working out 8.4 × 63 ÷ 42, which is true for arcs of a circle but never for the sides of a triangle.
- (d) 8.8 km — Method: turn each bearing into an angle of triangle ABC, find the third angle from the angle sum, then use the sine rule. Working: B is due east of A, so AB itself lies on a bearing of 090°, and the angle at A between AB and AC is 090° − 062° = 28°. From B, station A lies due west on a bearing of 270°, and C lies on 315°, so the angle at B is 315° − 270° = 45°. The third angle is 180° − 28° − 45° = 107°. The side BC faces the 28° angle and AB = 18 km faces the 107° angle, so BC = 18 × sin 28° ÷ sin 107° = 8.4505 ÷ 0.95630 = 8.8366. Answer: the boat is 8.8 km from B, to 1 decimal place. The distractors: 13.3 km comes from pairing BC with the 45° angle at B instead of the 28° angle it faces, which gives the distance AC; 12.0 km comes from never working out the third angle and dividing by sin 45° instead of sin 107°; 16.6 km comes from using the bearing 062° itself as the angle at A, instead of the 28° between AC and AB.
- (c) 152.2 cm² — Method: find the area of one of the 5 isosceles triangles with 1/2ab sin C, then multiply by 5 for the whole pentagon. Working: one triangle has area 1/2 × 8 × 8 × sin 72° = 30.4338 cm², so the pentagon's area = 5 × 30.4338 = 152.2 cm² — keep the unrounded triangle area, since 5 × 30.4 would give 152.0. Reporting the area of a single triangle and forgetting to multiply by 5 gives 30.4 cm²; multiplying by 6 instead of 5, as for a hexagon, gives 182.6 cm²; and leaving out the 1/2 from each triangle before multiplying by 5 gives 304.3 cm². A regular pentagon splits into exactly 5 triangles at its centre, each with a 72° angle, since 360° ÷ 5 = 72°.
- (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) 47.2° — Method: two sides and the angle between them are given, so find the third side with the cosine rule and then use the sine rule for the angle. Working: BC² = 12² + 9² − 2 × 12 × 9 × cos 55° = 144 + 81 − 216 × 0.57358 = 225 − 123.89 = 101.11, so BC = 10.0552 cm. Angle ABC faces AC = 9 cm and the 55° angle faces BC = 10.0552 cm, so sin ABC = 9 × sin 55° ÷ 10.0552 = 7.3724 ÷ 10.0552 = 0.73318, and the inverse sine of 0.73318 is 47.154°. Since AC is not the longest side, angle ABC is not the largest angle and the acute value is the one that fits. Answer: angle ABC = 47.2° to 1 decimal place. The distractors: 77.8° comes from pairing the angle at B with AB = 12 cm, the side beside it, instead of AC = 9 cm, the side it faces, which actually produces angle ACB; 23.2° comes from a sign slip in the cosine rule step, 225 + 123.89 = 348.89, giving BC = 18.68 cm before the sine rule is applied; 37.9° comes from dividing by AB = 12 cm rather than by the side facing the 55° angle.
- (a) 33 — Method: total crates = (number of floor positions that actually have crates on them) × (the stack height). Working: there are 4 × 3 = 12 floor positions in the whole arrangement, but one corner position is left empty, leaving 11 filled positions; each filled position is stacked 3 crates high, so 11 × 3 = 33. Answer: 33. The distractors: 36 comes from forgetting to remove the empty corner and using all 12 positions (12 × 3). 35 comes from removing only one crate for the empty corner instead of the full stack of 3 (36 − 1). 11 comes from counting the filled floor positions and stopping there, forgetting that each one carries a stack 3 crates high.
- (c) 1 : 300 — First convert both lengths to the same unit: 15 m = 1500 cm. The scale compares 5 cm on the drawing to 1500 cm in real life, so dividing both parts by 5 gives a scale of 1 : 300. A candidate who compares 5 to 15 without converting units gets 1 : 3. A candidate who converts 15 m to 150 cm, using the wrong conversion factor, gets 1 : 30. A candidate who converts 15 m to 15000 cm, again using the wrong conversion factor, gets 1 : 3000. The scale of the drawing is 1 : 300.
- (d) (−1, 3) — Method: two 90° rotations about the SAME centre, applied one after another, combine into a single 180° rotation about that same centre: use the shortcut (x, y) → (2a − x, 2b − y) for a half-turn about (a, b). Working: with centre (2, 3), doubling each coordinate gives 2 × 2 = 4 and 2 × 3 = 6, so the rule is (x, y) → (4 − x, 6 − y). Applying it to (5, 3) gives 4 − 5 = −1 and 6 − 3 = 3, so the coin ends at (−1, 3). Answer: (−1, 3). Rotate about the centre (2, 3) stated in the game, not about the origin, and remember the button is pressed TWICE: stopping after one press, or rotating about the wrong centre, both leave the coin somewhere else.
