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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- (d) 63.6 — Method: the three angles inside the triangle formed by the two ladders and the ground add up to 180°. Working: 180 − 58.2 − 58.2 = 63.6. Answer: 63.6°. A candidate who assumes the top angle equals the base angles gives 58.2. A candidate who subtracts only one base angle from 180°, working out 180 − 58.2, gets 121.8. A candidate who doubles the base angle instead of subtracting it twice from 180°, working out 2 × 58.2, gets 116.4.
- (c) 17.71 m — The perimeter is the two long sides (5 m + 5 m = 10 m) plus the one remaining short side (3 m) plus the curved semicircular edge. The 3 m side is the diameter of the semicircle, so its radius is 3 ÷ 2 = 1.5 m and the curved length is half the circumference: (1/2) × 2 × 3.14 × 1.5 = 4.71 m. Total: 10 + 3 + 4.71 = 17.71 m.
- (d) Yes, because 45 ÷ 360 = 1/8, and this fraction applies to any radius. — Priti is correct: 45 ÷ 360 = 1/8, and because sector area = (angle ÷ 360) × full circle area, this fraction depends only on the angle, not on the radius, so it holds for every circle. (The statement giving 45 ÷ 360 = 1/6 comes from an arithmetic slip when simplifying the fraction; the statement that the fraction depends on the radius is wrong because the radius cancels out of the angle fraction in the sector-area formula; the statement adding that it only works for this particular radius is wrong because the 1/8 fraction is true for every radius, not a special case.)
- (a) 188.4 cm² — Curved surface area of a cone = πrl. With r = 6 cm, l = 10 cm and π = 3.14, curved surface area = 3.14 × 6 × 10 = 188.4 cm². A student who uses the cylinder's curved surface area formula, 2πrl, instead of the cone's gets 2 × 3.14 × 6 × 10 = 376.8 cm². A student who uses the circle-area formula πr² instead of πrl gets 3.14 × 36 = 113.04 cm². A student who multiplies r × l but leaves out π entirely gets 6 × 10 = 60 cm².
- (d) 21.2 m — Method: the cable is the hypotenuse of a right-angled triangle whose vertical side is the drop from the roof to the bracket and whose horizontal side is 15 m, so use Pythagoras' theorem. Working: the drop is 20 − 5 = 15 m, so c² = 15² + 15² = 225 + 225 = 450 and c = √450 = 21.213…, which is 21.2 m to 1 decimal place. Answer: 21.2 m. The distractors: 25.0 m comes from using the whole 20 m height of the roof as the vertical side and forgetting that the bracket is already 5 m up; 30.0 m comes from adding the two sides of the triangle, 15 + 15, instead of using Pythagoras' theorem; 15.0 m is the horizontal distance on its own, which would be the length of the cable only if it ran level.
- (d) 42.4 cm² — Sector area = (angle ÷ 360) × π × r². First, 60 ÷ 360 = 1/6. Next, π × r² = 3.14 × 81 = 254.34 cm². So the area = (1/6) × 254.34 = 42.39 cm², which rounds to 42.4 cm². (4.7 cm² comes from forgetting to square the radius: using π × r = 3.14 × 9 = 28.26, then (1/6) × 28.26 = 4.71 cm²; 254.3 cm² comes from finding the area of the whole circle, 254.34 cm², and forgetting the angle fraction; 9.4 cm² comes from using the arc length formula instead: 2 × π × r = 2 × 3.14 × 9 = 56.52 cm, then (1/6) × 56.52 = 9.42 cm.)
- (c) 8 cm — Area of a trapezium = (sum of parallel sides) ÷ 2 × height, so 45 = (a + 10) ÷ 2 × 5. Dividing 45 by 5 gives 9, so (a + 10) ÷ 2 = 9, meaning a + 10 = 18, so a = 18 − 10 = 8 cm. A pupil who correctly finds that the two parallel sides add up to 18 but forgets to subtract the known side of 10 cm gives the sum of both parallel sides, 18 cm, as the answer. A pupil who then adds 10 again by mistake instead of subtracting gets 18 + 10 = 28 cm. A pupil who halves the correct answer by mistake gets 8 ÷ 2 = 4 cm. The correct length of the other parallel side is 8 cm.
