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.
Geometry and measures worksheet — GCSE Foundation
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- 1.A cuboid has three different edge lengths: a length, a width and a height, all different from each other. How many rectangles make up its net in total?
- 2.A line segment is measured with a ruler and found to be 6.5 cm long. What is this length in millimetres?
- 3.Point B is at (3, 5). It is reflected in the line y = 2. Work out the coordinates of the image of point B.
- 4.A packaging designer is told a box has 6 faces, 12 edges and 8 vertices, and every face is a rectangle. To fit exactly onto a shelf, the designer needs to know whether the box must be a cube. Based only on this information, which solid must the box be?
- 5.Triangles LMN and XYZ have LM = XY, MN = YZ and LN = XZ. Give a reason why triangle LMN is congruent to triangle XYZ.
- 6.Triangle ABC has AB = 5 cm and BC = 7 cm. Triangle DEF has DE = 5 cm and EF = 7 cm. Which extra fact would prove the triangles are congruent by SAS?
- 7.In kite WXYZ, WX = WZ and XY = ZY (so at each of X and Z, one side from each of the two unequal pairs meets). Angle X = 100°. Work out angle Z.
- 8.A garden flower bed is in the shape of a trapezium. Its two parallel edges are 3.5 m and 6.5 m long, and the perpendicular distance between them is 4 m. Bedding plants cost £2.50 per square metre to cover the bed. Work out the total cost of the bedding plants.
- 9.Write down the name given to a quadrilateral with exactly one pair of parallel sides.
- 10.A running track has a straight section and a semicircular bend. The bend is a semicircle with radius 32 m. Work out the length of the curved part of the bend, to 1 decimal place. (Use π = 3.14.)
- 11.A solid pyramid has a square base of side 10 cm and four identical triangular faces. Its apex is directly above the centre of the base, at a vertical height of 12 cm, and the slant height of each triangular face — the distance from the apex to the midpoint of a base edge — is 13 cm. Work out the total surface area of the pyramid.
- 12.A pop-up canopy has two sloping supports that meet at the top. Each support makes an angle of 30° with the ground and is 6 m long. Using the exact value of cos 30°, work out the total width of the canopy's base.
- 13.A cylindrical drum and a cuboid box are being wrapped for a display. A shop assistant wraps only the flat faces of each item, not any curved surface. How many flat faces in total does the assistant wrap across both items?
- 14.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?
- 15.In a construction, a point P is marked at the foot of the perpendicular drawn from point X to a straight line. Which statement correctly compares the distance XP with the distance from X to any other point on the line?
Answer key
- (a) 6 — A cuboid has six faces in total: a top, a bottom, a front, a back, and two ends — each one is a rectangle in the net, giving six rectangles altogether. Choosing 3 counts only the three PAIRS of congruent rectangles (top/bottom, front/back, two ends) rather than all six individual faces. Choosing 5 forgets one face, as if the net were missing its lid. Choosing 12 is the number of edges of a cuboid, not the number of rectangles in its net.
- (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.
- (b) (3, −1) — Method: find the distance from the point to the mirror line, then place the image the same distance on the other side of the line. Working: B is 5 − 2 = 3 units above the line y = 2, so its image is 3 units below the line, at y = 2 − 3 = −1, giving (3, −1); the x-coordinate does not change, since the mirror line is horizontal. Options: (3, −5) comes from reflecting in the x-axis (y = 0) instead of the line y = 2; (−3, 5) comes from reflecting in the y-axis instead of a horizontal line; (3, −4) comes from doubling the distance from the line instead of reflecting it, giving 2 − 2×3 = −4. Answer: (3, −1).
- (b) Cuboid — not necessarily a cube — A solid with 6 faces, 12 edges and 8 vertices in which every face is a rectangle is a cuboid, but nothing here confirms that all the edges are the same length, so the box could be a cube or a non-cube cuboid; the most that can be concluded is that it is a cuboid, making 'Cuboid — not necessarily a cube' correct. 'Cube — only a cube fits this' is wrong because a cube is just one particular cuboid; a general cuboid with different length, width and height has exactly the same face, edge and vertex counts and rectangular faces. 'Triangular prism' is wrong because a triangular prism has 5 faces, 9 edges and 6 vertices, and two of its faces are triangles, so it matches neither the counts nor the face shape. 'Not enough information' is wrong because rectangular faces with these counts do pin the solid down to the cuboid family, even though they cannot pin down a cube specifically.
