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.
Answer key: Geometry and measures worksheet — GCSE Higher
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- (c) (1/3)a + (2/3)b — Method: OP = OA + AP, and since AP is twice PB, AP is 2/3 of the whole of AB, with AB = b − a. Working: OP = a + 2/3(b − a) = a − (2/3)a + (2/3)b = (1/3)a + (2/3)b. Answer: OP = (1/3)a + (2/3)b. Measuring 2/3 of AB from B's end instead of A's swaps the fractions round, giving (2/3)a + (1/3)b; adding (2/3)b onto the whole of a without first subtracting a inside the bracket gives a + (2/3)b; and treating the ratio as though AP and PB were equal gives the midpoint, (1/2)a + (1/2)b. Convert the ratio to a fraction of AB measured from the point named first in the ratio, subtract before you scale, and then add the result to OA.
- (a) 10.1 cm — Method: the area formula 1/2 × a × b × sin C contains the unknown side, so substitute what is known and rearrange. Working: 42 = 1/2 × 9.5 × AC × sin 61°. Multiplying both sides by 2 gives 84 = 9.5 × AC × sin 61°, and 9.5 × sin 61° = 9.5 × 0.87462 = 8.3089, so AC = 84 ÷ 8.3089 = 10.1096. Answer: AC = 10.1 cm to 1 decimal place. The distractors: 5.1 cm comes from forgetting to double the area when clearing the factor 1/2 and working out 42 ÷ 8.3089; 18.2 cm comes from using cos 61° in place of sin 61° in the denominator; 7.7 cm comes from multiplying by sin 61° instead of dividing by it, 84 × sin 61° ÷ 9.5, the standard slip when the unknown is inside a product.
- (b) (−1, −1) — An enlargement by scale factor −1, centre (2, 1), sends a point P to the point on the opposite side of the centre, the same distance away: the image is 2 × centre − P. For the vertex (5, 3), this gives (2 × 2 − 5, 2 × 1 − 3) = (4 − 5, 2 − 3) = (−1, −1). Treating the centre as though it were the origin, and simply negating the point's coordinates, gives (−5, −3) — this ignores that the true centre is (2, 1), not (0, 0). Using scale factor +1 instead of −1 leaves the point exactly where it started, at (5, 3). Adding the point's displacement from the centre instead of subtracting it gives (2 × 2 + 5, 2 × 1 + 3) = (9, 5). Double the centre and subtract the point, and the image is (−1, −1).
- (d) Wrong - the given angle is not the included angle — Sides AB and BC meet at vertex B, so the included angle needed for SAS is angle B, not angle A — the information given is SSA. SSA does not prove congruence: with AB = 10 cm, BC = 7 cm and angle A = 40° there are two different triangles that fit, one with angle C ≈ 74.6° and one with angle C ≈ 105.4°, so Sam's triangles need not be the same shape and size at all. Sam is not correct just because two sides and an angle are equal, since the angle must be the one INCLUDED between those two sides. The condition is not ASA either, because ASA needs two angles, and only one angle is given here. It is also not true that nothing matches — the stated lengths and angle DO match between the two triangles; the problem is which angle was given, not whether the values agree.
- (d) 12 — Method: rearrange F + V − E = 2 so that E is on its own: E = F + V − 2. Working: E = 6 + 8 − 2 = 12. A student who answers 14 has added F and V but forgotten to subtract 2 at all. A student who answers 16 has added 2 instead of subtracting it. A student who answers 10 has subtracted 2 twice by mistake. Answer: 12 edges.
- (b) 128.2° — Method: rearrange the area formula for the sine of the enclosed angle, then remember that the inverse sine key returns only the acute angle, so the obtuse angle must be found by subtracting from 180°. Working: 33 = 1/2 × 12 × 7 × sin BAC, so sin BAC = 2 × 33 ÷ (12 × 7) = 66 ÷ 84 = 0.78571. The inverse sine of 0.78571 is 51.787°, and the obtuse angle with the same sine is 180° − 51.787° = 128.213°. Answer: angle BAC = 128.2° to 1 decimal place. The distractors: 51.8° is the acute angle straight off the calculator, given by a candidate who never acts on the instruction that the angle is obtuse; 38.2° comes from pressing the inverse cosine key on 0.78571 instead of the inverse sine key; 156.9° comes from forgetting to double the area, so that sin BAC is taken as 33 ÷ 84 = 0.39286, and then subtracting the resulting 23.1° from 180°.
- (d) (−2, 0) — A 180° rotation about a centre (a, b) maps (x, y) to (2a − x, 2b − y). Here that gives (2 × 1 − 4, 2 × 1 − 2) = (−2, 0). A pupil who rotates about the origin instead of (1, 1) gets (−4, −2). A pupil who adds the centre's coordinates instead of applying the rotation formula gets (4 + 1, 2 + 1) = (5, 3). A pupil who just subtracts the centre's coordinates from E's, without doubling and reversing, gets (4 − 1, 2 − 1) = (3, 1). The correct image is (−2, 0).
