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.
Answer key: Geometry and measures worksheet — GCSE Higher
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- (c) It is a parallelogram but not a rectangle — Method: match the given properties to the definition of each named quadrilateral. Working: having both pairs of opposite sides parallel and equal in length is exactly the definition of a parallelogram; since none of the angles are right angles, it cannot also be a rectangle, which needs four right angles. Options: 'must be a rectangle' wrongly assumes every parallelogram has right angles; 'must be a rhombus' wrongly assumes equal opposite sides means all four sides are equal, but only the opposite pairs are stated as equal here; 'trapezium' is wrong because a trapezium has exactly one pair of parallel sides, while this shape has two pairs, so it is a parallelogram and not a trapezium. Answer: it is a parallelogram but not a rectangle.
- (c) (3, 2) — For an enlargement centred on the origin, multiply every coordinate by the scale factor: (9 × 1/3, 6 × 1/3) = (3, 2). A pupil who multiplies by 3 instead of by 1/3 gets (27, 18). A pupil who subtracts a third of each coordinate instead of scaling by a third gets (9 − 3, 6 − 2) = (6, 4). A pupil who applies the scale factor to the x-coordinate only gets (3, 6). The correct image is (3, 2).
- (d) (−2, 0) — A 180° rotation about a centre (a, b) maps (x, y) to (2a − x, 2b − y). Here that gives (2 × 1 − 4, 2 × 1 − 2) = (−2, 0). A pupil who rotates about the origin instead of (1, 1) gets (−4, −2). A pupil who adds the centre's coordinates instead of applying the rotation formula gets (4 + 1, 2 + 1) = (5, 3). A pupil who just subtracts the centre's coordinates from E's, without doubling and reversing, gets (4 − 1, 2 − 1) = (3, 1). The correct image is (−2, 0).
- (b) 5 cm by 3 cm — The plan view looks straight down on the cuboid's footprint, so it shows the length (5 cm, left to right) and the depth (3 cm, front to back) — the two dimensions that do not involve height. "5 cm by 2 cm" repeats the front elevation's dimensions, pairing the length with the height instead of the depth. "3 cm by 2 cm" repeats the side elevation's dimensions, again pairing the depth with the height rather than with the length. "5 cm by 5 cm" comes from mistakenly assuming the plan must be a square, pairing the length with itself instead of with the depth.
- (b) (−2, −1) — Method: apply the reflection to the point first, then rotate the image about the given centre, in the order the question states them. Working: reflecting (3, 4) in the line y = 1 keeps x = 3 and puts the image as far below the line as the point is above it: 4 is 3 units above y = 1, so the image is 3 units below, at 1 − 3 = −2 (the same as 2 × 1 − 4 = −2). The reflected point is (3, −2). Rotating (3, −2) by 90° clockwise about (1, 1): subtracting the centre gives 3 − 1 = 2 and −2 − 1 = −3, the clockwise rule swaps and negates these to give −3 and −2, and adding the centre back gives 1 + (−3) = −2 and 1 + (−2) = −1. The final image is (−2, −1). Answer: (−2, −1). Reflect before you rotate, exactly as the design process is described, and rotate about the CENTRE (1, 1) given in the question rather than the origin: either mistake, or reversing the two steps, sends the tile to a different point.
- (d) (6, 2) — Applying the first vector: (2, 1) + (5, −3) = (7, −2), which is the warehouse. Applying the second vector: (7, −2) + (−1, 4) = (6, 2), the delivery address. '(7, −2)' stops at the warehouse and forgets the second flight. '(8, −6)' comes from adding (1, −4) instead of (−1, 4) for the second vector, getting both signs wrong. '(11, −3)' comes from swapping the components of the second vector to (4, −1) before adding.
- (b) (0, 2) — Method: to rotate about a point that is not the origin, first subtract the centre's coordinates, apply the rotation rule to the shifted point, then add the centre's coordinates back on; only after that do you apply the translation, in the order the question states them. Working: shifting P relative to the centre gives (6 − 1, 4 − 2) = (5, 2); rotating 90° clockwise sends (x, y) to (y, −x), giving (2, −5); adding the centre back on gives (2 + 1, −5 + 2) = (3, −3); applying the translation (−3, 5) gives (3 − 3, −3 + 5) = (0, 2). Answer: (0, 2). Applying the translation BEFORE the rotation, reversing the order the question gives them in, gives (8, 0); stopping after the rotation and forgetting the translation altogether gives (3, −3); and rotating anticlockwise instead of clockwise, using (x, y) → (−y, x), gives (−4, 12). Always carry out the two transformations in the order stated — rotate about the given centre first, then translate — and check each step before moving to the next.
