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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- (c) (1, 0) — To enlarge about a centre other than the origin, find the vector from the centre to the point, scale that vector, then add it back to the centre. The vector from (2, 2) to A(4, 6) is (2, 4). Scaling by −1/2 gives (−1, −2). Adding this to the centre (2, 2) gives the image point (1, 0). (3, 4) comes from using +1/2 instead of −1/2, so the image lands on the same side as A instead of the opposite side. (−2, −3) comes from scaling A's coordinates directly about the origin, ignoring that the centre is (2, 2). (−2, −6) comes from using a scale factor of −2 instead of −1/2.
- (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.
- (d) 1.1 — Method: multiply the drawing length by the scale factor to get the real length, then convert to the units asked for. Working: 4.4 cm × 25 = 110 cm = 1.1 m. A student who answers 4.4 has forgotten to use the scale at all. A student who answers 110 has correctly worked out the real length in centimetres but forgotten to convert it to metres. A student who answers 11 has used a scale factor of 2.5 instead of 25 by misreading the scale. Answer: 1.1 m.
- (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) 180 cm³ — Method: the volume of a right prism is the area of its cross-section multiplied by its length, and the area of a triangle is half the base multiplied by the perpendicular height. Working: the cross-section has area (6 × 5) ÷ 2 = 15 cm², and 15 × 12 = 180. Answer: 180 cm³. The distractors: 360 cm³ comes from taking the cross-section as 6 × 5 = 30 and never halving it, which measures the rectangle around the triangular face rather than the face itself; 66 cm³ comes from adding the base and the perpendicular height and halving, (6 + 5) ÷ 2 = 5.5, which is the trapezium rule used where the triangle rule is needed, and then multiplying by the 12 cm length; 15 cm³ comes from working out the triangular cross-section correctly and stopping there, so the 12 cm length is never used and an area is handed in as a volume.
- (d) 63.6 — Method: the three angles inside the triangle formed by the two ladders and the ground add up to 180°. Working: 180 − 58.2 − 58.2 = 63.6. Answer: 63.6°. A candidate who assumes the top angle equals the base angles gives 58.2. A candidate who subtracts only one base angle from 180°, working out 180 − 58.2, gets 121.8. A candidate who doubles the base angle instead of subtracting it twice from 180°, working out 2 × 58.2, gets 116.4.
- (c) 0.5 — The real distance is 2 × 25000 = 50000 cm. Converting to metres, by dividing by 100, gives 500 m, and converting to kilometres, by dividing by 1000, gives 0.5 km. A candidate who divides the 50000 cm by 1000 in one go, applying the metres-to-kilometres factor straight to the centimetres, gets 50 km. A candidate who divides by 100 twice, treating 100 m as 1 km, gets 5 km. A candidate who slips one extra decimal place when converting 500 m to kilometres gets 0.05 km. The real distance is 0.5 km.
- (b) (12, −9) — Two enlargements one after the other combine into a single scale factor: 1.5 × 2 = 3. Multiplying a vector by a scalar means multiplying both the top number and the bottom number by it: top = 4 × 3 = 12, bottom = −3 × 3 = −9, giving (12, −9). A candidate who adds the scale factor to each number instead of multiplying gets (4 + 3, −3 + 3) = (7, 0). A candidate who multiplies the top number but leaves the bottom number unchanged gets (12, −3). A candidate who multiplies the bottom number but leaves the top number unchanged gets (4, −9). The correct column vector for the poster is (12, −9).
- (c) No — 50° is angle P, not between PQ and QR. — Method: 1/2ab sin C only gives the area when C is the angle between the two named sides. Working: PQ and QR meet at vertex Q, so the angle between them is angle PQR, not angle QPR (which is the angle at vertex P, between sides PQ and PR). Since angle QPR is not included between PQ and QR, the student's calculation 1/2 × 8 × 11 × sin 50° does not give the area of the triangle — the verdict is No. Claiming that any angle works with any two sides ignores that the formula is only true for the INCLUDED angle. Claiming the formula needs all three sides is a different, incorrect statement about the formula itself — Area = 1/2ab sin C is valid with exactly two sides and their included angle, when that angle is actually known.
- (c) 1/4 — A full turn at the centre of a circle is 360°, so a sector's fraction of the circle is its angle divided by 360°: 90 ÷ 360 = 1/4. 1/2 would be the fraction for a sector with an angle of 180°, not 90°. 3/4 is the fraction of the rest of the circle, the major sector left over from the 270° that is not part of this sector. 1/8 would be the fraction for a sector with an angle of 45°, half of 90°.
- (c) (2, −3) — Method: apply the transformations in the order given, translation first, then reflection, to the point itself. Working: translating (5, 1) by the vector (−3, 2) moves the x-coordinate by −3 and the y-coordinate by 2, giving 5 − 3 = 2 and 1 + 2 = 3, so the image after the translation is (2, 3). Reflecting this point in the x-axis leaves the x-coordinate unchanged and reverses the sign of the y-coordinate, giving (2, −3). Answer: (2, −3). Do the two maps in the stated order and carry the point all the way through both of them: swapping the order, or stopping after the first map, lands you on a different point altogether.
- (a) Triangular-based pyramid (tetrahedron) — A solid with 4 triangular faces, 4 vertices and 6 edges is a triangular-based pyramid, also called a tetrahedron. A triangular prism also has triangular faces, but it has 2 triangular faces plus 3 rectangular faces, 6 vertices and 9 edges — the extra rectangular faces and edges rule it out here. A square-based pyramid has 5 faces (one square, four triangles), 5 vertices and 8 edges, which does not match. A cube has 6 faces, 8 vertices and 12 edges, all much higher than the numbers given. The solid described is a triangular-based pyramid.
- (c) Swapped the x and y components — The student's vector has the same two numbers, 5 and −2, but in swapped positions, so the error is swapping the x and y components rather than an error with signs or size. 'Reversed both signs' is wrong because the numbers 5 and −2 have not changed sign, only position. 'Reversed only the y sign' is wrong for the same reason — no sign has actually changed. 'Doubled the x component' is wrong because neither number has changed in size.
- (a) 8,000,000 cm³ — Method: change the edge length into centimetres first and then cube it, because 1 m = 100 cm and a volume needs that conversion applied to all three dimensions. Working: 2 m = 2 × 100 = 200 cm, so the volume is 200 × 200 × 200. 200 × 200 = 40,000 and 40,000 × 200 = 8,000,000. Answer: 8,000,000 cm³. The distractors: 8,000 cm³ comes from converting 2 m to 20 cm and cubing that; 80,000 cm³ comes from cubing in metres to get 8 m³ and then multiplying by 10,000, the conversion factor for an area rather than the 1,000,000 a volume needs; 8 cm³ comes from cubing the 2 without converting at all and simply writing cm³ because the question asked for that unit.
- (d) DE = 8 cm — Method: use the stated correspondence ABC ≅ DEF to work out which side in DEF matches the known side AB in ABC. Working: the correspondence sends A to D, B to E and C to F, so AB corresponds to DE; the right angles at B and E and the equal hypotenuses AC = DF = 17 cm are already given, so DE = 8 cm supplies the third ingredient — Right angle, Hypotenuse, Side. Options: EF = 8 cm matches AB to the wrong side, since EF corresponds to BC, and BC = √(17² − 8²) = 15 cm, not 8 cm; BC = 8 cm states something about triangle ABC rather than the missing fact about DEF, and it is false as well, since BC = 15 cm; angle D = angle A does follow once the triangles are congruent, but RHS is completed by a matching side, not by a matching angle. Answer: DE = 8 cm.
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