- (b) 10 cm — Arc length = (angle ÷ 360) × 2 × π × r, so 15.7 = (90 ÷ 360) × 2 × 3.14 × r = 1.57r, giving r = 15.7 ÷ 1.57 = 10 cm. (2.5 cm comes from forgetting the angle fraction and dividing by 2π alone; 20 cm comes from leaving out the factor of 2 in the arc length formula before dividing; 5 cm comes from dividing the arc length by π alone, ignoring both the angle fraction and the factor of 2.)
- (a) 118° — When two straight lines cross, the angles that are vertically opposite each other are always equal. So the angle vertically opposite 118° is also 118°. A candidate who instead finds the angle next to it on the straight line, using 180° − 118° = 62°, has found the adjacent angle, not the vertically opposite one. A candidate who answers 180° has confused the rule with angles on a straight line. A candidate who doubles the angle, giving 236°, has applied no valid angle rule at all. The vertically opposite angle is 118°.
- (d) (11, 0.75) — Add the moves to the starting point one component at a time. x: 12.5 + (−3.75) + 2.25 = 11; y: −4.25 + 6.5 + (−1.5) = 0.75, giving (11, 0.75). (8.75, 2.25) stops after the first move only and never applies the second vector. (6.5, 3.75) comes from subtracting the second vector instead of adding it. (0.75, 11) comes from swapping the final x-coordinate and y-coordinate.
- (a) 32 m — The scale factor from the larger pond to the smaller pond is 4 ÷ 11, so the smaller perimeter is 88 × 4 ÷ 11 = 32 m. The distractor 242 m comes from using the ratio the wrong way round, 88 × 11 ÷ 4 = 242. The distractor 84 m comes from subtracting the smaller ratio number, 88 − 4 = 84, instead of scaling. The distractor 121 m comes from multiplying 11 × 11 = 121, ignoring the given perimeter altogether.
- (a) 33 — Method: total crates = (number of floor positions that actually have crates on them) × (the stack height). Working: there are 4 × 3 = 12 floor positions in the whole arrangement, but one corner position is left empty, leaving 11 filled positions; each filled position is stacked 3 crates high, so 11 × 3 = 33. Answer: 33. The distractors: 36 comes from forgetting to remove the empty corner and using all 12 positions (12 × 3). 35 comes from removing only one crate for the empty corner instead of the full stack of 3 (36 − 1). 11 comes from counting the filled floor positions and stopping there, forgetting that each one carries a stack 3 crates high.
- (b) 3 — The real width is 0.6 × 500 = 300 cm, which converts to 3 m by dividing by 100. A candidate who uses the wrong side of the rectangle, 1.2 cm, instead of the 0.6 cm width, gets 1.2 × 500 = 600 cm = 6 m. A candidate who multiplies correctly but converts the 300 cm to metres by dividing by 1000 instead of 100 gets 0.3 m. A candidate who converts by dividing by 10 instead of 100 gets 30 m. The real width of the bay is 3 m.
- (c) 52° — When two straight lines cross, the angles opposite each other at the crossing point are vertically opposite and are equal. Ray OQ is opposite ray OP and ray OS is opposite ray OR, so angle QOS is vertically opposite angle POR. Therefore angle QOS = 52°.
- (a) 75° — The three angles in the triangle formed by the two supports and the ground add up to 180°. Two of the angles are 34° and 71°, so the third angle is 180 − 34 − 71 = 75, which is 75°. 105° comes from adding the two given angles together instead of subtracting them from 180° (34 + 71 = 105). 146° comes from computing 180 − 34 = 146 and forgetting to also subtract 71°. 37° comes from subtracting the two given angles from each other instead of using the triangle's angle sum (71 − 34 = 37).
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