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) 282.6 cm³ — Volume of a cylinder = πr²h = 3.14 × 3² × 10 = 3.14 × 9 × 10 = 282.6 cm³. (94.2 cm³ comes from using πrh and forgetting to square the radius; 90 cm³ comes from using r²h and leaving π out altogether; 1130.4 cm³ comes from using the diameter, 6 cm, in place of the radius.)
- (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.)
- (c) 3π cm — Circumference = πd. With d = 3 cm, circumference = π × 3 = 3π cm. A student who uses the radius (1.5 cm) instead of the diameter in the formula gets 1.5π cm.
- (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) 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.
- (c) 36 cm² — The area of a parallelogram is base × height = 9 × 4 = 36 cm². 26 cm² is 2 × (9 + 4), the perimeter you would get by treating the perpendicular height as though it were a side length — and a perimeter is a length, not an area. 18 cm² uses the triangle formula, half of base × height, by mistake. 72 cm² doubles the correct area.
- (b) 12.56 cm² — Sector area = (angle ÷ 360) × π × r² = (90 ÷ 360) × 3.14 × 4² = 0.25 × 3.14 × 16 = 12.56 cm². (3.14 cm² comes from forgetting to square the radius; 50.24 cm² comes from finding the area of the whole circle and forgetting the angle fraction; 6.28 cm² comes from using the arc length formula instead of the sector area formula.)
- (d) 2π cm — Arc length is the fraction θ/360 of the full circumference, 2πr. Substitute θ = 60 and r = 6: 60 out of 360 is one sixth, and the full circumference is 2π × 6 = 12π. One sixth of 12π is 2π, so the arc length is 2π cm. Choosing 12π cm uses the full circumference without scaling it down by the fraction θ/360 first. Choosing 6π cm applies the SECTOR AREA formula, (θ/360) × πr², instead of the arc length formula — that calculation actually gives 6π, which is the area in cm², not a length. Choosing π cm uses πr instead of 2πr, missing the factor of 2 in the circumference formula.
- (b) 56.52 cm — Arc length = (216 ÷ 360) × 2 × 3.14 × 15 = 0.6 × 94.2 = 56.52 cm. (28.26 cm comes from leaving out the factor of 2, using πr instead of 2πr; 94.2 cm comes from finding the full circumference and forgetting the angle fraction; 113.04 cm comes from using the diameter, 30 cm, instead of the radius.)
- (c) 155° — Turning clockwise adds to the bearing. Starting on a bearing of 065° and turning clockwise through 90° gives 065° + 90° = 155°. A candidate who instead subtracts, working out 90° − 65° = 25°, has performed the wrong operation, giving 025°. A candidate who turns anticlockwise instead of clockwise works out 065° − 90°, which gives a negative number, and adding 360° to fix this gives 335° — the bearing for turning the other way. A candidate who thinks turning does not change the bearing at all keeps the answer as 065°. The new bearing, turning clockwise, is 155°.
- (b) Wrong: 1 litre = 1000 ml, so 3.2 l = 3200 ml. — 1 litre = 1000 ml, so 3.2 litres = 3.2 × 1000 = 3200 ml — Sam is wrong because he multiplied by 100 instead of 1000. Saying Sam is correct accepts the wrong multiplier. Saying Sam is wrong only because 3.2 should be rounded first misses the real error, which is the multiplier, not the starting number. Saying '1 litre is 100 ml' misstates the basic fact and blames the wrong part of Sam's working.
- (c) 27 cm² — Area of a rectangle = length × width = 4.5 × 6 = 27 cm². A pupil who adds the two sides instead of multiplying gets 4.5 + 6 = 10.5 cm². A pupil who works out the perimeter instead of the area gets 2 × (4.5 + 6) = 21 cm². A pupil who rounds 4.5 up to 5 before multiplying gets 5 × 6 = 30 cm². The correct area is 27 cm².
- (b) SAS — Two pairs of matching sides are equal (AB = DE and BC = EF) and the angle between them is equal (angle ABC = angle DEF), which is exactly the Side-Angle-Side congruence test: SAS. SSS is wrong because only two pairs of sides are given, not three. ASA is wrong because only one angle is given, and it sits between the two sides rather than the side sitting between two angles. RHS is wrong because no right angle is mentioned in either triangle.
- (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.)
- (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.
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