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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- (a) Parallel, since both gradients are 2 — Gradient of line 1 = (5 − 1) ÷ (2 − 0) = 4 ÷ 2 = 2. Gradient of line 2 = (6 − 2) ÷ (5 − 3) = 4 ÷ 2 = 2. The two gradients are equal, so the lines are parallel. "Not parallel, since the gradients are 2 and 4" uses 4 for line 2, which comes from dividing the change in y, 4, by 1 rather than by the change in x, 5 − 3 = 2. "Not parallel, since the two lines cross the y-axis at different points" confuses parallel lines with the same line repeated — parallel lines are distinct lines, and where a line crosses the y-axis has no effect on its gradient. "Parallel, since both lines pass through positive coordinates" reaches the right conclusion for the wrong reason: lines are parallel because their gradients are equal, not because the coordinates given happen to be positive.
- (b) A semicircle of radius 4 m, away from the wall. — Every point the dog can reach is at most 4 m from the fixed ring, so without any wall the region would be a full circle of radius 4 m. The wall runs straight through the ring and blocks the dog from crossing it, and since the wall extends further than the lead in both directions, exactly half of that circle is cut off — leaving a semicircle of radius 4 m on the side of the wall the dog is tied on. (A full circle of radius 4 m ignores that the wall blocks half of the region; a quarter circle of radius 4 m would only be correct if the ring were fixed at a corner where two walls met, not along a single straight wall; a rectangle 4 m wide along the wall ignores that the lead lets the dog swing round in a curve, not stay a fixed distance out from the wall.)
- (a) 245 m² — Method: for similar figures the ratio of the areas is the square of the ratio of the lengths, so multiply the smaller area by the square of the length scale factor. Working: the length scale factor is 7 ÷ 3, so the area scale factor is 49 ÷ 9, and the larger area is 45 × 49 ÷ 9 = 5 × 49 = 245. Answer: 245 m². The distractors: 105 m² comes from multiplying by the length scale factor 7 ÷ 3 instead of by its square, the commonest slip on this topic; 315 m² comes from multiplying by 7 and forgetting to divide by 3; 405 m² comes from multiplying by 3² = 9, squaring the wrong part of the ratio.
- (c) 5√3 cm — The space diagonal of a cube with edge a satisfies d² = a² + a² + a² = 3a², using Pythagoras' theorem in three dimensions. With a = 5, d² = 3 × 5² = 3 × 25 = 75, so d = √75 = √(25 × 3) = 5√3 cm. Finding the diagonal of one face instead, using only two of the three edges, gives d = √(5² + 5²) = √50 = 5√2 cm, which leaves out the third dimension. Adding the three edges directly, 5 + 5 + 5 = 15 cm, ignores that Pythagoras' theorem works with squares of lengths, not the lengths themselves. Squaring the edge and multiplying by 3 correctly, 3 × 5² = 75, but then forgetting to take the square root, leaves 75 cm — the squared length, not the diagonal itself.
- (b) a rectangle — Lying on its side, the cylinder's curved surface touches the table along a straight line, and the two flat circular ends face sideways rather than up or down; viewed from directly above, the outline traced is a rectangle — as long as the cylinder and as wide as its diameter. "a circle" would be correct if the cylinder stood upright on one of its circular ends instead of lying on its side. "a triangle" belongs to a cone lying or standing so that it narrows to a point in that view, which a cylinder never does. "an oval" is a common guess from picturing the round ends, but from directly above those ends are edge-on and contribute to the rectangle's short sides, not a curved outline.
- (a) 2 — Gradient = (change in y) ÷ (change in x) = (11 − 3) ÷ (6 − 2) = 8 ÷ 4 = 2. "0.5" comes from dividing the change in x by the change in y the wrong way round: 4 ÷ 8. "8" is only the change in y, forgetting to divide by the change in x at all. "−2" comes from a sign error, as if the y-coordinate had decreased rather than increased.
- (d) 13 — Method: OP and PQ meet at a right angle because of the tangent–radius fact, so triangle OPQ is right-angled at P; use Pythagoras' theorem. Working: OQ² = OP² + PQ² = 5² + 12² = 25 + 144 = 169; OQ = √169 = 13. A student who answers 17 has simply added the two given lengths (5 + 12) instead of using Pythagoras' theorem. A student who answers 7 has subtracted the two given lengths (12 − 5) instead of using Pythagoras' theorem. A student who answers 144 has correctly squared 12 but stopped there, forgetting to add 5² and take the square root. Answer: 13 cm.
- (c) A rotation of 180° about the origin — An enlargement by scale factor −1 sends every point (x, y) to (−x, −y) — both coordinates change sign. A rotation of 180° about the origin does exactly the same thing to every point, so the two transformations have identical effect. A reflection in the x-axis only changes the sign of the y-coordinate, sending (x, y) to (x, −y), leaving the x-coordinate untouched. A reflection in the y-axis only changes the sign of the x-coordinate, sending (x, y) to (−x, y), leaving the y-coordinate untouched. Treating a negative scale factor as though it behaves like a positive one gives no transformation at all, but the minus sign is not decorative — it reverses both coordinates. Both signs flip together, which is exactly what a 180° rotation about the origin does.
