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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Answer key: Geometry and measures worksheet — GCSE Higher
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- (b) 1 point — A rotation about a point P always leaves P itself unchanged, and — unless the shape has rotational symmetry about that exact point — no other point of the shape maps onto itself. The centre of rotation here is the vertex (2, 1), so exactly one point of the trapezium, that vertex, is invariant. Thinking that a rotation always has no invariant points ignores the centre of rotation itself, which is always fixed, and gives 0 points. Assuming every vertex of the shape is invariant confuses a rotation with the identity transformation and gives 4 points, the total number of vertices. Believing that the point diametrically opposite the centre is also fixed applies point symmetry of the whole coordinate grid rather than checking whether that point actually lies on this particular trapezium, and gives 2 points.
- (b) 5 — Method: the two points have the same y-coordinate, so the segment joining them runs horizontally and its length is the gap between the two x-coordinates; a length is a distance, so it is never negative. Working: the x-coordinates are 2 and −3, so the gap is 2 − (−3) = 2 + 3 = 5. Both points have y = −4, so there is no vertical part to add on. Answer: 5. The distractors: 1 comes from dropping the minus sign on −3 and working out 3 − 2 instead; 0 comes from subtracting the y-coordinates, which are equal, in place of the x-coordinates; 6 comes from counting the grid lines from −3 across to 2 inclusive, which counts one more than the number of gaps between them.
- (a) 180 cm³ — Method: the volume of a right prism is the area of its cross-section multiplied by its length, and the area of a triangle is half the base multiplied by the perpendicular height. Working: the cross-section has area (6 × 5) ÷ 2 = 15 cm², and 15 × 12 = 180. Answer: 180 cm³. The distractors: 360 cm³ comes from taking the cross-section as 6 × 5 = 30 and never halving it, which measures the rectangle around the triangular face rather than the face itself; 66 cm³ comes from adding the base and the perpendicular height and halving, (6 + 5) ÷ 2 = 5.5, which is the trapezium rule used where the triangle rule is needed, and then multiplying by the 12 cm length; 15 cm³ comes from working out the triangular cross-section correctly and stopping there, so the 12 cm length is never used and an area is handed in as a volume.
- (b) (3, −1) — Method: find the distance from the point to the mirror line, then place the image the same distance on the other side of the line. Working: B is 5 − 2 = 3 units above the line y = 2, so its image is 3 units below the line, at y = 2 − 3 = −1, giving (3, −1); the x-coordinate does not change, since the mirror line is horizontal. Options: (3, −5) comes from reflecting in the x-axis (y = 0) instead of the line y = 2; (−3, 5) comes from reflecting in the y-axis instead of a horizontal line; (3, −4) comes from doubling the distance from the line instead of reflecting it, giving 2 − 2×3 = −4. Answer: (3, −1).
- (b) 6√3 m — The horizontal distance covered by one support is adjacent to the 30° angle, so it equals 6 × cos 30° = 6 × √3/2 = 3√3 m. The total base width is made up of both supports, so it is 2 × 3√3 = 6√3 m. '3√3 m' gives only one support's horizontal distance and forgets to double it for the total width. '6 m' comes from using sin 30° instead of cos 30° for the horizontal distance (6 × sin 30° = 3, doubled to 6). '12 m' comes from doubling the full sloping length of 6 m without using any trigonometry at all.
- (d) 1687.5 ml — The length scale factor from the standard bottle to the giant bottle is 1.5, so the volume scale factor is 1.5³ = 3.375. The capacity of the giant bottle is 500 × 3.375 = 1687.5 ml. 750 ml comes from multiplying by the length factor 1.5 directly, without cubing it. 1125 ml comes from using the area scale factor 1.5² = 2.25 instead of the volume scale factor. 2250 ml comes from treating 'cubed' as 'multiplied by 3', giving 1.5 × 3 = 4.5 as the scale factor instead of 1.5³.
- (a) a line parallel to both, 3 cm from each — Being equidistant from two parallel lines 6 cm apart means being exactly halfway between them all along their length, tracing out a third line, parallel to both, at 3 cm from each — half of the 6 cm gap. "a line parallel to both, 6 cm from each" repeats the full gap instead of halving it, which puts those points past one of the lines entirely. "a circle of radius 3 cm, centred midway" applies to a locus equidistant from a single fixed POINT, not from two parallel lines running the full length. "the perpendicular bisector of the gap" crosses the gap at right angles and meets each line at only one point — it is not the whole locus, which runs parallel to the lines, not across them.
