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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- (a) 8 — A cylinder has two flat circular faces (the top and the base) and one curved surface, so only its 2 flat faces are wrapped. A cuboid has 6 faces and all of them are flat, so all 6 are wrapped. Adding these: 2 + 6 = 8, so 8 is correct. 9 comes from wrongly counting the cylinder's curved surface as a flat face. 6 comes from counting only the cuboid and forgetting the cylinder's two flat circular faces. 3 comes from counting only the cylinder and including its curved surface in that total.
- (b) 7 — Method: each coordinate of a midpoint is the mean of the matching pair of coordinates, so the y-coordinate of the midpoint depends on the two y-coordinates alone. Working: the y-coordinates are 4 and 10, so the mean is (4 + 10) ÷ 2 = 14 ÷ 2 = 7. Answer: 7. The distractors: 5 comes from working out the x-coordinate of the midpoint, (2 + 8) ÷ 2 = 5, and writing that down in place of the y-coordinate the question asked for; 3 comes from halving the difference of the y-coordinates, (10 − 4) ÷ 2 = 3, which is half the vertical gap rather than a position; 14 comes from adding the two y-coordinates and forgetting to halve the total.
- (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°.
- (c) (−35, −55) — Add the three stages component by component to find the drone's position relative to base: (30+(−10)+15, 40+20+(−5)) = (35, 55). The flight back to base is the negative of this vector, reversing both numbers: (−35, −55). (35, 55) is the vector from base to the drone's position — it forgets to reverse direction for the return flight. (−35, 55) only reverses the top number. (35, −55) only reverses the bottom number.
- (a) 1023 m² — Method: split the quadrilateral along the diagonal AC into two triangles. Triangle ABC has two sides and the angle between them, so the cosine rule gives AC and the area formula gives its area; triangle ACD then has three known sides, so the cosine rule gives an angle and the area formula gives its area. Working: AC² = 40² + 32² − 2 × 40 × 32 × cos 95° = 1600 + 1024 + 223.12 = 2847.12, so AC = 53.358 m. The area of triangle ABC is 1/2 × 40 × 32 × sin 95° = 640 × 0.99619 = 637.56 m². In triangle ACD, cos ADC = (25² + 36² − 2847.12) ÷ (2 × 25 × 36) = (1921 − 2847.12) ÷ 1800 = −0.51451, so angle ADC = 120.965° and sin ADC = 0.85748, giving an area of 1/2 × 25 × 36 × 0.85748 = 385.87 m². The total is 637.56 + 385.87 = 1023.43. Answer: the field has an area of 1023 m² to the nearest square metre. The distractors: 1071 m² comes from taking cos 95° as positive, so the diagonal is found as 49.00 m instead of 53.358 m and the second triangle comes out too large; 2047 m² comes from leaving the factor 1/2 out of both area calculations; 638 m² is the area of triangle ABC alone, written down by a candidate who finds the diagonal and then forgets that the second triangle is part of the field.
- (c) 12 cm — By Pythagoras' Theorem, QR² = PQ² + PR², so PR² = QR² − PQ² = 15² − 9² = 225 − 81 = 144. Square root: √144 = 12 cm.
- (d) 46.2 cm — The perimeter of a sector is the arc length plus its two straight radii. The circumference is 2 × 3.14 × 10 = 62.8 cm, and the arc is 150 ÷ 360 of that: 62.8 × 150 ÷ 360 = 26.2 cm (1 d.p.). Adding the two radii, 26.2 + 10 + 10 = 46.2 cm. Giving just the arc length, without adding the straight edges, gives 26.2 cm. Adding only ONE radius instead of two gives 36.2 cm. Using 150 ÷ 180 instead of 150 ÷ 360 for the fraction gives an arc of 52.3 cm and a perimeter of 72.3 cm.
- (d) 49° — Method: first combine the two given angles at the centre to find the whole central angle FOM, then apply the angle-at-the-centre theorem, which halves the central angle to give the angle at the circumference on the major arc. Working: angle FOM = angle FOP + angle POM = 55 + 43 = 98 degrees. Since S is on the major arc FM, angle FSM = 98 ÷ 2 = 49 degrees. Answer: 49°. Add BOTH given angles to find the whole angle FOM before you halve anything: halving only one of them, doubling the total instead of halving it, or subtracting from 180° all give the wrong angle for the programme.