- (c) ∠XYZ — The angle at a named vertex is written with that vertex's letter in the middle, flanked by its two neighbouring vertices. The angle at Y sits between X and Z, its neighbours in quadrilateral WXYZ, so it is written ∠XYZ. ∠WXY names the angle at X, since X is the middle letter, not Y. ∠YZW names the angle at Z, since Z is the middle letter. ∠ZWX names the angle at W, since W is the middle letter.
- (c) 5 — The front elevation shows one square for every cube visible from the front, column by column: the left-hand column is 2 cubes high, so it contributes 2 squares; the middle column is 2 cubes high, so it contributes 2 more; the right-hand column is 1 cube high, so it contributes 1. The total is 2 + 2 + 1 = 5 squares. "6" comes from drawing a full 3 by 2 rectangle, treating every column as if it reached the greatest height. "4" comes from losing a square from one of the two tall columns, counting 2 + 1 + 1. "3" comes from counting one square per column — the width of the solid — and ignoring the heights altogether.
- (b) 166.3 cm² — Method: split the regular hexagon into 6 identical triangles meeting at the centre, each with two sides of 8 cm and a 60° angle between them, and use Area = (1/2)ab sin C on just one of them. Working: one triangle's area = 1/2 × 8 × 8 × sin 60° = 27.7 cm² (1 d.p.); the hexagon is 6 of these, so its area is 6 × 27.7 = 166.3 cm² (1 d.p.). Answer: 166.3 cm². Reporting just one triangle's area, without multiplying by 6, gives 27.7 cm²; treating the angle at the centre as a right angle instead of 60°, using 1/2 × 8 × 8 with no sine factor at all, gives 6 × 32 = 192.0 cm²; and multiplying by 5 instead of 6, miscounting the triangles in the hexagon, gives 5 × 27.7 = 138.6 cm². A regular hexagon always splits into exactly 6 triangles at its centre — count them before you multiply.
- (d) 1 — sin 30° = 1/2, so (sin 30°)² = 1/4. cos 30° = √3/2, so (cos 30°)² = 3/4. Adding these gives 1/4 + 3/4 = 1. '1/4' only calculates (sin 30°)² and forgets to add the cos 30° term. '3/4' only calculates (cos 30°)² and forgets to add the sin 30° term. '−1/2' comes from subtracting the two squared values instead of adding them: 1/4 − 3/4 = −1/2.
- (a) 120° — Method: three sides are known, so use the cosine rule rearranged as cos A = (b² + c² − a²) ÷ 2bc, with a the side facing the angle wanted. Working: angle ABC lies between AB = 5 cm and BC = 3 cm and faces AC = 7 cm, so cos ABC = (5² + 3² − 7²) ÷ (2 × 5 × 3) = (25 + 9 − 49) ÷ 30 = −15 ÷ 30 = −0.5. The angle between 0° and 180° whose cosine is −0.5 is 180° − 60°. Answer: angle ABC = 120°. The distractors: 60° comes from taking the subtraction the other way round, (49 − 25 − 9) ÷ 30 = 0.5, which loses the minus sign that makes the angle obtuse; 90° comes from the instinct that three known sides always mean Pythagoras, and 5² + 3² = 34 is not 49, so the triangle is not right-angled; 150° comes from knowing the cosine is −0.5 but subtracting 30° from 180°, using the angle whose sine is 0.5 rather than the angle whose cosine is 0.5.
- (a) Reflex + non-reflex angle BOD = 360°, so 2a + 2c = 360°. — A correct proof must build in a strictly logical order: the reflex and non-reflex angles at O, which together make a complete turn about the centre, must be added to 360° before 2a and 2c can be combined and divided. The statement 'Reflex + non-reflex angle BOD = 360°, so 2a + 2c = 360°' is the only one that follows directly from having 2a and 2c already established, and it is exactly what is needed before the final division step. Jumping straight to 'Divide by 2 throughout to get a + c = 180°' skips the step that justifies why 2a + 2c equals 360° in the first place — there is nothing yet to divide. Repeating 'Angle BOD is 2a and 2c, by the centre theorem' does not move the proof forward at all, since that fact has already been established in the sentence given. Introducing 'OB = OD are radii, so triangle OBD is isosceles' brings in an unrelated triangle and an unrelated method that plays no part in this particular proof.
- (b) It is the longest of the three chords — Method: in any circle the length of a chord is decided by how far the chord lies from the centre, because a chord passing nearer the centre cuts further across the circle. Working: the chord through O lies at a distance of zero from the centre, and no chord can lie closer than that, so no chord of the circle can be longer than it; a chord through the centre is a diameter. The other two chords lie at some distance greater than zero, so each of them falls short of that maximum. Answer: It is the longest of the three chords. The distractors: It is the shortest of the three chords comes from reversing the rule and picturing a chord near the centre as a short line tucked inside; It is the same length as the other two chords comes from carrying the fact that all radii of a circle are equal across to chords, which are not all equal; It is half the length of each of the other two chords comes from confusing a chord through the centre with a radius, which really is half a diameter.
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