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
MathsUKwww.geekhero.co.uk
- (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) 320.3 cm³ — Cylinder volume = πr²h = 3.14 × 3² × 10 = 3.14 × 9 × 10 = 282.6 cm³. Cone volume = (1/3)πr²h = (1/3) × 3.14 × 9 × 4 = (1/3) × 113.04 = 37.68 cm³. Total = 282.6 + 37.68 = 320.28 cm³, which rounds to 320.3 cm³.
- (a) 71° — Method: first use the straight line through T to find the tangent-chord angle RTC, then apply the alternate segment theorem, which states that this angle equals the angle subtended by the chord in the alternate segment. Working: the two rails form a straight line through T, so angle RTC = 180 − 109 = 71 degrees. D lies in the alternate segment of the chord TC, so angle TDC = angle RTC = 71°. Answer: 71°. Find angle RTC FIRST from the straight line before applying the theorem: using the given 109° directly, doubling the tangent-chord angle, or taking its complement from 90° all give the wrong angle at D.
- (a) 49 cm² — Method: two sides and the angle between them are given, so use the cosine rule a² = b² + c² − 2bc cos A with BC facing the 60° angle. By hand, cos 60° = 0.5. Working: BC² = 8² + 5² − 2 × 8 × 5 × cos 60° = 64 + 25 − 80 × 0.5 = 89 − 40 = 49. Answer: BC² = 49 cm². The distractors: 129 cm² comes from adding the final term instead of subtracting it, 89 + 40; 89 cm² comes from leaving the cosine term out altogether and treating the 60° as though it were a right angle, so that Pythagoras applies; 69 cm² comes from forgetting the factor 2 in 2bc cos A and subtracting only 8 × 5 × 0.5 = 20.
- (c) 6 squares — Method: the plan view shows the footprint of the solid; a second layer stacked on top of floor positions that are already covered does not create any new squares in the plan. Working: the row of 4 cubes and the row of 2 cubes attached at the end do not overlap, so the footprint has 4 + 2 = 6 distinct squares. Answer: 6 squares. The distractors: 12 squares comes from counting the total number of cubes used, including the second layer (6 floor positions × 2 layers = 12), instead of the footprint. 4 squares comes from counting only the row of four and forgetting the attached row of two. 5 squares comes from wrongly treating the corner square as shared between the two rows (4 + 1 instead of 4 + 2).
- (d) A prism — A prism has two identical, parallel polygon faces (its cross-section) joined by flat rectangular side faces, so the same cross-sectional shape runs all the way along its length — this description matches a prism. A pyramid instead narrows from one polygon base up to a single point (the apex), so it does not have two identical parallel faces. A cone has one curved surface and a single circular base narrowing to a point — no flat rectangular sides at all. A cylinder has two identical circular faces, but they are joined by a curved surface, not by flat rectangles.
- (d) 6 m — The horizontal distance is adjacent to the 60° angle and the zip-wire is the hypotenuse, so horizontal distance = 12 × cos 60° = 12 × 1/2 = 6 m. 6√3 m comes from using sin 60° = √3/2 instead of cos 60°, which would give the vertical drop, not the horizontal distance. 4√3 m comes from treating 12 as the side adjacent to a tangent ratio and dividing by tan 60° = √3. 24 m comes from dividing 12 by cos 60° instead of multiplying by it.
- (b) (1/2)b − (1/2)a — Method: MN runs from M to N, so MN = ON − OM, with OM = (1/2)a and ON = (1/2)b. Working: MN = (1/2)b − (1/2)a. Answer: MN = (1/2)b − (1/2)a. Subtracting the other way round gives (1/2)a − (1/2)b, the reverse vector from N to M; subtracting the wrong way round AND forgetting to halve gives a − b, which is BA, not MN; and adding the two halved vectors instead of subtracting them gives (1/2)a + (1/2)b, which is the position vector of the midpoint of AB. Always subtract the START point's vector from the END point's vector, and halve OA and OB before you combine them, not after.