- (d) Triangle 2, by 3.6 cm² — Method: find both areas with 1/2ab sin C, then compare them. Working: area of triangle 1 = 1/2 × 10 × 13 × sin 64° = 58.4 cm²; area of triangle 2 = 1/2 × 11 × 12 × sin 70° = 62.0 cm²; triangle 2 is larger, by 62.0 − 58.4 = 3.6 cm². Getting the right difference but naming triangle 1 as the larger one, the subtraction done the wrong way round, gives 'Triangle 1, by 3.6 cm²'; leaving out the 1/2 when finding triangle 1's area (giving 116.8 cm² instead of 58.4 cm²) and then subtracting gives 'Triangle 1, by 54.8 cm²'; and using cos 64° instead of sin 64° for triangle 1 (giving 28.5 cm² instead of 58.4 cm²) gives 'Triangle 2, by 33.5 cm²'. Work out both areas fully and correctly before comparing which is bigger.
- (b) 128° — Angle AOC and angle ABC stand on the same arc AC (the minor arc, which does not contain B), so by the angle at the centre theorem angle ABC = 104° ÷ 2 = 52°. ABCD is a cyclic quadrilateral, so its opposite angles ABC and ADC sum to 180°: angle ADC = 180° − 52° = 128°. A candidate who finds angle ABC = 52° correctly but then treats opposite angles as equal, as in a parallelogram, instead of supplementary, writes down 52° and stops there. Skipping the halving step and using 104° as angle ABC gives 180° − 104° = 76°. Reading off the given centre angle itself as the final answer, without applying either theorem, gives 104°. Work through both theorems in order and you land on 128°.
- (a) 5√2 cm — Method: the angles of a triangle add to 180°, so the third angle is 45° as well and the two shorter sides are equal. Take one of them as the side opposite a 45° angle and use sin 45° = opposite ÷ hypotenuse. Working: the exact value of sin 45° is √2/2, so the shorter side = 10 × √2 ÷ 2, and half of 10 is 5. Answer: 5√2 cm, which is about 7.07 cm. Remembering sin 45° as √2 rather than as √2 halved gives 10√2 cm, which is longer than the hypotenuse. Halving the hypotenuse because 45° is half of 90° gives 5 cm. Taking the value from the other special triangle, sin 60° = √3/2, gives 5√3 cm.
- (c) −1.5 — Since n = k × m, dividing a number in n by the matching number in m gives k: k = −6 ÷ 4 = −1.5 (check with the bottom numbers: −9 ÷ 6 = −1.5, the same value, confirming n is a scalar multiple of m). 1.5 has the correct size but is missing the negative sign. −10 comes from subtracting the top numbers, −6 − 4, instead of dividing them. −24 comes from multiplying the top numbers, −6 × 4, instead of dividing them.
- (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.
- (a) 35.9 cm² — Method: the area of any triangle is 1/2 × a × b × sin C, where a and b are two sides and C is the angle between them. Working: the 47° angle lies between AB = 8.6 cm and AC = 11.4 cm, so the area is 1/2 × 8.6 × 11.4 × sin 47° = 49.02 × 0.73135 = 35.851. Answer: the area is 35.9 cm² to 1 decimal place. The distractors: 71.7 cm² comes from leaving out the factor 1/2 and working out 8.6 × 11.4 × sin 47°; 33.4 cm² comes from pressing cos instead of sin, 49.02 × cos 47°, which is the same as using the complement 43° in place of 47°; 52.6 cm² comes from pressing tan instead of sin, 49.02 × tan 47°.
- (d) (−14, 5) — To undo a composition, reverse the order and invert each transformation: undo the rotation first, then undo the translation. The inverse of 'rotate 90° clockwise about the origin' is 'rotate 90° anticlockwise about the origin', which maps (x, y) to (−y, x); applied to (3, 9) this gives (−9, 3). Then undo the translation by subtracting the vector (5, −2), i.e. adding (−5, 2): (−9 − 5, 3 + 2) = (−14, 5). Undoing the two inverse steps in the same order as the original composition, rather than reversing it, gives (−11, −2). Rotating 90° clockwise again instead of inverting the rotation's direction gives (4, −1). Adding the translation vector again instead of subtracting it, after correctly inverting the rotation, gives (−4, 1).
- (b) 13 — Using the sum of interior angles formula, (n − 2) × 180° = 1980°, so n − 2 = 1980 ÷ 180 = 11, and n = 11 + 2 = 13. 11 stops after the division, forgetting to add 2 back to find n. 15 adds 2 twice by mistake, giving 11 + 2 + 2. 22 divides 1980 by 90 instead of 180.
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