Printable · GCSE Higher · ages 14-16
Geometry and measures worksheet — GCSE Higher
Fifteen questions across the geometry and measures statements at Higher 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 Higher
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- (b) (3, −3) — Subtracting column vectors means subtracting the top numbers and subtracting the bottom numbers, in the order given: top = 5 − 2 = 3, bottom = 1 − 4 = −3, giving (3, −3). A candidate who works out v − u instead of u − v, reversing the order, gets (−3, 3). A candidate who makes a sign error on the bottom number, treating 1 − 4 as 3 instead of −3, gets (3, 3). A candidate who adds the vectors instead of subtracting gets (7, 5). Because the question asks for u − v, not v − u, the correct answer is (3, −3).
- (d) 77 — Method: angles that meet at a point add up to 360°. Working: 82 + 105 + 96 = 283; 360 − 283 = 77. Answer: 77°. A candidate who gives the sum of the three known angles and forgets to subtract it from 360° gets 283. A candidate who leaves out the 105° angle, working out 360 − 82 − 96, gets 182. A candidate who leaves out the 82° angle, working out 360 − 105 − 96, gets 159.
- (b) Rotate 180° about the origin, then translate by (6, 0). — Rotating 180° about the origin sends (x, y) to (−x, −y); applied to T's vertices (1, 1), (3, 1) and (1, 4) this gives (−1, −1), (−3, −1) and (−1, −4). Translating this image by the vector (6, 0) adds 6 to every x-coordinate, giving (5, −1), (3, −1) and (5, −4), which matches T′ exactly. Reflecting in the x-axis first changes the sign of the y-coordinate only, and translating that image by (6, 0) gives (7, −1), (9, −1) and (7, −4) — the wrong triangle. Using the correct rotation but translating by (4, 0) instead of (6, 0) gives (3, −1), (1, −1) and (3, −4), shifted 2 units too far left. Reflecting in the y-axis first changes the sign of the x-coordinate only, so translating that image by (6, 0) leaves every y-coordinate positive, giving (5, 1), (3, 1) and (5, 4) — the correct x-coordinates but the wrong sign throughout on y.
- (b) 7 — Method: each coordinate of a midpoint is the mean of the matching pair of coordinates, so the y-coordinate of the midpoint depends on the two y-coordinates alone. Working: the y-coordinates are 4 and 10, so the mean is (4 + 10) ÷ 2 = 14 ÷ 2 = 7. Answer: 7. The distractors: 5 comes from working out the x-coordinate of the midpoint, (2 + 8) ÷ 2 = 5, and writing that down in place of the y-coordinate the question asked for; 3 comes from halving the difference of the y-coordinates, (10 − 4) ÷ 2 = 3, which is half the vertical gap rather than a position; 14 comes from adding the two y-coordinates and forgetting to halve the total.
- (a) 4700 cm² — Method: split the T-shaped outline into the two rectangles it is made from, find the area of each, then add them together. Working: the wide base gives a rectangle 90 cm × 30 cm = 2700 cm²; the narrower block on top gives a rectangle 40 cm × 50 cm = 2000 cm²; adding these, 2700 + 2000 = 4700 cm². Answer: 4700 cm². The distractors: 2700 cm² comes from finding only the area of the base rectangle and forgetting to add the block on top. 2000 cm² comes from finding only the area of the top block and forgetting the base. 7200 cm² comes from treating the whole outline as one large rectangle, 90 cm wide by (30 + 50) = 80 cm tall, instead of splitting it into the two separate rectangles that actually make up the shape.
- (b) (3, 10) — Method: multiply every part of p by 2, then add the matching parts of q. Working: 2p = (4, 6); adding q gives top 4 + (−1) = 3 and bottom 6 + 4 = 10. Answer: 2p + q = (3, 10). A candidate who forgets to double p first, working out p + q instead, gets (1, 7). A candidate who doubles q instead of p, working out p + 2q, gets (0, 11). A candidate who subtracts q instead of adding it, working out 2p − q, gets (5, 2).
- (b) 1 point — A rotation about a point P always leaves P itself unchanged, and — unless the shape has rotational symmetry about that exact point — no other point of the shape maps onto itself. The centre of rotation here is the vertex (2, 1), so exactly one point of the trapezium, that vertex, is invariant. Thinking that a rotation always has no invariant points ignores the centre of rotation itself, which is always fixed, and gives 0 points. Assuming every vertex of the shape is invariant confuses a rotation with the identity transformation and gives 4 points, the total number of vertices. Believing that the point diametrically opposite the centre is also fixed applies point symmetry of the whole coordinate grid rather than checking whether that point actually lies on this particular trapezium, and gives 2 points.
- (c) 15/17 — Method: cos θ = adjacent ÷ hypotenuse, so find the hypotenuse with Pythagoras' theorem first and then decide which short side is next to θ. Working: the hypotenuse is √(8² + 15²) = √(64 + 225) = √289 = 17 cm. The angle θ is opposite the 8 cm side, so the side next to it is the 15 cm side, and cos θ = 15 ÷ 17. Answer: 15/17. The distractors: 8/17 is sin θ, opposite over hypotenuse, used in place of the cosine; 8/15 is tan θ, opposite over adjacent; 17/15 comes from writing the cosine ratio upside down, as hypotenuse over adjacent.