- (d) (−14, 5) — To undo a composition, reverse the order and invert each transformation: undo the rotation first, then undo the translation. The inverse of 'rotate 90° clockwise about the origin' is 'rotate 90° anticlockwise about the origin', which maps (x, y) to (−y, x); applied to (3, 9) this gives (−9, 3). Then undo the translation by subtracting the vector (5, −2), i.e. adding (−5, 2): (−9 − 5, 3 + 2) = (−14, 5). Undoing the two inverse steps in the same order as the original composition, rather than reversing it, gives (−11, −2). Rotating 90° clockwise again instead of inverting the rotation's direction gives (4, −1). Adding the translation vector again instead of subtracting it, after correctly inverting the rotation, gives (−4, 1).
- (b) Hexagon — Cutting straight across a prism, at right angles to its length, always gives a cross-section that is the same shape as its end faces. The end faces of a hexagonal prism are hexagons (6-sided), so the cross-section is a hexagon. Choosing Pentagon comes from miscounting the sides of the hexagonal end as five instead of six. Choosing Rectangle comes from cutting along the LENGTH of the prism instead of across it, which gives a rectangular face, not the cross-section asked for. Choosing Triangle comes from confusing a hexagonal prism with a triangular prism.
- (b) (6, −8) — Method: a scalar multiple of m has the same ratio between its top and bottom numbers as m does. Working: m = (3, −4); multiplying both parts by 2 gives 2 × 3 = 6 and 2 × (−4) = −8, so (6, −8) is a scalar multiple of m. Answer: (6, −8). The vector (6, −4) needs a multiplier of 2 for the top number but only 1 for the bottom number, so it is not a multiple. The vector (−6, −8) needs a multiplier of −2 for the top number but 2 for the bottom number, so it is not a multiple. The vector (9, −8) needs a multiplier of 3 for the top number but 2 for the bottom number, so it is not a multiple.
- (d) 15 cm — Method: any two sides of a triangle must together be longer than the third, so the third side must be longer than the difference of the two given sides and shorter than their sum. Working: the difference is 20 − 8 = 12 cm and the sum is 20 + 8 = 28 cm, so the third side must be between 12 cm and 28 cm, and 15 cm lies inside that range. Answer: 15 cm. The distractors: 12 cm is exactly the difference, so the three lengths would lie flat along a straight line and never close into a triangle; 5 cm is shorter than the difference — 5 + 8 = 13 cm cannot reach across the 20 cm side — and is chosen by candidates who check no lower limit at all; 30 cm is longer than the sum of the other two, so those two sides could never meet, and it is chosen by candidates who check no upper limit.
- (d) sector — The region enclosed by two radii and the arc between them is called a sector. A segment is a different region, enclosed by a CHORD and an arc — it does not touch the centre at all. A chord is a straight line, not a region, so it cannot be the answer to a question asking for a region. A semicircle is only correct when the angle between the two radii is exactly 180°, which is not stated here, so it is too specific to be the general answer.
- (a) 3 m — Height = sloping length × sin 45° = 3√2 × √2/2 = (3 × 2)/2 = 3 m, since √2 × √2 = 2. 3√2 m comes from forgetting to multiply by sin 45° at all. 3√2/2 m comes from using sin 30° = 1/2 instead of sin 45° = √2/2. 6 m comes from using √2 instead of √2/2 for sin 45°, dropping the denominator of the exact value: 3√2 × √2 = 6.
- (a) 32.2° — Method: all three sides are known, so rearrange the cosine rule as cos A = (b² + c² − a²) ÷ 2bc, where a is the side facing the angle you want. Working: angle ABC sits between AB = 8 cm and BC = 11 cm and faces AC = 6 cm, so cos ABC = (8² + 11² − 6²) ÷ (2 × 8 × 11) = (64 + 121 − 36) ÷ 176 = 149 ÷ 176 = 0.84659, and the inverse cosine of 0.84659 is 32.157°. Answer: angle ABC = 32.2° to 1 decimal place. The distractors: 102.6° comes from subtracting the square of the longest side, 11, rather than the square of the side the angle actually faces, giving (64 + 36 − 121) ÷ 96; 57.8° comes from taking the inverse sine of 0.84659 instead of the inverse cosine; 147.8° comes from rearranging with the subtraction the wrong way round, (36 − 64 − 121) ÷ 176 = −0.84659, which turns an acute angle into its supplement.
- (b) 46.8 cm² — Method: for a parallelogram, not a triangle, the two sides and the angle between them give Area = ab sin C — there is no 1/2. Working: Area = 6 × 9 × sin 60° = 46.8 cm² (1 d.p.). Answer: 46.8 cm². Using the triangle formula, (1/2)ab sin C, on a parallelogram by mistake gives half the true area, 23.4 cm²; using cos 60° instead of sin 60° gives 27.0 cm²; and adding the two sides before multiplying by sin 60° gives 13.0 cm². A parallelogram is exactly two of the triangles this formula was built for, so never carry the 1/2 across from the triangle version.
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