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) 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) 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) SSA - not sufficient to prove congruence — The angle given, angle A, is not the angle between sides AB and BC — it is not the included angle — so this data is SSA (side, side, angle), which is not sufficient to prove congruence on its own; two triangles can share this SSA information without being congruent. SAS is wrong because the given angle is not the one included between the two given sides. ASA is wrong because only one angle (A) is given, not two. AAS is wrong for the same reason — only one angle is given, not two.
- (a) 59° — Method: the three angles of a triangle add up to 180°, and in an isosceles triangle the two angles opposite the equal sides are equal, so take the known angle away from 180° and share what is left equally between the other two. Working: 180° − 62° = 118°, and 118° ÷ 2 = 59°. Answer: 59°. The distractors: 118° comes from taking 62° from 180° and stopping there, which gives the two angles together rather than one of them; 31° comes from halving the 62° that is given instead of halving what is left; 62° comes from assuming that all three angles of the triangle are equal to the one that is given.
- (a) 1/2 — cos 0° = 1 and sin 30° = 1/2, so cos 0° − sin 30° = 1 − 1/2 = 1/2. 1 comes from writing down cos 0° alone and forgetting to subtract sin 30°. 3/2 comes from adding the two values instead of subtracting. −1/2 comes from working out sin 30° − cos 0°, the two terms the wrong way round.
- (d) 20 — Alternate angles between parallel lines are equal, so 3x + 10 = 5x − 30. Rearranging, 10 + 30 = 5x − 3x, so 40 = 2x, and x = 20. −10 comes from a sign error when rearranging, moving a term to the wrong side and getting −20 = 2x instead. 25 comes from wrongly treating the two angles as co-interior and adding them to 180°: (3x + 10) + (5x − 30) = 180 gives 8x − 20 = 180, so x = 25. 47.5 makes the same co-interior mistake but sets the sum equal to 360° instead of 180°, giving 8x − 20 = 360 and x = 47.5.
- (c) 24 cm² — Method: the front elevation of a cuboid is a rectangle formed by the cuboid's length and its height, so its area is length × height. Working: 6 cm × 4 cm = 24 cm². Answer: 24 cm². The distractors: 12 cm² comes from using width × height (3 × 4) instead of length × height, mistaking the side elevation's dimensions for the front's. 18 cm² comes from using length × width (6 × 3), which gives the area of the plan view instead of the front elevation. 20 cm² comes from finding the perimeter of the front face instead of its area: 2 × (6 + 4) = 20.
- (b) 3 — Method: the scale factor is a length of the second triangle divided by the corresponding length of the first, and corresponding means shortest paired with shortest. Working: the shortest side of the first triangle is 3 cm and the shortest side of the second is 9 cm, so the scale factor is 9 ÷ 3 = 3. Answer: 3. The distractors: 1/3 comes from dividing 3 by 9, which is the scale factor from the second triangle back to the first; 6 comes from subtracting, 9 − 3, treating the scale factor as a difference rather than a multiplier; 1.8 comes from dividing 9 by 5, pairing the second triangle's shortest side with the first triangle's longest side.
- (c) 38° — The interior angles of a hexagon sum to (6 − 2) × 180° = 720°. 130° + 142° + 125° + 150° + 135° = 682°. x = 720° − 682° = 38°. A student who uses 6 × 180° instead of (6 − 2) × 180°, forgetting to subtract the 2, gets 1080° − 682° = 398°. A student who mis-adds the five given angles as 680° instead of 682° gets x = 720° − 680° = 40°. A student who answers with one of the given angles instead of solving for x might write 142°.
- (a) (2, 6) — To translate by the vector (−3, 4), add −3 to the x-coordinate and add 4 to the y-coordinate: (5 − 3, 2 + 4) = (2, 6). A pupil who adds 3 instead of subtracting for the x-coordinate gets (8, 6). A pupil who swaps the x- and y-components of the vector gets (5 + 4, 2 − 3) = (9, −1). A pupil who subtracts both components instead of adding the y-component gets (5 − 3, 2 − 4) = (2, −2). The correct image is (2, 6).
- (d) Yes, since 3² + 4² = 5² — AB is horizontal with length 5 − 1 = 4, BC is vertical with length 4 − 1 = 3, and CA = √(4² + 3²) = √25 = 5. Since the two shorter sides satisfy 3² + 4² = 5², the triangle is right-angled, with the right angle at B. "No, since 3 + 4 ≠ 5" wrongly tests Pythagoras' theorem by adding the sides instead of squaring them first. "No, since AB, BC and CA are not all equal" confuses a right-angled triangle with an equilateral one — a triangle does not need equal sides to have a right angle. "Yes, since 4² + 5² = 3²" reaches the correct conclusion but puts the longest side, 5, on the wrong side of the equation, as if it were one of the two shorter sides instead of the hypotenuse.
- (b) (7, 3) — Reflecting in the line y = x swaps the x- and y-coordinates: (3, 7) → (7, 3). A pupil who reflects in the x-axis instead gets (3, −7). A pupil who reflects in the y-axis instead gets (−3, 7). A pupil who confuses y = x with y = −x, swapping the coordinates and changing both signs, gets (−7, −3). The correct image is (7, 3).
- (d) 56.52 cm — Circumference = πd. Substitute d = 18: circumference = 3.14 × 18 = 56.52 cm.
- (b) (1, 6) — Method: when a square is set square-on to the grid, so that its sides run parallel to the axes, each vertex shares its x-coordinate with one neighbour and its y-coordinate with the other, and the missing vertex then borrows one coordinate from each of the two vertices it is joined to; so the first job is to check from the given points that the sides really do run parallel to the axes. Working: A(1, 2) and B(5, 2) share y = 2, so AB is a horizontal side; B(5, 2) and C(5, 6) share x = 5, so BC is a vertical side, which confirms that this square lies square-on to the axes and that the rule may be used. In square ABCD the vertex D is joined to C and to A. DC must be horizontal like AB, so D takes the y-coordinate of C, which is 6; DA must be vertical like CB, so D takes the x-coordinate of A, which is 1. D is therefore (1, 6), and checking confirms every side is 4 long. Answer: (1, 6). The distractors: (1, 5) comes from lifting the first number out of each of A and C, pairing the x-coordinate of A with the x-coordinate of C; (6, 1) comes from finding the right two numbers but writing them the wrong way round, height before sideways position; (9, 6) comes from stepping a further 4 to the right from C instead of closing the square back to the column A stands in.
- (d) 1.2 m² — Method: an area in square metres needs lengths in metres, so convert first and then multiply. Working: 100 cm = 1 m, so 150 cm = 1.5 m and 80 cm = 0.8 m, and the area = 1.5 × 0.8 = 1.2 m². Answer: 1.2 m². The same result comes from working in centimetres: 150 × 80 = 12 000 cm², and a square metre is a square of side 100 cm, so 100 × 100 = 10 000 cm² make one square metre and 12 000 ÷ 10 000 = 1.2. Dividing the 12 000 cm² by 100 instead, as though a square metre held only 100 square centimetres, gives 120 m²; dividing by 1000 gives 12 m². Working out the perimeter rather than the area gives 1.5 + 0.8 + 1.5 + 0.8 = 4.6, which is a length and not an area.
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