- (a) SSS – all three corresponding sides are equal — All three pairs of corresponding sides are stated as equal — LM = XY, MN = YZ and LN = XZ — with no angle mentioned. This matches the SSS condition, so triangle LMN is congruent to triangle XYZ.
- (b) angle B = angle E — AB and BC meet at vertex B, so the angle INCLUDED between them is angle B; making angle B = angle E completes SAS. Angle A sits between AB and AC, not between AB and BC, so it is not the included angle needed for SAS here. Angle C sits between BC and CA, not between AB and BC, so it is not included either. AC = DF would give a third pair of equal sides, which proves congruence by SSS instead of SAS.
- (a) 100° — In a kite, the pair of angles between the unequal sides are equal to each other. Angle X and angle Z are both between one side from the WX/WZ pair and one side from the XY/ZY pair, so angle Z = angle X = 100°.
- (d) £50 — Area = 1/2 × (3.5 + 6.5) × 4 = 1/2 × 10 × 4 = 20 m². Cost = 20 × £2.50 = £50. (£100 comes from forgetting to halve the trapezium area, giving 40 m² instead of 20 m²; £65 comes from using only the longer parallel side, 6.5 × 4 = 26 m², instead of the trapezium formula; £35 comes from adding all three given lengths, 3.5 + 6.5 + 4, and treating that total as the area in square metres.)
- (a) Trapezium — A trapezium is defined as a quadrilateral with exactly one pair of parallel sides. A parallelogram has two pairs, and a kite and rhombus are defined by side lengths, not by having only one pair of parallel sides.
- (c) 100.5 m — The curved part of a semicircular bend is half of a full circle's circumference. The full circumference would be 2 × 3.14 × 32 = 200.96 m, and half of that is 200.96 ÷ 2 = 100.48 m, which rounds to 100.5 m. Choosing 201.0 m uses the FULL circumference, forgetting to halve it for a semicircle. Choosing 50.2 m halves the radius as well as taking a semicircle, using 3.14 × 16 = 50.24 m instead of the correct radius of 32 m. Choosing 64.0 m uses the diameter, 2 × 32 = 64, as if it were the curved length, ignoring π and the semicircle shape entirely.
- (b) 360 cm² — Method: the total surface area is the square base plus the four triangular faces, and the height used for a triangular face is the slant height of 13 cm, not the vertical height of 12 cm. Working: the base is 10 × 10 = 100 cm²; one triangular face is (10 × 13) ÷ 2 = 65 cm², so four faces give 4 × 65 = 260 cm²; the total is 100 + 260 = 360. Answer: 360 cm². The distractors: 260 cm² comes from adding the four triangular faces and leaving out the base; 340 cm² comes from using the vertical height of 12 cm as the height of each triangle, 100 + 4 × 60; 620 cm² comes from working out each face as 10 × 13 without halving, 100 + 4 × 130.
- (b) 6√3 m — The horizontal distance covered by one support is adjacent to the 30° angle, so it equals 6 × cos 30° = 6 × √3/2 = 3√3 m. The total base width is made up of both supports, so it is 2 × 3√3 = 6√3 m. '3√3 m' gives only one support's horizontal distance and forgets to double it for the total width. '6 m' comes from using sin 30° instead of cos 30° for the horizontal distance (6 × sin 30° = 3, doubled to 6). '12 m' comes from doubling the full sloping length of 6 m without using any trigonometry at all.
- (a) 8 — A cylinder has two flat circular faces (the top and the base) and one curved surface, so only its 2 flat faces are wrapped. A cuboid has 6 faces and all of them are flat, so all 6 are wrapped. Adding these: 2 + 6 = 8, so 8 is correct. 9 comes from wrongly counting the cylinder's curved surface as a flat face. 6 comes from counting only the cuboid and forgetting the cylinder's two flat circular faces. 3 comes from counting only the cylinder and including its curved surface in that total.
- (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.
- (d) XP is the shortest distance from X to the line — The perpendicular from a point to a line always gives the shortest possible distance to that line — joining X to any other point on the line forms the hypotenuse of a right-angled triangle with XP as one of the shorter sides, and a hypotenuse is always longer than either of the other two sides. So XP is shorter than the distance to every other point on the line. "XP is the longest distance from X to the line" reverses this relationship. "XP equals every other distance from X to the line" would only be true if X were equidistant from every point on the line, which is impossible for a point and a straight line. "XP cannot be compared without knowing the line's length" is false — the shortest-distance fact holds whatever the line's length, since only the local right angle matters.
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