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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- (b) $\binom{-0.4}{-7}$ — The overall movement is the sum of the two vectors: (−1.2 + 0.8, −4.5 + (−2.5)) = (−0.4, −7). '$\binom{-2}{-2}$' comes from subtracting the second vector from the first instead of adding them. '$\binom{-0.4}{7}$' gets the top number right but drops the negative sign on the bottom number. '$\binom{2}{-7}$' comes from treating −1.2 + 0.8 as if the signs did not matter, giving +2 instead of −0.4.
- (d) 2 — Method: if f is 3 times e, then each part of f equals 3 times the matching part of e. Working: using the bottom numbers, 6 = 3 × k, so k = 2. Answer: k = 2. A candidate who multiplies instead of dividing, working out 6 × 3, gets 18. A candidate who uses the top numbers' ratio instead, 15 ÷ 5, and gives that ratio as k gets 3. A candidate who adds instead of using the multiple relationship, working out 6 + 3, gets 9.
- (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.
- (b) 3 — Method: the scale factor is a length of the second triangle divided by the corresponding length of the first, and corresponding means shortest paired with shortest. Working: the shortest side of the first triangle is 3 cm and the shortest side of the second is 9 cm, so the scale factor is 9 ÷ 3 = 3. Answer: 3. The distractors: 1/3 comes from dividing 3 by 9, which is the scale factor from the second triangle back to the first; 6 comes from subtracting, 9 − 3, treating the scale factor as a difference rather than a multiplier; 1.8 comes from dividing 9 by 5, pairing the second triangle's shortest side with the first triangle's longest side.
- (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) Cuboid — not necessarily a cube — A solid with 6 faces, 12 edges and 8 vertices in which every face is a rectangle is a cuboid, but nothing here confirms that all the edges are the same length, so the box could be a cube or a non-cube cuboid; the most that can be concluded is that it is a cuboid, making 'Cuboid — not necessarily a cube' correct. 'Cube — only a cube fits this' is wrong because a cube is just one particular cuboid; a general cuboid with different length, width and height has exactly the same face, edge and vertex counts and rectangular faces. 'Triangular prism' is wrong because a triangular prism has 5 faces, 9 edges and 6 vertices, and two of its faces are triangles, so it matches neither the counts nor the face shape. 'Not enough information' is wrong because rectangular faces with these counts do pin the solid down to the cuboid family, even though they cannot pin down a cube specifically.
- (d) 6 cm — Volume = πr²h, so height = volume ÷ (πr²) = 471 ÷ (3.14 × 25) = 471 ÷ 78.5 = 6 cm. (30 cm comes from dividing by πr instead of πr², missing one factor of the radius; 150 cm comes from dividing by π only, without using r² at all; 24 cm comes from treating the given 5 cm as a diameter and using a radius of 2.5 cm instead.)
- (c) (1, 0) — To enlarge about a centre other than the origin, find the vector from the centre to the point, scale that vector, then add it back to the centre. The vector from (2, 2) to A(4, 6) is (2, 4). Scaling by −1/2 gives (−1, −2). Adding this to the centre (2, 2) gives the image point (1, 0). (3, 4) comes from using +1/2 instead of −1/2, so the image lands on the same side as A instead of the opposite side. (−2, −3) comes from scaling A's coordinates directly about the origin, ignoring that the centre is (2, 2). (−2, −6) comes from using a scale factor of −2 instead of −1/2.
- (c) (2, 1) — A point that lies on both mirror lines is fixed by each reflection individually, and so is fixed by the combination of the two — it is the intersection point of l1 and l2 that is invariant. Substituting x = 2 into y = x − 1 gives y = 2 − 1 = 1, so the intersection point is (2, 1). Forgetting the '− 1' in l2's equation and using y = x instead gives (2, 2). Making a sign error and computing y = x − (−1) = x + 1 instead gives (2, 3). Solving for x from an assumed y = 0 instead of substituting the given x = 2 gives (1, 0). Substitute x = 2 into l2's equation correctly, and the invariant point is (2, 1).
- (d) 38 m² — The two fences meet at a right angle, so the rope sweeps a quarter of a circle: 90 ÷ 360 = 1/4. Grazing area = (90 ÷ 360) × 3.14 × 7² = 0.25 × 153.86 = 38.465 m², which rounds to 38 m². (5 m² comes from forgetting to square the rope length; 154 m² comes from finding the area of a full circle and forgetting the angle fraction; 11 m² comes from using the arc length formula instead of the sector area formula.)
- (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 $\binom{-5}{-5}$.
- (b) 60 units² — Method: the area of a triangle is half the base times the perpendicular height, so choose a side to act as the base and measure the perpendicular distance from the opposite vertex to it. Working: A(0, 0) and B(10, 0) both lie on the x-axis, so AB is horizontal and AB = 10 − 0 = 10. The perpendicular height is the distance of C from the x-axis, which is its y-coordinate, 12. Area = (10 × 12) ÷ 2 = 120 ÷ 2 = 60. Answer: 60 units². The distractors: 120 units² comes from multiplying base by height and forgetting to halve; 65 units² comes from using the slanting side AC, which is 13 long, as the height in place of the perpendicular distance 12; 30 units² comes from halving the base to 5 before multiplying and then halving the product as well, so the halving is done twice.
- (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) 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) 44.7 m — Method: the line joining the two tops is the hypotenuse of a right-angled triangle whose horizontal side is the gap between the masts and whose vertical side is the difference in their heights, so Pythagoras' theorem applies. Working: the difference in heights is 50 − 30 = 20 m, so d² = 40² + 20² = 1600 + 400 = 2000 and d = √2000 = 44.721…, which is 44.7 m to 1 decimal place. Answer: 44.7 m. The distractors: 34.6 m comes from subtracting the squares, √(40² − 20²), instead of adding them; 60.0 m comes from adding the two sides of the triangle, 40 + 20, rather than using Pythagoras' theorem; 50.0 m is the height of the taller mast, copied from the question in place of the distance asked for.
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