Printable · GCSE Foundation · ages 14-16
Geometry and measures worksheet — GCSE Foundation
Fifteen questions across the geometry and measures statements at Foundation 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 Foundation
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- (a) 8,000,000 cm³ — Method: change the edge length into centimetres first and then cube it, because 1 m = 100 cm and a volume needs that conversion applied to all three dimensions. Working: 2 m = 2 × 100 = 200 cm, so the volume is 200 × 200 × 200. 200 × 200 = 40,000 and 40,000 × 200 = 8,000,000. Answer: 8,000,000 cm³. The distractors: 8,000 cm³ comes from converting 2 m to 20 cm and cubing that; 80,000 cm³ comes from cubing in metres to get 8 m³ and then multiplying by 10,000, the conversion factor for an area rather than the 1,000,000 a volume needs; 8 cm³ comes from cubing the 2 without converting at all and simply writing cm³ because the question asked for that unit.
- (d) sector — The region enclosed by two radii and the arc between them is called a sector. A segment is a different region, enclosed by a CHORD and an arc — it does not touch the centre at all. A chord is a straight line, not a region, so it cannot be the answer to a question asking for a region. A semicircle is only correct when the angle between the two radii is exactly 180°, which is not stated here, so it is too specific to be the general answer.
- (d) 68 cm² — Method: the area of a parallelogram is base × perpendicular height. Working: 4 × 17 = 68. Answer: 68 cm². The distractors: 34 cm² comes from halving the product, which is the rule for a triangle and not for a parallelogram; 42 cm comes from treating the two given lengths as the sides of the shape and working out a perimeter, 2 × (4 + 17), which is a length and not an area; 21 cm comes from adding the base and the height, 4 + 17, instead of multiplying them.
- (d) 60 — Method: write the two supplementary angles as x and 2x, since one is twice the other, and solve x + 2x = 180. Working: 3x = 180, so x = 60. Answer: the smaller angle is 60°. A candidate who gives the larger angle, 2x, instead of the smaller angle gets 120. A candidate who uses a complementary sum of 90° instead of a supplementary sum of 180° gets 30. A candidate who divides 180 by 2 instead of by 3 gets 90.
- (c) 2 hours 15 minutes — Divide the total minutes by 60: 135 ÷ 60 = 2 remainder 15, so that is 2 hours 15 minutes. Treating 100 minutes as one hour, a metric-style mistake, gives 135 − 100 = 35, so 1 hour 35 minutes. Subtracting 60 twice to reach the 15 minutes left over, but losing count and recording only one of the two hours removed, gives 1 hour 15 minutes. Writing 135 ÷ 60 = 2.25 and then reading the '25' as minutes, instead of converting the 0.25 of an hour into 15 minutes, gives 2 hours 25 minutes.
- (a) 314 cm² — Method: the area of a circle is πr², and the radius is half the diameter, so halve the 20 cm before squaring. Working: r = 20 ÷ 2 = 10 cm, so the area is 3.14 × 10² = 3.14 × 100 = 314. Answer: 314 cm². The distractors: 1256 cm² comes from putting the diameter straight into πr² without halving it, 3.14 × 20²; 628 cm² comes from halving correctly but then using 2πr², a mixture of the circumference and area formulae; 62.8 cm² comes from working out πd = 3.14 × 20, which is the circumference of the plate rather than its area.
- (b) 2 — A rhombus has two lines of symmetry — along each of its two diagonals.
- (b) 60 cm² — Method: the area of a triangle is half the base multiplied by the perpendicular height, and here the perpendicular height is the 12 cm, not the sloping side. Working: 10 × 12 = 120, then 120 ÷ 2 = 60. Answer: 60 cm². The distractors: 65 cm² comes from using the sloping side of 13 cm as the height, (10 × 13) ÷ 2; 120 cm² comes from using the right two lengths but forgetting to halve, 10 × 12; 78 cm² comes from taking 13 cm and 12 cm as the base and the height and ignoring BC altogether, (13 × 12) ÷ 2.
