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 sailmaker cuts two triangular sail panels from a pattern. Panel X has a 10 m side, with angles of 50° and 75° at its two ends. Panel Y also has a 10 m side, with angles of 50° and 75° at its two ends, in the same arrangement. The sailmaker wants to check the panels will match exactly, without cutting a third measurement. Which condition proves the two panels are congruent?
- 2.A triangular flag has a base of 40 cm and a perpendicular height of 25 cm. Work out its area.
- 3.A footpath sign shows a distance of 0.45 km to the next village. Work out this distance in metres.
- 4.A kitchen scale is marked in equal divisions of 20 g. The pointer rests three divisions past the 400 g mark. Work out the reading on the scale.
- 5.A sector of a circle has radius 10 cm and angle 150°. Using π = 3.14, work out the perimeter of the sector, to 1 decimal place.
- 6.A robotic arm's tip starts at the point (12.5, −4.25) on a grid measured in centimetres. It moves by the vector to pick up a component, then by the vector to place it. Work out the coordinates of the tip after both moves.
- 7.A sector of a circle has radius 6 cm and an arc length of 15.7 cm. Using π = 3.14, work out the angle of the sector.
- 8.Two similar ponds have perimeters in the ratio 4 : 11. The perimeter of the larger pond is 88 m. Work out the perimeter of the smaller pond.
- 9.A market stall sells apples at £2.40 per kilogram. Which of these expressions gives the cost, in pounds, of buying m kilograms of apples?
- 10.A rectangular tile measures 4.5 cm by 6 cm. Work out the area of the tile.
- 11.A right-angled triangle has a hypotenuse of 10 cm and one of its other angles is 45°. Work out the length of one of the two shorter sides. Give your answer to 1 decimal place.
- 12.A cyclist rides around a perfectly circular track with centre O. At every moment of the ride, the distance from O to the cyclist is exactly the same. Which term names this fixed distance?
- 13.Two vertical masts stand on level ground, 40 m apart. One mast is 30 m tall and the other is 50 m tall. Work out the straight-line distance between the tops of the two masts. Give your answer to 1 decimal place.
- 14.A trapezium has parallel sides of 5.6 cm and 8.4 cm, and a perpendicular height of 4 cm. Work out the area of the trapezium.
- 15.A warehouse stores identical cube-shaped crates. Its plan view is a 2 by 4 rectangle of crate positions, and every position is filled to a height of 3 crates, except one corner position, which has only 2 crates stacked on it because a delivery was incomplete. How many crates are there in total?
Answer key
- (c) ASA — Method: check which condition matches two angles and the side between them, since that is all the sailmaker has measured. Working: the 10 m side lies between the 50° and 75° angles in both panels, so this is two Angles and the included Side, ASA. Options: SAS would need two sides and the angle between them, but only one side has been measured here; SSS would need three sides, but only one is known; RHS needs a right angle and a hypotenuse, and neither panel has a stated right angle. Answer: ASA.
- (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.
- (c) 450 m — To convert km to m, multiply by 1000: 0.45 × 1000 = 450 m. Multiplying by 100 instead of 1000 gives 45 m. Multiplying by 10 000 instead of 1000 gives 4500 m. Not converting at all, just relabelling the number, gives 0.45 m.
- (c) 460 g — Each division is 20 g, so three divisions past the mark is 3 × 20 = 60 g. Adding this to the 400 g mark gives 400 + 60 = 460 g. Treating each division as worth 1 g instead of 20 g gives 400 + 3 = 403 g. Working out the extra amount correctly but forgetting to add the 400 g mark gives just 60 g. Treating each division as worth 10 g instead of 20 g gives 400 + 30 = 430 g.
- (d) 46.2 cm — The perimeter of a sector is the arc length plus its two straight radii. The circumference is 2 × 3.14 × 10 = 62.8 cm, and the arc is 150 ÷ 360 of that: 62.8 × 150 ÷ 360 = 26.2 cm (1 d.p.). Adding the two radii, 26.2 + 10 + 10 = 46.2 cm. Giving just the arc length, without adding the straight edges, gives 26.2 cm. Adding only ONE radius instead of two gives 36.2 cm. Using 150 ÷ 180 instead of 150 ÷ 360 for the fraction gives an arc of 52.3 cm and a perimeter of 72.3 cm.
- (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) 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°.
- (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) 2.40m — Method: a cost found from a rate is the rate multiplied by the amount bought. Working: the rate is £2.40 per kilogram and the amount is m kilograms, so the cost is 2.40 × m. Answer: 2.40m. A candidate who divides the amount by the rate instead of multiplying writes m/2.40. A candidate who adds the rate to the amount instead of multiplying writes 2.40 + m. A candidate who subtracts the rate from the amount instead of multiplying writes m − 2.40.
- (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.
- (c) 7.1 cm — Method: a shorter side is opposite the 45° angle and the hypotenuse is known, so sin θ = opposite ÷ hypotenuse gives that side directly. Working: sin 45° = x ÷ 10, so x = 10 × sin 45° = 7.071…, which is 7.1 to 1 decimal place. Answer: 7.1 cm. The distractors: 5.0 cm comes from halving the hypotenuse, which is the rule for the side opposite a 30° angle and not a 45° one; 14.1 cm comes from dividing by sin 45° instead of multiplying by it, which makes a shorter side longer than the hypotenuse; 10.0 cm comes from taking tan 45° = 1 and concluding that the shorter side matches the hypotenuse.
- (a) radius — The distance from the centre O to any point on the circle is the radius. The diameter is the distance right across the circle, through the centre, from one side to the other — twice the radius, not the same measurement. The circumference is the total distance around the circle, not from the centre. A chord is a line joining two points on the circle that does not have to pass through the centre at all.
- (a) 44.7 m — Method: the line joining the two tops is the hypotenuse of a right-angled triangle whose horizontal side is the gap between the masts and whose vertical side is the difference in their heights, so Pythagoras' theorem applies. Working: the difference in heights is 50 − 30 = 20 m, so d² = 40² + 20² = 1600 + 400 = 2000 and d = √2000 = 44.721…, which is 44.7 m to 1 decimal place. Answer: 44.7 m. The distractors: 34.6 m comes from subtracting the squares, √(40² − 20²), instead of adding them; 60.0 m comes from adding the two sides of the triangle, 40 + 20, rather than using Pythagoras' theorem; 50.0 m is the height of the taller mast, copied from the question in place of the distance asked for.
- (b) 28 cm² — Area of a trapezium = (sum of parallel sides) ÷ 2 × height. Sum of parallel sides = 5.6 + 8.4 = 14 cm. Half of that is 14 ÷ 2 = 7 cm. Area = 7 × 4 = 28 cm². A pupil who forgets to halve gets 14 × 4 = 56 cm². A pupil who uses only the longer parallel side, as if this were a rectangle, gets 8.4 × 4 = 33.6 cm². A pupil who subtracts the parallel sides instead of adding them gets (8.4 − 5.6) ÷ 2 × 4 = 1.4 × 4 = 5.6 cm². The correct area is 28 cm².
- (b) 23 — If every position were filled to the full height of 3, the total would be 2 × 4 × 3 = 24 crates. One corner position has only 2 crates instead of 3, one crate short of full height there, so the actual total is 24 − 1 = 23. "24" comes from using the full height everywhere and forgetting the one incomplete corner. "22" comes from removing 2 crates for the incomplete corner instead of the 1 that is actually missing (3 − 2 = 1, not 2). "21" comes from removing all 3 crates at that corner, as though the position were completely empty rather than 2 crates short.
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