- (c) 72° — Method: opposite angles in a cyclic quadrilateral always sum to 180°. Working: angle DAB and angle BCD are opposite angles of the cyclic quadrilateral ABCD, so angle BCD = 180 − 108 = 72 degrees. Answer: 72°. Opposite angles in a cyclic quadrilateral are SUPPLEMENTARY, not equal, so do not simply copy the given angle; and do not subtract from 360°, which is the rule for angles round a point, or confuse this with the angle-in-a-semicircle theorem, which gives a fixed 90° that has nothing to do with this quadrilateral.
- (a) 24 — The display is 4 tins wide, 3 tins high and 2 tins deep, and every position in that block is filled, so the total number of tins is the product of all three measurements: 4 × 3 × 2 = 24. "12" comes from multiplying only the width and height shown in the front elevation (4 × 3), forgetting the depth entirely. "9" comes from adding the three measurements (4 + 3 + 2) instead of multiplying them. "6" comes from multiplying only the height and depth (3 × 2), forgetting the width shown by the front elevation.
- (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.
- (c) 13 cm — Method: the two given sides meet at the right angle, so they are the shorter pair and the hypotenuse comes from Pythagoras' theorem, a² + b² = c². Working: c² = 5² + 12² = 25 + 144 = 169, so c = √169 = 13. Answer: 13 cm. The distractors: 17 cm comes from adding the two sides, 5 + 12, rather than adding their squares; 60 cm comes from multiplying them, 5 × 12, which is twice the area of the triangle and not a length; 7 cm comes from subtracting, 12 − 5, as though the hypotenuse were the difference of the two shorter sides.
- (d) (9, −4) — Reflecting (1, 3) in the line x = 4 gives (2 × 4 − 1, 3) = (7, 3). Reflecting (7, 3) in the line y = 1 gives (7, 2 × 1 − 3) = (7, −1). Translating (7, −1) by the vector (2, −3) gives (7 + 2, −1 + (−3)) = (9, −4). Stopping after the two reflections and forgetting the translation gives (7, −1). Applying the translation's y-component with the wrong sign, adding 3 instead of subtracting it, gives (9, 2). Forgetting that the two reflections are centred on (4, 1) rather than the origin, and instead rotating (1, 3) by 180° about the origin to (−1, −3) before translating, gives (1, −6). Do all three steps in order, each one correctly, and the final image is (9, −4).
- (d) 2/3 — Method: since A is the midpoint of OE, OE = 2a, so E = 2a. Since FB is twice AF, F is 1/3 of the way along AB from A, so F = a + 1/3(b − a) = (2/3)a + (1/3)b. G is the midpoint of OB, so G = (1/2)b. Working: EF = F − E = (2/3)a + (1/3)b − 2a = −(4/3)a + (1/3)b, and EG = G − E = −2a + (1/2)b. Comparing term by term, 2/3 × (−2a + (1/2)b) = −(4/3)a + (1/3)b, which matches EF exactly — the same scalar works on both the a-term and the b-term, so the two vectors are parallel, and since they share the point E the three points are collinear. Answer: k = 2/3, so E, F and G lie on a straight line. Giving 1/3 instead is the scalar linking F to G (FG = (1/3)EG), not E to F; giving 3/2 is the reciprocal — it is EG that equals 3/2 × EF, not the other way round, since EF = k × EG was what was asked for; and giving 4/3 is EF's a-coefficient read off raw, without ever dividing it by EG's a-coefficient to form the comparison. Always match the direction of the scalar to the vectors exactly as the question states them.
- (b) 2.8 m — Use Pythagoras' Theorem: the plank is the hypotenuse (3.5 m) of a right-angled triangle formed with the wall and the ground (2.1 m). height² = 3.5² − 2.1² = 12.25 − 4.41 = 7.84. height = √7.84 = 2.8 m. A student who subtracts the two given lengths directly instead of using Pythagoras gets 3.5 − 2.1 = 1.4 m. A student who doubles the distance from the wall by mistake gets 2.1 × 2 = 4.2 m.
- (b) £157 — Method: round the hours worked up to the next whole hour, multiply by the hourly rate, then add the call-out fee. Working: 3 hours 30 minutes rounds up to 4 hours; 28 × 4 = 112; 112 + 45 = 157. Answer: £157. A candidate who uses the unrounded time of 3.5 hours, working out 28 × 3.5 = 98 and adding 45, gets £143. A candidate who forgets the call-out fee, giving only 28 × 4, gets £112. A candidate who rounds down to 3 hours instead of up, working out 28 × 3 = 84 and adding 45, gets £129.
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