- (a) 7.2 cm — The distance around the mat is the circumference, π × diameter = 3.14 × 20 = 62.8 cm. Yasmin started with 70 cm, so the ribbon left over is 70 − 62.8 = 7.2 cm. 62.8 cm is the circumference itself, the amount of ribbon used, not what is left over. 50 cm comes from subtracting the diameter instead of the circumference: 70 − 20 = 50. 20 cm is simply the diameter of the mat and involves no calculation with the ribbon length at all.
- (a) Square-based pyramid — A solid with one square base and four triangular faces meeting at a single apex above the base is a square-based pyramid. A triangular prism has two triangular faces and three rectangular faces, not a square base with four triangles, so that is a different solid. A cube has six square faces, and a cuboid has six rectangular faces — neither has any triangular faces at all. The solid described is a square-based pyramid.
- (d) 1 — An isosceles trapezium has one line of symmetry, running through the midpoints of the two parallel sides. 0 would be true for a scalene trapezium, whose non-parallel sides are unequal, but this trapezium's non-parallel sides are equal, so it is symmetrical. 2 is the number of lines of symmetry of a rectangle, not a trapezium. 4 comes from wrongly applying the 'number of sides equals number of lines of symmetry' rule, which only holds for REGULAR polygons — a trapezium is not regular.
- (a) Triangular prism — Count the given properties against each solid. A triangular prism has 5 faces (2 triangular ends + 3 rectangular sides), 9 edges and 6 vertices — this matches exactly, so the solid is a triangular prism. A square-based pyramid also has 5 faces, but 8 edges and 5 vertices, and only one face is a square rather than three rectangles — the edge and vertex counts don't match. A cuboid has 6 faces, 12 edges and 8 vertices, none of which match the numbers given. A tetrahedron has 4 faces, all triangles, with 6 edges and 4 vertices — too few faces, and no rectangular faces at all.
- (d) 10 — Method: the distance between two points is the hypotenuse of a right-angled triangle whose shorter sides are the horizontal and vertical gaps, so work out both gaps first, handling the negative coordinates carefully, and then apply Pythagoras' theorem. Working: the horizontal gap is 5 − (−3) = 5 + 3 = 8 and the vertical gap is 4 − (−2) = 4 + 2 = 6. Then d² = 8² + 6² = 64 + 36 = 100, so d = √100 = 10. Answer: 10. The distractors: 14 comes from adding the two gaps, 8 + 6, instead of adding their squares and taking the root; 100 comes from stopping at the sum of the squares and never taking the square root; 50 comes from reaching 100 correctly and then halving it instead of taking its square root, a candidate who has read the last step as “halve” rather than “root”.
- (c) Two angles and one side are known; x is another side — The sine rule needs a matching pair, a side and the angle opposite it, that you already know, so you can set up a ratio with the unknown. When two angles and one side are known, you can find the third angle from the angle sum, giving you an angle opposite the known side and an angle opposite x: the sine rule applies directly. When all three sides are known and x is an angle, there is no side-angle pair available at all, so the cosine rule, rearranged for an angle, is what's needed instead. When two sides and the included angle are known and x is the third side, again there is no matching side-angle pair yet, so the cosine rule finds the third side directly. When two sides and the included angle are known and x is one of the other angles, you still have no side-angle pair to start from — the cosine rule has to be used first, to find the third side, before any angle can be found. Only the two-angles-and-a-side case hands you a ready-made pair, which is exactly what the sine rule needs.
- (b) $\binom{-5}{5}$ — Method: add the two flights to get the single vector of the whole journey out, then reverse that vector to get the journey home. Working: across, 6 − 1 = 5; up, 4 − 9 = −5. So the drone finishes at (7, −8), which is 5 to the right of its start and 5 below it. The way home is therefore 5 to the left and 5 up. Answer: $\binom{-5}{5}$. Giving the combined journey itself, $\binom{5}{-5}$, describes the flight out rather than the flight home. Subtracting the second vector instead of adding it makes the journey out 7 across and 13 up, and reversing that gives $\binom{-7}{-13}$. Reversing the horizontal movement but leaving the vertical one alone gives $\binom{-5}{-5}$.
Build your own mix at the worksheet builder.