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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- (d) Two angles and a side: sine rule finds other sides. — Method: match the data you are given to the rule that needs it. Working: the sine rule a/sin A = b/sin B = c/sin C needs a complete angle-side pair to set up its ratio, so the statement that two angles and a side (AAS or ASA) let the sine rule find the other sides is the correct one — the third angle comes from the angle sum, and each unknown side is then opposite a known angle. The statement that two sides and the angle between them (SAS) call for the sine rule is wrong: no angle-side pair is complete, so the cosine rule is what works there. The statement that the sine rule finds any angle from three sides (SSS) is wrong for the same reason in reverse — no angle is known at all, so the cosine rule must find the first one. The statement that the cosine rule finds a missing angle directly from two sides and a non-included angle (SSA) is wrong: the cosine rule reports the angle enclosed by the two sides it uses, so with SSA it is the sine rule that reaches the missing angle, and the ambiguous case is then settled from the wording of the question.
- (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.
- (b) $\binom{-5}{5}$ — Method: add the two flights to get the single vector of the whole journey out, then reverse that vector to get the journey home. Working: across, 6 − 1 = 5; up, 4 − 9 = −5. So the drone finishes at (7, −8), which is 5 to the right of its start and 5 below it. The way home is therefore 5 to the left and 5 up. Answer: $\binom{-5}{5}$. Giving the combined journey itself, $\binom{5}{-5}$, describes the flight out rather than the flight home. Subtracting the second vector instead of adding it makes the journey out 7 across and 13 up, and reversing that gives $\binom{-7}{-13}$. Reversing the horizontal movement but leaving the vertical one alone gives $\binom{-5}{-5}$.
- (a) Reflect in the x-axis, then translate by (0, 7). — Reflecting in the x-axis sends (x, y) to (x, −y); applied to S's vertices (2, 2), (5, 2) and (2, 5) this gives (2, −2), (5, −2) and (2, −5). Translating this image by the vector (0, 7) adds 7 to every y-coordinate, giving (2, 5), (5, 5) and (2, 2), which matches S′ exactly. Using the correct reflection but translating by (7, 0) instead moves the image sideways rather than upwards, giving (9, −2), (12, −2) and (9, −5) — nowhere near S′. Reflecting in the y-axis instead of the x-axis changes the sign of the x-coordinate rather than the y-coordinate, so translating that image by (0, 7) gives (−2, 9), (−5, 9) and (−2, 12), the wrong shape entirely. Rotating 180° about the origin instead of reflecting sends every coordinate to its negative, so translating by (0, 7) gives (−2, 5), (−5, 5) and (−2, 2) — the y-coordinates match S′ but the x-coordinates do not.
- (b) (5, −3) — Method: rotate the point about the given centre first, then translate the image, in the stated order. Working: rotating (0, 2) by 180° about (2, 2) uses the rule (x, y) → (4 − x, 4 − y), since the centre doubles in each coordinate. This gives 4 − 0 = 4 and 4 − 2 = 2, so (0, 2) maps to (4, 2). Translating (4, 2) by the vector (1, −5) gives 4 + 1 = 5 and 2 − 5 = −3, so the final image is (5, −3). Answer: (5, −3). Rotate about the centre GIVEN in the question, (2, 2), not about the origin, and translate the rotated image afterwards, in that order: rotating about the wrong centre, swapping the order, or stopping after one step all give a different point.
- (a) (5, 4) — Method: find the vector from the centre to the point, multiply it by the scale factor, then add the result back to the centre. Working: the vector from (2, 4) to (8, 4) is (6, 0); multiplying by 1/2 gives (3, 0); adding this to the centre (2, 4) gives (5, 4). Options: (4, 2) comes from multiplying the original coordinates by 1/2 directly, ignoring the centre of enlargement; (14, 4) comes from using a scale factor of 2 instead of 1/2, giving (2, 4) + 2×(6, 0) = (14, 4); (8, 2) comes from halving only the y-coordinate and leaving the x-coordinate unchanged. Answer: (5, 4).
- (a) No — it's a 180° rotation, not a reflection. — A negative scale factor scales the shape by the size of the factor and turns it through 180° about the centre of enlargement — every point of the image lands on the opposite side of the centre from the object. A 180° turn is a rotation, and a rotation never reverses a shape's handedness (its sense), whereas a reflection always does, so the image is not a mirror image: Priti is wrong. The option claiming negative numbers always mean a reflection confuses the sign of a scale factor with the transformation of reflection — they are not the same thing. The option saying only the size changes ignores the sign altogether: the 180° turn does change the way the shape faces, even though it leaves the handedness unchanged. The option equating 'upside down' with 'mirrored' assumes the two always mean the same thing, but a 180° rotation and a reflection in a horizontal line can both look 'upside down' while being different transformations — only the reflection reverses the shape's handedness.
- (b) 70 cm³ — Area of the triangular cross-section = base × height ÷ 2. Base × height = 3.5 × 4 = 14 cm², and half of that is 14 ÷ 2 = 7 cm². Volume of the prism = cross-sectional area × length = 7 × 10 = 70 cm³. A pupil who forgets to halve when finding the triangle's area gets 3.5 × 4 × 10 = 140 cm³. A pupil who correctly finds the cross-sectional area but forgets to multiply by the length stops at 7 cm³. A pupil who ignores the height altogether, multiplying base by length, gets 3.5 × 10 = 35 cm³. The correct volume is 70 cm³.
- (c) (6, −5) — Method: the translation has already happened, so it must be undone: reverse the vector and apply the reverse to the point that is given. Working: reversing $\binom{-2}{6}$ gives $\binom{2}{-6}$, so the x-coordinate is 4 + 2 = 6 and the y-coordinate is 1 − 6 = −5. Check: from (6, −5) the given vector gives 6 − 2 = 4 and −5 + 6 = 1, which is the point named in the question. Answer: (6, −5). Applying the vector forwards instead of backwards gives (2, 7). Reversing the horizontal movement but not the vertical one gives (6, 7), and reversing the vertical movement but not the horizontal one gives (2, −5).
- (d) 600 cm² — Enlarging by scale factor 2 makes the new dimensions 10 × 2 = 20 cm and 15 × 2 = 30 cm, so the poster's area = 20 × 30 = 600 cm². A pupil who scales the original area, 150 cm², by the scale factor itself instead of by its square gets 150 × 2 = 300 cm². A pupil who adds the scale factor to each dimension instead of multiplying gets (10 + 2) × (15 + 2) = 204 cm². A pupil who forgets to enlarge the postcard at all just uses the original area, 150 cm². The correct area of the poster is 600 cm².
- (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.
- (c) cos 45° — cos 45° = √2/2, and √2 cannot be written as an exact fraction, so this value is irrational. tan 0° = 0, sin 90° = 1 and sin 30° = 1/2 are all rational, since each can be written as an exact whole number or fraction.
- (c) I is the same distance from all three sides. — The angle bisector from A is the locus of points equidistant from sides AB and AC, and the angle bisector from B is the locus of points equidistant from sides AB and BC. Point I lies on both bisectors, so I is equidistant from AB and AC, and also equidistant from AB and BC — meaning I is the same distance from all three sides. (Being the same distance from all three vertices instead describes the circumcentre, found from the perpendicular bisectors of the sides, not the angle bisectors; I being the midpoint of AB confuses the angle bisector construction with the perpendicular bisector of a side; I lying on side AC is wrong because the angle bisectors meet inside the triangle, not on one of its sides.)
- (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) 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.
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