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.
Answer key: Geometry and measures worksheet — GCSE Foundation
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- (b) Parallel, since AB and DC are both horizontal lines — A and B both have y-coordinate 1, so AB is horizontal; D and C both have y-coordinate 4, so DC is horizontal too. Two horizontal lines are always parallel, so AB and DC are parallel. "Not parallel, since AB and DC have different lengths" is wrong twice over: AB runs from x = 1 to x = 5 and DC from x = 2 to x = 6, so both are in fact 4 units long, and in any case length has no bearing on whether two lines are parallel — only direction does. "Not parallel, since AB is horizontal and DC is vertical" misreads the coordinates of D and C, which share the y-coordinate 4 and so give a horizontal line, not a vertical one. "Parallel, since A and D have the same x-coordinate" rests on a false claim: A has x-coordinate 1 and D has x-coordinate 2, so they do not share an x-coordinate — and even if they did, it would say nothing about whether AB is parallel to DC.
- (d) A rectangle — Method: to find the plan view, work out the outline traced when looking straight down onto the solid from directly above, not the outline shown in the angled sketch. Working: this solid has two identical flat round ends joined by one curved surface, and it is lying on its side rather than standing upright; viewed from above, the curved surface gives two straight edges running the full length of the solid, the width of the round ends apart, and each flat round end — seen edge-on from directly above — also becomes a straight edge of that same width, with no curve remaining. Four straight edges, with opposite sides equal and meeting at right angles, form a rectangle. Answer: a rectangle. The distractors: a circle comes from picturing the solid as if it were standing upright on one of its flat ends, giving the plan of an upright version instead of working out the plan of the solid as it actually lies. An oval comes from copying the foreshortened shape of a round end as it is drawn in the angled sketch, instead of working out the true shape seen from directly above, which has no such foreshortening. A rectangle with rounded ends comes from carrying the curve of the round ends over into the plan view, when in fact a flat round end viewed edge-on from directly above shows no curve at all, only a straight edge.
- (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.
- (a) (2, 6) — Adding the vector to the point gives the new position: (5 + (−3), 2 + 4) = (2, 6). '(8, 6)' comes from adding 3 instead of subtracting it, treating the top number as positive. '(9, −1)' comes from swapping the two components of the vector before adding them. '(2, 2)' applies the horizontal movement correctly but forgets to add the vertical movement.
- (d) 1.5 m — By Pythagoras' theorem, diagonal² = 1.2² + 0.9² = 1.44 + 0.81 = 2.25, so diagonal = √2.25 = 1.5 m. 2.1 m comes from simply adding the two sides (1.2 + 0.9) instead of using Pythagoras' theorem. 0.3 m comes from subtracting the two sides (1.2 − 0.9) instead. 2.25 m comes from correctly finding 1.2² + 0.9² = 2.25 but forgetting to take the square root at the end.
- (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) Triangular prism — Count the given properties against each solid. A triangular prism has 5 faces (2 triangular ends + 3 rectangular sides), 9 edges and 6 vertices — this matches exactly, so the solid is a triangular prism. A square-based pyramid also has 5 faces, but 8 edges and 5 vertices, and only one face is a square rather than three rectangles — the edge and vertex counts don't match. A cuboid has 6 faces, 12 edges and 8 vertices, none of which match the numbers given. A tetrahedron has 4 faces, all triangles, with 6 edges and 4 vertices — too few faces, and no rectangular faces at all.
- (b) (5, −2) — Method: subtract column vectors by subtracting the bottom vector's top number from the top vector's top number, and doing the same for the bottom numbers. Working: top numbers 7 − 2 = 5; bottom numbers 3 − 5 = −2. Answer: u − v = (5, −2). A candidate who drops the negative sign gets (5, 2). A candidate who works out v − u instead of u − v gets (−5, 2). A candidate who adds instead of subtracts gets (9, 8).
- (a) 24 — The display is 4 tins wide, 3 tins high and 2 tins deep, and every position in that block is filled, so the total number of tins is the product of all three measurements: 4 × 3 × 2 = 24. "12" comes from multiplying only the width and height shown in the front elevation (4 × 3), forgetting the depth entirely. "9" comes from adding the three measurements (4 + 3 + 2) instead of multiplying them. "6" comes from multiplying only the height and depth (3 × 2), forgetting the width shown by the front elevation.
- (b) 18.84 m² — The pond has radius 2.5 m (half of the 5 m diameter), so the outer edge of the path has radius 2.5 + 1 = 3.5 m. Path area = area of outer circle − area of pond = 3.14 × 3.5² − 3.14 × 2.5² = 38.465 − 19.625 = 18.84 m².
- (d) (6, 8) — Method: add column vectors by adding the top numbers together and the bottom numbers together. Working: top numbers 5 + 1 = 6; bottom numbers 2 + 6 = 8. Answer: p + q = (6, 8). A candidate who subtracts instead of adds gets (4, −4). A candidate who swaps the top and bottom of the correct answer gets (8, 6). A candidate who only adds the top numbers and leaves the bottom number of p unchanged gets (6, 2).
- (b) 5 — Method: the two points have the same y-coordinate, so the segment joining them runs horizontally and its length is the gap between the two x-coordinates; a length is a distance, so it is never negative. Working: the x-coordinates are 2 and −3, so the gap is 2 − (−3) = 2 + 3 = 5. Both points have y = −4, so there is no vertical part to add on. Answer: 5. The distractors: 1 comes from dropping the minus sign on −3 and working out 3 − 2 instead; 0 comes from subtracting the y-coordinates, which are equal, in place of the x-coordinates; 6 comes from counting the grid lines from −3 across to 2 inclusive, which counts one more than the number of gaps between them.
- (c) 5 — The front elevation shows one square for every cube visible from the front, column by column: the left-hand column is 2 cubes high, so it contributes 2 squares; the middle column is 2 cubes high, so it contributes 2 more; the right-hand column is 1 cube high, so it contributes 1. The total is 2 + 2 + 1 = 5 squares. "6" comes from drawing a full 3 by 2 rectangle, treating every column as if it reached the greatest height. "4" comes from losing a square from one of the two tall columns, counting 2 + 1 + 1. "3" comes from counting one square per column — the width of the solid — and ignoring the heights altogether.
- (a) $\binom{5}{−6}$ — The vector is (image − original) in each coordinate: (2 − (−3), −1 − 5) = (5, −6). $\binom{−5}{6}$ comes from working out original − image instead of image − original. $\binom{5}{6}$ gets the x-component right but makes a sign error on the y-component. $\binom{−5}{−6}$ gets the y-component right but makes a sign error on the x-component, working out −3 − 2 = −5 instead of 2 − (−3) = 5.
- (c) 2 hours 15 minutes — Divide the total minutes by 60: 135 ÷ 60 = 2 remainder 15, so that is 2 hours 15 minutes. Treating 100 minutes as one hour, a metric-style mistake, gives 135 − 100 = 35, so 1 hour 35 minutes. Subtracting 60 twice to reach the 15 minutes left over, but losing count and recording only one of the two hours removed, gives 1 hour 15 minutes. Writing 135 ÷ 60 = 2.25 and then reading the '25' as minutes, instead of converting the 0.25 of an hour into 15 minutes, gives 2 hours 25 minutes.
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