- (b) 13 — Method: the straight-line distance between two points on a grid is the hypotenuse of a right-angled triangle whose shorter sides are the horizontal gap and the vertical gap between them, so Pythagoras' theorem applies. Working: one point is the origin, so the horizontal gap is 5 and the vertical gap is 12. Then d² = 5² + 12² = 25 + 144 = 169, so d = √169 = 13. Answer: 13. The distractors: 17 comes from adding the two gaps, 5 + 12, instead of adding their squares and taking the root; 12 comes from reading off the vertical gap alone and offering that as the whole distance; 7 comes from subtracting the gaps, 12 − 5, as though a distance were a difference.
- (c) 5 cm — Volume of a cylinder = πr²h, so r² = V ÷ (πh) = 942 ÷ (3.14 × 12) = 942 ÷ 37.68 = 25, and r = √25 = 5 cm. A pupil who finds r² = 25 but forgets to take the square root gives 25 cm. A pupil who forgets to divide by π, using r² = 942 ÷ 12 = 78.5, gets r = √78.5 ≈ 8.9 cm. A pupil who forgets to divide by the height, using r² = 942 ÷ 3.14 = 300, gets r = √300 ≈ 17.3 cm. The correct radius is 5 cm.
- (b) 25 cm² — Method: a square has four equal sides, so divide the perimeter by 4 to recover the side length, then square that side to get the area. Working: 20 ÷ 4 = 5 cm, then 5 × 5 = 25. Answer: 25 cm². The distractors: 400 cm² comes from squaring the perimeter itself, 20 × 20, treating the 20 cm as though it were the side length; 100 cm² comes from dividing the perimeter by 2 rather than by 4, giving a side of 10 cm, and squaring that; 5 cm is the side length, from stopping as soon as the perimeter has been divided by 4 and never squaring it, which also leaves a length where an area was asked for.
- (c) 14 units — Method: the perimeter of a rectangle is the distance all the way round its outside, 2 × (length + width), so the two side lengths must be found first; on a coordinate grid a side's length is the difference between the coordinates that change along it. Working: along AB, from (1, 1) to (4, 1), only x changes, so AB = 4 − 1 = 3. Along BC, from (4, 1) to (4, 5), only y changes, so BC = 5 − 1 = 4. Perimeter = 2 × (3 + 4) = 2 × 7 = 14. Answer: 14 units. The distractors: 18 units comes from reading the vertex numbers 4 and 5 as the side lengths instead of subtracting, giving 2 × (4 + 5); 12 units comes from working out the area, 3 × 4, in place of the perimeter; 7 units comes from adding one length to one width and stopping there, without doubling for the opposite pair of sides.
- (b) Nonagon — Method: match the number of sides given to the correct polygon name. Working: a polygon with 9 sides is called a nonagon. An octagon has 8 sides, a decagon has 10 sides, and a heptagon has 7 sides, so each names a different number of sides from the 9 given. Answer: nonagon.
- (c) (6, 2) — Method: add the vector's components to the point's coordinates, x-component to x, y-component to y. Working: A′ = (2 + 4, 3 + (−1)) = (6, 2). Options: (6, 4) comes from adding the size of the y-component, 1, instead of subtracting it; (1, 7) comes from swapping the components, adding the y-component (−1) to the x-coordinate and the x-component (4) to the y-coordinate; (−2, 4) comes from reversing the sign of both components, using the vector (−4, 1) instead of (4, −1). Answer: (6, 2).
- (c) Swapped the x and y components — The student's vector has the same two numbers, 5 and −2, but in swapped positions, so the error is swapping the x and y components rather than an error with signs or size. 'Reversed both signs' is wrong because the numbers 5 and −2 have not changed sign, only position. 'Reversed only the y sign' is wrong for the same reason — no sign has actually changed. 'Doubled the x component' is wrong because neither number has changed in size.
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