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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Geometry and measures worksheet — GCSE Foundation
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- 1.A triangular sail for a small boat has a base of 2.4 m and a height of 1.75 m. Sailcloth costs £12.50 per m². Work out the total cost of the sailcloth needed for the sail.
- 2.A flagpole is 12 m tall. Freya stands 9 m from the base of the flagpole on level ground. Work out the angle of elevation of the top of the flagpole from where Freya stands. Give your answer correct to 1 decimal place.
- 3.In triangle ABC and triangle DEF, AB = DE, angle ABC = angle DEF and BC = EF. Write down the congruence condition that proves the two triangles are congruent.
- 4.A circular coaster has a radius of 4 cm. Work out the circumference of the coaster. Give your answer in terms of π.
- 5.A solid cone has a base radius of 6 cm and a vertical height of 10 cm. Work out the volume of the cone. Use π = 3.14 and give your answer correct to 1 decimal place.
- 6.A sector of a circle has radius 9 cm and angle 120°. Using π = 3.14, work out the arc length of the sector, to 1 decimal place.
- 7.The top of a clock tower is 25 m above level ground. Oliver stands on the ground 25 m from the foot of the tower. Work out the angle of elevation of the top of the tower from the point where Oliver stands.
- 8.A sprinkler waters a sector-shaped patch of a garden with radius 8 m and angle 90°. Work out the perimeter of the watered sector, to 1 decimal place. (Use π = 3.14.)
- 9.A shop stacks identical storage crates in a display. The diagram shows the plan of the floor positions used, with one corner position marked. Every floor position is filled with crates stacked 3 high, except the marked corner position, which is left completely empty. Work out how many crates are used in total.
- 10.A sector of a circle has radius 9 cm and angle 40°. Work out the area of the sector, in terms of π.
- 11.A trapezium has an area of 45 cm², a perpendicular height of 5 cm, and one parallel side of length 10 cm. Work out the length of the other parallel side.
- 12.A sector of a circle has radius 6 cm and angle 60°. Work out the arc length of the sector, in terms of π.
- 13.A surveyor is marking out field ABCD for a new crop. The plan states that side AB is parallel to side DC, but AB and DC are not equal in length, and sides BC and AD are not parallel to each other. Which term correctly describes the shape of field ABCD?
- 14.Two identical ladders lean against the same vertical wall from opposite sides, each making an angle of 58.2° with the ground. The two ladders and the ground form a triangle. Work out the size of the angle between the two ladders at the top, where they meet.
- 15.The angle between north and a footpath is 35°, measured clockwise from north. What is the three-figure bearing of the footpath?
Answer key
- (b) £26.25 — First multiply the base and height: 2.4 × 1.75 = 4.2. The area of the triangular sail is half of that: half of 4.2 is 2.1 m². Then multiply by the cost per m²: 2.1 × £12.50 = £26.25. £52.50 forgets to halve in the area formula, giving an area of 4.2 m² and doubling the true cost. £30.00 multiplies the base length by the cost per m² (2.4 × £12.50) without ever finding the area. £25.00 rounds the area to 2 m² before multiplying by the cost, losing accuracy.
- (b) 53.1° — The angle of elevation is opposite the height of the flagpole, 12 m, and adjacent to the distance from its base, 9 m, so tan θ = 12/9 = 1.333..., giving θ = tan⁻¹(1.333...) = 53.13...° ≈ 53.1°. "36.9°" finds the OTHER acute angle of the triangle, 90° − 53.1°, the angle at the top of the flagpole rather than the angle of elevation at Freya's position. "48.6°" comes from wrongly treating 9/12 as a sine ratio and finding sin⁻¹(0.75) = 48.6°, when neither side here is the hypotenuse. "41.4°" comes from wrongly treating 9/12 as a cosine ratio and finding cos⁻¹(0.75) = 41.4°, again without a hypotenuse in the ratio at all.
- (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.
