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 student draws a net using 5 identical squares arranged in a row of four with one extra square attached to the side of one of them. Can this net be folded to make a closed cube?
- 2.A line segment is measured with a ruler and found to be 6.5 cm long. What is this length in millimetres?
- 3.A parallelogram has an area of 54 cm² and a base of 9 cm. Work out its perpendicular height.
- 4.A plan of a car park is drawn to a scale of 1 : 500. A parking bay is drawn as a rectangle measuring 1.2 cm by 0.6 cm. The real width of the bay corresponds to the 0.6 cm side. What is the real width of the bay, in metres?
- 5.The bearing of a harbour B from a ferry's position A is 260°. What is the bearing of A from B?
- 6.A gardener wants to buy edging trim to go all the way around a circular flower bed with a diameter of 8 m. Work out the length of edging trim needed, using π = 3.14. Give your answer to the nearest metre.
- 7.A square has sides of length 7 cm. Work out its area.
- 8.A cylinder has a radius of 3 cm and a height of 10 cm. Using π = 3.14, work out the volume of the cylinder. Give your answer to the nearest whole number of cubic centimetres.
- 9.A circular decoration has a radius of 10 cm. A ribbon costs £0.50 per centimetre and is fixed along the curved edge (the arc) of a sector-shaped section with an angle of 90°. Using π = 3.14, work out the cost of the ribbon for this section.
- 10.A garden plan is drawn to a scale of 1 cm : 1.8 m. A path on the plan measures 4.5 cm. What is the real length of the path, in metres, to 1 decimal place?
- 11.A sector of a circle has radius 6 cm and an area of 4.71 cm². Using π = 3.14, work out the angle of the sector.
- 12.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.
- 13.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.
- 14.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.
- 15.A line segment joins A(1, 1) and B(4, 7). Write down the sign of the gradient of this line.
Answer key
- (b) No, it needs one more square — A closed cube has exactly 6 faces, so its net must be made of exactly 6 identical squares, arranged so each one unfolds to a separate face with none overlapping. This net has only 5 squares, so it is one square short and cannot be folded into a closed cube. Choosing 'Yes, it folds into a cube' ignores that a cube needs 6 faces, not 5. Choosing 'No, it has one square too many' miscounts in the wrong direction — 5 is one too FEW, not one too many. Choosing 'Yes, but only if two squares overlap' is not a valid net: a net's faces must not overlap when folded.
- (a) 65 — There are 10 millimetres in every centimetre, so to convert from cm to mm, multiply by 10: 6.5 × 10 = 65 mm. A candidate who forgets to convert at all writes down the original number, 6.5. A candidate who multiplies by 100 instead of 10, confusing cm-to-mm with m-to-cm, gets 650. A candidate who divides by 10 instead of multiplying gets 0.65. The correct length in millimetres is 65.
- (c) 6 cm — For a parallelogram, area = base × height, so height = area ÷ base = 54 ÷ 9 = 6 cm. 12 cm comes from using the triangle's reverse formula, height = 2 × area ÷ base, which does not apply to a parallelogram. 45 cm comes from subtracting the base from the area, 54 − 9, instead of dividing. 486 cm comes from multiplying the area by the base, 54 × 9, instead of dividing.
- (b) 3 — The real width is 0.6 × 500 = 300 cm, which converts to 3 m by dividing by 100. A candidate who uses the wrong side of the rectangle, 1.2 cm, instead of the 0.6 cm width, gets 1.2 × 500 = 600 cm = 6 m. A candidate who multiplies correctly but converts the 300 cm to metres by dividing by 1000 instead of 100 gets 0.3 m. A candidate who converts by dividing by 10 instead of 100 gets 30 m. The real width of the bay is 3 m.
- (a) 080° — The back bearing differs from the given bearing by 180°. Because 260° is greater than 180°, subtract 180°: 260 − 180 = 80°, so the bearing of A from B is 080°. Choosing 440° adds 180° instead of subtracting it, even though the result would be more than a full turn (260 + 180 = 440). Choosing 180° assumes the back bearing is always exactly 180°, ignoring the original bearing altogether. Choosing 100° comes from measuring the reflex angle the other way round the circle (360 − 260 = 100) instead of applying the 180° back-bearing rule.
- (c) 25 m — The gardener needs the circumference, since the trim goes around the flower bed. Circumference = πd = 3.14 × 8 = 25.12 m, which rounds to 25 m. A student who mistakes the diameter (8 m) for the radius, then doubles it before multiplying by π, gets 3.14 × 16 = 50.24 m, rounding to 50 m. A student who makes the same mistake but uses the area formula instead of the circumference gets 3.14 × 8² = 200.96 m², rounding to 201. A student who correctly halves the diameter to find the radius (4 m) but then uses πr instead of πd gets 3.14 × 4 = 12.56 m, rounding to 13 m.
- (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.
- (d) 283 cm³ — Volume of a cylinder = πr²h. Using π = 3.14: V = 3.14 × 3² × 10 = 3.14 × 9 × 10 = 282.6, which rounds to 283 cm³. A pupil who uses the diameter, 6 cm, instead of the radius gets 3.14 × 6² × 10 = 1130.4 ≈ 1130 cm³. A pupil who forgets to square the radius gets 3.14 × 3 × 10 = 94.2 ≈ 94 cm³. A pupil who uses the circumference formula 2πr instead of πr² gets 2 × 3.14 × 3 × 10 = 188.4 ≈ 188 cm³. The correct volume is 283 cm³.
- (c) £7.85 — Arc length = (90 ÷ 360) × 2 × 3.14 × 10 = 0.25 × 62.8 = 15.7 cm. Cost = 15.7 × £0.50 = £7.85. (£31.40 comes from finding the full circumference and forgetting the angle fraction; £15.70 comes from using the diameter, 20 cm, in place of the radius; £157.00 comes from multiplying the arc length by the radius instead of by the cost per centimetre.)
- (a) 8.1 m — Multiply the length on the plan by the scale factor: 4.5 × 1.8 = 8.1, so the real length is 8.1 m. Choosing 6.3 m adds the two numbers instead of multiplying them (4.5 + 1.8 = 6.3). Choosing 2.5 m divides the plan length by the scale factor the wrong way round (4.5 ÷ 1.8 = 2.5) instead of multiplying. Choosing 9.0 m rounds the scale factor up to 2 before multiplying (4.5 × 2 = 9.0), losing the accuracy the 1.8 was giving.
- (d) 15° — Sector area = (angle ÷ 360) × π × r², so 4.71 = (angle ÷ 360) × 3.14 × 36 = (angle ÷ 360) × 113.04. Dividing gives angle ÷ 360 = 4.71 ÷ 113.04 = 1/24, so angle = 360 ÷ 24 = 15°. (90° comes from forgetting to square the radius, using (angle ÷ 360) × 3.14 × 6 = 18.84 in place of 113.04; 3.75° comes from using the diameter, 12 cm, in place of the radius, giving (angle ÷ 360) × 3.14 × 144 = 452.16; 45° comes from using the arc length formula, (angle ÷ 360) × 2 × 3.14 × 6 = 37.68, instead of the sector area formula.)
- (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.
- (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.
- (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) positive — Moving from A to B, the x-coordinate increases from 1 to 4 and the y-coordinate also increases from 1 to 7, so the line rises as it goes from left to right, which means the gradient is positive. "negative" would need y to decrease as x increases, which is not the case here. "zero" would need the y-coordinate to stay the same, but it changes from 1 to 7. "cannot be determined" is wrong because the coordinates of both points are known, so the direction of the line can always be found.
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