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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- (c) 6.3 m — By Pythagoras' theorem, the height = √(7² − 3²) = √(49 − 9) = √40 = 6.32...≈ 6.3 m. "6.4 m" rounds 6.32...m up to 6.4 instead of correctly rounding it down to 6.3. "4.0 m" comes from subtracting the two given lengths directly, 7 − 3 = 4, instead of subtracting their squares. "10.0 m" comes from adding the two given lengths, 7 + 3 = 10, instead of using Pythagoras' theorem at all.
- (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.
- (b) £157 — Method: round the hours worked up to the next whole hour, multiply by the hourly rate, then add the call-out fee. Working: 3 hours 30 minutes rounds up to 4 hours; 28 × 4 = 112; 112 + 45 = 157. Answer: £157. A candidate who uses the unrounded time of 3.5 hours, working out 28 × 3.5 = 98 and adding 45, gets £143. A candidate who forgets the call-out fee, giving only 28 × 4, gets £112. A candidate who rounds down to 3 hours instead of up, working out 28 × 3 = 84 and adding 45, gets £129.
- (b) 70 cm³ — Area of the triangular cross-section = base × height ÷ 2. Base × height = 3.5 × 4 = 14 cm², and half of that is 14 ÷ 2 = 7 cm². Volume of the prism = cross-sectional area × length = 7 × 10 = 70 cm³. A pupil who forgets to halve when finding the triangle's area gets 3.5 × 4 × 10 = 140 cm³. A pupil who correctly finds the cross-sectional area but forgets to multiply by the length stops at 7 cm³. A pupil who ignores the height altogether, multiplying base by length, gets 3.5 × 10 = 35 cm³. The correct volume is 70 cm³.
- (b) 18.84 m² — The pond has radius 2.5 m (half of the 5 m diameter), so the outer edge of the path has radius 2.5 + 1 = 3.5 m. Path area = area of outer circle − area of pond = 3.14 × 3.5² − 3.14 × 2.5² = 38.465 − 19.625 = 18.84 m².
- (b) 16.747 m³ — Volume of a sphere = (4/3)πr³, so a hemisphere is half that: (2/3)πr³. Substitute r = 2: (2/3) × 3.14 × 2³ = (2/3) × 3.14 × 8 = (2/3) × 25.12 = 16.7467 m³, which rounds to 16.747 m³.
- (c) 0.5 — The real distance is 2 × 25000 = 50000 cm. Converting to metres, by dividing by 100, gives 500 m, and converting to kilometres, by dividing by 1000, gives 0.5 km. A candidate who divides the 50000 cm by 1000 in one go, applying the metres-to-kilometres factor straight to the centimetres, gets 50 km. A candidate who divides by 100 twice, treating 100 m as 1 km, gets 5 km. A candidate who slips one extra decimal place when converting 500 m to kilometres gets 0.05 km. The real distance is 0.5 km.
- (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.
- (a) 150° — The arc length is the same fraction of the circumference as the angle is of 360°. The full circumference is 2 × 3.14 × 6 = 37.68 cm, so the angle is 15.7 ÷ 37.68 × 360 = 150°. Treating the 6 cm as a diameter instead of a radius gives a circumference of 18.84 cm and an angle of 300°. Using π = 3 instead of the given 3.14 gives a circumference of 36 cm and an angle of 157°. Treating the fraction 15.7 ÷ 37.68 as being out of 100 rather than 360 gives about 42°.
- (b) 320° — A bearing is always measured clockwise from north. An angle measured anticlockwise must be converted by subtracting it from 360°: 360 − 40 = 320°, so the bearing is 320°. Choosing 040° treats the anticlockwise angle as if it were already a clockwise bearing, without converting it. Choosing 220° adds 180° to the angle, mixing this up with a back-bearing calculation (40 + 180 = 220). Choosing 400° adds the angle to 360° instead of subtracting it (40 + 360 = 400), giving a bearing greater than a full turn.
- (d) 12.56 cm — Arc length = (angle ÷ 360) × 2 × π × r = (72 ÷ 360) × 2 × 3.14 × 10 = 0.2 × 62.8 = 12.56 cm. (62.8 cm comes from finding the full circumference and forgetting to take the angle fraction; 6.28 cm comes from leaving out the factor of 2, using πr instead of 2πr; 25.12 cm comes from using the diameter, 20 cm, in place of the radius.)
- (c) 500 cm² — The area of a triangle is half of base × height. First, base × height = 40 × 25 = 1,000. Half of 1,000 is 500 cm². 1,000 cm² forgets to halve and just gives base × height. 65 cm² adds the base and height together instead of multiplying them. 2,000 cm² doubles base × height instead of halving it.
- (a) arc — A curved part of a circle's circumference, between two points, is called an arc. A chord is the straight line joining those two points, not the curved part. A segment is the region enclosed between a chord and an arc, and a sector is the region enclosed between two radii and an arc — both segment and sector are regions, not curved lengths.
- (d) £50 — Area = 1/2 × (3.5 + 6.5) × 4 = 1/2 × 10 × 4 = 20 m². Cost = 20 × £2.50 = £50. (£100 comes from forgetting to halve the trapezium area, giving 40 m² instead of 20 m²; £65 comes from using only the longer parallel side, 6.5 × 4 = 26 m², instead of the trapezium formula; £35 comes from adding all three given lengths, 3.5 + 6.5 + 4, and treating that total as the area in square metres.)
- (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.
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