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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- (b) y-axis — A point lies on the y-axis when its x-coordinate is 0; here the x-coordinate of (0, −5) is 0, so it lies on the y-axis. "x-axis" would need the y-coordinate to be 0 instead, which is not the case here. "the origin" is the single point (0, 0), not a whole axis, and this point is not (0, 0). "both axes" would only be true for the origin itself, where both coordinates are 0.
- (b) circumference — The distance all the way around the outside edge of a circle is called the circumference. The diameter is a straight line across the circle through the centre, so it is a length through the circle, not around it. The radius is a straight line from the centre to the edge, again a length across, not around. The area is the amount of surface inside the circle, a region, not a length at all.
- (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.)
- (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.
- (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.
- (a) 49 cm² — The area of a square is side × side = 7 × 7 = 49 cm². 28 cm² is the perimeter of the square (4 × 7), not its area. 14 cm² adds just two of the sides together. 21 cm² multiplies the side length by 3 instead of squaring it.
- (b) 10.5 m² — The area of a trapezium is half of the sum of the parallel sides, multiplied by the width. Add the parallel sides: 2.5 + 4.5 = 7. Multiply by the width: 7 × 3 = 21. Half of 21 is 10.5 m². 21 m² forgets to halve, using the full (2.5+4.5)×3. 3.5 m² averages the two parallel sides, (2.5+4.5)÷2 = 3.5, but forgets to multiply by the width. 6.75 m² treats the bed as a triangle, using only the longer parallel side: half of 4.5 × 3.
- (d) 80° — The four angles of any quadrilateral add up to 360°. Three of the angles add up to 100 + 85 + 95 = 280°, so the fourth angle is 360 − 280 = 80°. "70°" comes from using 350° as the total instead of 360°, an easy slip on the quadrilateral angle sum. "280°" comes from stopping at the sum of the three given angles and writing that total down as the answer, instead of subtracting it from 360°. "260°" comes from subtracting only the one angle 100° from 360°, instead of subtracting all three given angles.
- (a) 615.44 cm² — Method: the surface area of a sphere is 4πr². Square the radius, multiply by π, then multiply by 4. Working: r² = 7² = 49, then 3.14 × 49 = 153.86, then 4 × 153.86 = 615.44. Answer: 615.44 cm². The distractors: 153.86 cm² comes from stopping at πr², which is the area of a flat circle of radius 7 cm and leaves out the factor of 4 that a curved surface needs; 87.92 cm² comes from 4 × 3.14 × 7, using the radius itself where the formula asks for its square; 1436.03 cm³ comes from working out the volume of the ball with (4 ÷ 3) × π × r³ instead of its surface area, which is a different measure and carries a cubic unit.
- (a) £157.00 — Area covered = (angle ÷ 360) × π × r² = (90 ÷ 360) × 3.14 × 400 = 0.25 × 1256 = 314 km². Charge = 314 × £0.50 = £157.00. (£7.85 comes from forgetting to square the radius, using 0.25 × 3.14 × 20 = 15.7 km² and then charging that; £628.00 comes from finding the area of a full circle, 3.14 × 400 = 1256 km², and forgetting the angle fraction before charging; £15.70 comes from using the arc length formula, 0.25 × 2 × 3.14 × 20 = 31.4, in place of the area, and charging that.)
- (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.)
- (d) 6.7 cm — Volume of the sphere = (4/3)πr³ = (4/3) × 3.14 × 5³ = (4/3) × 3.14 × 125 = 523.33 cm³. Set this equal to the cylinder's volume πr²h: 523.33 = 3.14 × 25 × h = 78.5h. Divide: h = 523.33 ÷ 78.5 = 6.667 cm, which rounds to 6.7 cm. 1.7 cm comes from using the sphere's diameter, 10 cm, as the cylinder's radius: 523.33 ÷ (3.14 × 10²) = 1.7 cm.
- (b) (22, −13) — The vector from the ship to the lighthouse is (12, −4) − (2, 5) = (10, −9). Sailing along this vector twice from the start gives (2, 5) + 2 × (10, −9) = (2 + 20, 5 − 18) = (22, −13). '(12, −4)' stops after the ship reaches the lighthouse and ignores the second identical leg. '(−18, 23)' comes from finding the vector the wrong way round, as (2, 5) − (12, −4) = (−10, 9), and then doubling that. '(2, 5)' comes from adding the vector and then subtracting it again, wrongly cancelling the two legs instead of adding them.
- (c) 14.7 cm — The scale factor from the smaller tile to the larger tile is 7 ÷ 4 = 1.75, so the larger side is 8.4 × 1.75 = 14.7 cm. '4.8 cm' comes from scaling by 4 ÷ 7 instead, using the ratio the wrong way round. '11.4 cm' comes from adding the difference between the ratio numbers, 7 − 4 = 3, onto 8.4, instead of scaling. '58.8 cm' comes from multiplying by 7 on its own, using a ratio number as the scale factor instead of working out 7 ÷ 4 = 1.75 first.
- (d) 6√3 cm — Method: AB lies alongside the 30° angle at A and AC is the hypotenuse, so the ratio needed is cosine: cos 30° = AB ÷ AC. Working: the exact value of cos 30° is √3/2, so AB = 12 × √3 ÷ 2, and half of 12 is 6. Answer: AB = 6√3 cm, which is about 10.4 cm. Using sine by mistake gives 12 × 1/2 = 6 cm, which is the length of BC rather than AB. Using tan 30° = 1/√3 gives 12 ÷ √3, which is 4√3 cm. Remembering cos 30° as √3 rather than as √3 halved gives 12√3 cm, longer than the hypotenuse and so impossible.
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