- (d) No — the volume factor is 2³ = 8, not the area factor 4. — For similar solids, area scales with the square of the length scale factor and volume scales with its cube — different powers of the same number, so they are not usually equal. The length scale factor here is 2, so the area scale factor is 2² = 4 (Priya's figure) but the volume scale factor is 2³ = 8. Priya is wrong: the correct volume scale factor is 8, not 4. The option claiming area and volume always scale by the same factor treats the two as interchangeable, which is only true for the length factor itself. The option giving 6 comes from multiplying the length factor by the number of dimensions (2 × 3) rather than cubing it. The option repeating '4' for volume simply reuses Priya's own area calculation.
- (d) 24 cm²; same way round as T — An enlargement scales area by the square of the scale factor, whatever its sign: area factor = (−2)² = 4, so the image's area is 6 × 4 = 24 cm². A negative scale factor is equivalent to an enlargement by the positive scale factor together with a rotation of 180°, so it does not create a mirror image — it is still a direct transformation, and combining it with a further rotation cannot create one either, since rotations never change a shape's orientation. Using the scale factor itself rather than its square gives area 6 × 2 = 12 cm², which is too small. Believing that a negative scale factor reflects the shape gives 'a mirror image of T' alongside the correct area of 24 cm², and combining both mistakes gives 12 cm² together with 'a mirror image of T'.
- (a) A square's diagonals are equal in length — A square has two pairs of parallel sides, four equal sides, and diagonals that are equal in length and cross at right angles. This tile's diagonals are not equal, so it is a rhombus that is not a square.
- (a) 100° — In a kite, the pair of angles between the unequal sides are equal to each other. Angle X and angle Z are both between one side from the WX/WZ pair and one side from the XY/ZY pair, so angle Z = angle X = 100°.
- (a) 5 — Method: BC faces the 60° angle, so put the two algebraic sides into the cosine rule and solve the equation that results. By hand, cos 60° = 0.5. Working: 7² = x² + (x + 3)² − 2 × x × (x + 3) × 0.5, so 49 = x² + x² + 6x + 9 − x² − 3x = x² + 3x + 9. That rearranges to x² + 3x − 40 = 0, which factorises as (x + 8)(x − 5) = 0, giving x = −8 or x = 5. A length cannot be negative, so the negative root is rejected. Answer: x = 5. The distractors: 8 comes from expanding (x + 3)² as x² + 9, the error of assuming that squaring a bracket squares each term, which turns the equation into x² − 3x − 40 = 0; −8 is the negative root, kept by a candidate who solves the quadratic but never checks that x has to be a length; 37 comes from applying the cosine rule as though the 60° angle were at C facing AB, which gives x² = (x + 3)² + 49 − 7(x + 3) and a single linear solution.
- (a) 9.3 cm — Method: the 40° angle is not between the two known sides, so call the unknown side x and put it into the cosine rule, which turns into a quadratic equation with two positive roots. Working: AC faces angle ABC, so 6² = 7² + x² − 2 × 7 × x × cos 40°, that is 36 = 49 + x² − 10.7246x, which rearranges to x² − 10.7246x + 13 = 0. The discriminant is 10.7246² − 4 × 13 = 115.02 − 52 = 63.02, whose square root is 7.9384, so x = (10.7246 + 7.9384) ÷ 2 = 9.3315 or x = (10.7246 − 7.9384) ÷ 2 = 1.3931. The longer of the two is wanted. Answer: BC = 9.3 cm to 1 decimal place. The distractors: 1.4 cm is the shorter root, taken by a candidate who solves the quadratic correctly but does not read which of the two lengths is wanted; 4.5 cm comes from treating the 40° as the angle between the two given sides and working out 7² + 6² − 2 × 7 × 6 × cos 40° directly, when 40° lies at B and faces AC; 3.6 cm comes from assuming the triangle is right-angled with AB as the hypotenuse and using the square root of 7² − 6².
- (a) 8.1 m — Multiply the length on the plan by the scale factor: 4.5 × 1.8 = 8.1, so the real length is 8.1 m. Choosing 6.3 m adds the two numbers instead of multiplying them (4.5 + 1.8 = 6.3). Choosing 2.5 m divides the plan length by the scale factor the wrong way round (4.5 ÷ 1.8 = 2.5) instead of multiplying. Choosing 9.0 m rounds the scale factor up to 2 before multiplying (4.5 × 2 = 9.0), losing the accuracy the 1.8 was giving.
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