- (b) £33.60 — The area of the parallelogram flower bed is base × height = 3.5 × 2 = 7 m². The cost is 7 × £4.80 = £33.60. £16.80 comes from using the triangle formula instead of the parallelogram formula: 3.5 × 2 = 7, and half of 7 is 3.5 m², then 3.5 × £4.80 = £16.80. £26.40 comes from adding the base and height, 3.5+2 = 5.5, instead of multiplying them, then multiplying by £4.80. £7.00 correctly finds the area, 7 m², but forgets to multiply it by the cost per m².
- (d) 706.5 cm² — Area of a circle = πr². With r = 15 cm and π = 3.14, area = 3.14 × 15² = 3.14 × 225 = 706.5 cm². A student who uses the circumference formula 2πr instead of the area formula gets 2 × 3.14 × 15 = 94.2 cm². A student who uses πr instead, forgetting to double, gets 3.14 × 15 = 47.1 cm². A student who squares the diameter (30 cm) instead of the radius gets 3.14 × 900 = 2826.0 cm².
- (c) 14.7 cm — The scale factor from the smaller tile to the larger tile is 7 ÷ 4 = 1.75, so the larger side is 8.4 × 1.75 = 14.7 cm. '4.8 cm' comes from scaling by 4 ÷ 7 instead, using the ratio the wrong way round. '11.4 cm' comes from adding the difference between the ratio numbers, 7 − 4 = 3, onto 8.4, instead of scaling. '58.8 cm' comes from multiplying by 7 on its own, using a ratio number as the scale factor instead of working out 7 ÷ 4 = 1.75 first.
- (a) 20 litres — Volume of water = length × width × depth of water = 40 × 25 × 20 = 20 000 cm³. Since 1000 cm³ = 1 litre, divide by 1000: 20 000 ÷ 1000 = 20 litres. A pupil who uses the full height of the tank, 30 cm, instead of the water depth, 20 cm, gets 40 × 25 × 30 = 30 000 cm³ = 30 litres. A pupil who forgets to convert cm³ to litres at all gives 20 000 litres. A pupil who divides by 1000 twice by mistake gets 20 000 ÷ 1000 ÷ 1000 = 0.02 litres. The correct volume of water is 20 litres.
- (b) 9.3 cm — Method: the area formula gives the second side that encloses the 38° angle, and once two sides and the angle between them are known the cosine rule gives the third side. Working: 61 = 1/2 × 15 × AC × sin 38°, so AC = 2 × 61 ÷ (15 × sin 38°) = 122 ÷ 9.2349 = 13.211 cm. Then BC² = 15² + 13.211² − 2 × 15 × 13.211 × cos 38° = 225 + 174.53 − 312.31 = 87.22, and the square root of 87.22 is 9.339. Answer: BC = 9.3 cm to 1 decimal place. The distractors: 10.6 cm comes from forgetting to double the area when rearranging, so AC is taken as 6.606 cm before the cosine rule is applied; 26.7 cm comes from adding the last term of the cosine rule instead of subtracting it, 399.53 + 312.31; 13.2 cm is the length of AC, written down by a candidate who completes the first step and stops there.
- (c) 40.5 kg — Since the statues are made from the same material, mass is proportional to volume, which scales with the cube of the length scale factor. The length scale factor is 45 ÷ 30 = 1.5, so the volume (and mass) scale factor is 1.5³ = 3.375. The larger statue's mass is 12 × 3.375 = 40.5 kg. 18 kg comes from multiplying by the length factor 1.5 directly. 27 kg comes from using the area scale factor 1.5² = 2.25 instead of the volume scale factor. 54 kg comes from treating 'cubed' as 'multiplied by 3', giving 1.5 × 3 = 4.5 instead of 1.5³.
- (d) No transformation — every point stays exactly where it was — The two vectors (5, −3) and (−5, 3) are opposites, so adding them gives (0, 0): every point ends up exactly where it started, and there is no transformation at all. Misreading the second vector's signs and effectively adding (5, −3) to itself instead of to its opposite gives a translation by the vector (10, −6). Assuming two translations must combine into a reflection gives a reflection in the x-axis — but a reflection reverses orientation, and translations never do. Assuming that two opposite vectors must mean a half turn gives a rotation of 180° about the origin — but a 180° rotation moves every point except its own centre, whereas this pair of translations leaves every single point exactly where it was. Two translations by opposite vectors always cancel exactly, leaving every point unmoved.
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