- (c) AB, AC and angle BAC: Area = 1/2 × AB × AC × sin(BAC) — Method: Area = 1/2ab sin C only works when the angle used is the one included between the two sides being multiplied. Working: AB and AC meet at A, and angle BAC is the angle at A between them, so the statement pairing AB, AC and angle BAC is the correct one. The statement that three sides with no angle can still go into 1/2 AB × AC × sin(BAC) is wrong: with no angle known, sin(BAC) cannot be evaluated, so a different method must find an angle first. The statement pairing AB and BC with sin(BAC) is wrong: AB and BC meet at B, so the angle between them is angle ABC, not angle BAC — it names the wrong angle for the sides it uses. The statement pairing AB and AC with sin(ABC) is wrong for the same reason: AB and AC meet at A, so their included angle is angle BAC, and angle ABC is not between them at all.
- (a) A smaller circle — Since the cutting plane is parallel to the circular base, the cross-section is also a circle, but smaller than the base because the cone narrows as it rises towards the apex, so 'a smaller circle' is correct. 'A triangle' wrongly describes the outline seen from the side of the cone, not a horizontal cross-section. 'An ellipse' would only result from a cut made at an angle to the base, not one parallel to it. 'The same size circle as the base' wrongly ignores that the cone tapers, so any parallel cross-section above the base must be smaller.
- (d) a rectangle — The shed is a prism, so its two triangular ends are identical and parallel, a fixed distance apart along the shed's length; the floor they stand on is therefore bounded by the base of one triangle, the base of the other, and the two straight edges joining them — a rectangle. Looking straight down, the sloping roof projects onto that same rectangle rather than outside it, so the plan view is a plain rectangle, as long as the shed and as wide as its gable end. "a triangle" is the shape of the END wall, seen from the front or back, not from above. "a triangle with a rectangle attached" wrongly combines a side-elevation feature with the plan — the plan does not show the triangular end wall at all, since looking down hides it completely. "two triangles joined at their bases" describes neither the plan nor any single elevation of this shed.
- (b) 62 — Method: co-interior (allied) angles between parallel lines add up to 180°. Working: 180 − 118 = 62. Answer: 62°. A candidate who treats co-interior angles as equal, like corresponding angles, gives 118. A candidate who uses 360° instead of 180°, working out 360 − 118, gets 242. A candidate who subtracts as if the angles were complementary, working out 118 − 90, gets 28.
- (a) 96 cm² — Method: a cube has six identical square faces, so the total surface area is six times the area of one face. Working: one face has area 4 × 4 = 16 cm², and 6 × 16 = 96. Answer: 96 cm². The distractors: 16 cm² is the area of a single face, from stopping before multiplying by the six faces; 64 cm³ comes from working out the volume, 4 × 4 × 4, which is a different measure and carries a different unit; 24 cm² comes from multiplying the six faces by the edge length, 6 × 4, instead of by the area of a face.
- (d) 2/3 — Method: since A is the midpoint of OE, OE = 2a, so E = 2a. Since FB is twice AF, F is 1/3 of the way along AB from A, so F = a + 1/3(b − a) = (2/3)a + (1/3)b. G is the midpoint of OB, so G = (1/2)b. Working: EF = F − E = (2/3)a + (1/3)b − 2a = −(4/3)a + (1/3)b, and EG = G − E = −2a + (1/2)b. Comparing term by term, 2/3 × (−2a + (1/2)b) = −(4/3)a + (1/3)b, which matches EF exactly — the same scalar works on both the a-term and the b-term, so the two vectors are parallel, and since they share the point E the three points are collinear. Answer: k = 2/3, so E, F and G lie on a straight line. Giving 1/3 instead is the scalar linking F to G (FG = (1/3)EG), not E to F; giving 3/2 is the reciprocal — it is EG that equals 3/2 × EF, not the other way round, since EF = k × EG was what was asked for; and giving 4/3 is EF's a-coefficient read off raw, without ever dividing it by EG's a-coefficient to form the comparison. Always match the direction of the scalar to the vectors exactly as the question states them.
- (c) $12\sqrt{3}$ cm² — Use Area = 1/2 × AB × AC × sin(angle BAC) = 1/2 × 6 × 8 × sin 60°. Since sin 60° = √3⁄2, this is 1/2 × 6 × 8 × √3⁄2 = 24 × √3⁄2 = 12√3 cm². Using AB × AB instead of AB × AC (taking 6 × 6 rather than 6 × 8) gives 1/2 × 6 × 6 × sin 60° = 18 × √3⁄2 = 9√3 cm². Using AC × AC instead of AB × AC (taking 8 × 8 rather than 6 × 8) gives 1/2 × 8 × 8 × sin 60° = 32 × √3⁄2 = 16√3 cm². Leaving out the 1/2 from the area formula gives 6 × 8 × sin 60° = 48 × √3⁄2 = 24√3 cm².
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