- (a) 8π cm — Circumference = 2πr. Substitute r = 4: circumference = 2 × π × 4 = 8π cm. Using r in place of 2r (halving the formula) gives 4π cm. Using the area formula πr² in place of the circumference formula gives π × 4² = 16π cm. Multiplying 2 × 4 without including π at all gives 8 cm.
- (a) 376.8 cm³ — Volume of a cone = (1/3)πr²h. Substitute r = 6 and h = 10: (1/3) × 3.14 × 6² × 10 = (1/3) × 3.14 × 36 × 10 = (1/3) × 1130.4 = 376.8 cm³.
- (d) 18.8 cm — Arc length is a fraction of the full circumference. Here 120 ÷ 360 = 1/3, and the full circumference is 2 × 3.14 × 9 = 56.52 cm, so the arc is 56.52 ÷ 3 = 18.84 cm, which rounds to 18.8 cm. Forgetting the fraction and giving the full circumference rounds to 56.5 cm. Using 9 cm as a diameter instead of a radius gives an arc of 9.4 cm. Using 120 ÷ 180 instead of 120 ÷ 360 for the fraction gives an arc of 37.7 cm.
- (c) 45° — Method: the tower, the ground and the line of sight form a right-angled triangle in which the 25 m height is opposite the angle of elevation and the 25 m along the ground is adjacent to it, so use tan θ = opposite ÷ adjacent. Working: tan θ = 25 ÷ 25 = 1, so θ = tan⁻¹(1). Answer: 45°. The distractors: 90° comes from using sin θ = 25 ÷ 25 = 1, which treats the 25 m along the ground as the hypotenuse when it is the side next to the angle; 1° comes from writing down the value of tan θ as though it were the angle itself; 50° comes from adding the two given lengths, 25 + 25, instead of comparing them.
- (a) 28.6 m — The perimeter of a sector is the two straight radii plus the curved arc. This sector's 90° angle is one quarter of a full turn, so its arc length is one quarter of the full circle's circumference. The full circumference is 2 × 3.14 × 8 = 50.24 m, and one quarter of that is 50.24 ÷ 4 = 12.56 m. Add the two 8 m radii: 12.56 + 8 + 8 = 28.56 m, which rounds to 28.6 m. Choosing 12.6 m gives the arc length alone (rounded), forgetting the two straight edges of the sector. Choosing 20.6 m adds only one radius to the arc length instead of two, missing one of the two straight sides. Choosing 41.1 m comes from using the diameter, 16 m, as if it were the radius when working out the arc length (2 × 3.14 × 16 = 100.48, one quarter of which is 25.12), then adding the two correct 8 m radii (25.12 + 8 + 8 = 41.12).
- (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) 9π cm² — Sector area is the fraction θ/360 of the full circle's area, πr². Substitute θ = 40 and r = 9: 40 out of 360 is one ninth, and the full circle's area is π × 9² = 81π. One ninth of 81π is 9π, so the sector area is 9π cm². Choosing 2π cm² is this sector's ARC LENGTH, not its area — a different formula and a different quantity. Choosing 81π cm² is the area of the WHOLE circle, forgetting to scale down by the sector's angle. Choosing 18π cm² is the CIRCUMFERENCE of the whole circle, 2π × 9, mixed up with an area.
- (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.
- (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) trapezium — A quadrilateral with exactly one pair of parallel sides is a trapezium; field ABCD has AB parallel to DC and no other pair of parallel sides, so trapezium is correct. Parallelogram requires BOTH pairs of opposite sides to be parallel, but the plan says only AB and DC are parallel. Rhombus requires all four sides to be equal in length, which is not stated here. A kite is defined by two pairs of adjacent equal sides, not by having a pair of parallel sides, so it does not match this description either.
- (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.
- (a) 035° — A bearing is measured clockwise from north, so an angle of 35° clockwise from north is a bearing of 035° (written with three figures). Choosing 325° measures the angle anticlockwise instead of clockwise (360 − 35 = 325). Choosing 215° adds 180° to the angle, mixing this up with a back-bearing calculation (35 + 180 = 215). Choosing 350° reorders the digits of 035, writing the ones digit before the tens digit by mistake.
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