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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- (a) 59.0° — Method: rearrange Area = (1/2)ab sin C to make sin C the subject: sin C = 2 × Area ÷ (a × b). Working: sin C = 2 × 24 ÷ (8 × 7) = 0.857, so C = sin⁻¹(0.857) = 59.0° (1 d.p.), which is acute as the question requires. Answer: 59.0°. Taking the obtuse angle instead of the acute one asked for, 180 − 59.0 = 121.0°; forgetting to double the area before dividing gives sin C = 24 ÷ (8 × 7) = 0.4286, whose acute angle is 25.4°; and combining that same forgotten-doubling error with the obtuse branch gives 180 − 25.4 = 154.6°. Always double the area first, and then pick the acute branch, since that is what this question asks for.
- (a) 18 — PQ lies along the x-axis with length 9, and PR lies along the y-axis with length 4, and these two sides meet at right angles at P, so they can be used as the base and height of the triangle. Area = 1/2 × base × height = 1/2 × 9 × 4 = 18. 36 comes from multiplying the base and height but forgetting to halve the result. 13 comes from adding the two lengths, 9 + 4, instead of multiplying them. 26 comes from the perimeter-style calculation 2 × (9 + 4) instead of the triangle area formula.
- (c) Equal sides and equal interior angles — Method: recall the full definition of 'regular' as applied to a polygon. Working: a regular polygon must have both equal side lengths and equal interior angles at the same time. Options: 'all sides equal' alone describes an equilateral but not necessarily equiangular shape, such as a rhombus, which is not regular; 'all angles equal' alone describes an equiangular but not necessarily equilateral shape, such as a rectangle, which is not regular; 'at least one line of symmetry' is a much weaker condition that many irregular shapes also satisfy. Answer: equal sides and equal interior angles.
- (c) A rotation of 180° about the origin — An enlargement by scale factor −1 sends every point (x, y) to (−x, −y) — both coordinates change sign. A rotation of 180° about the origin does exactly the same thing to every point, so the two transformations have identical effect. A reflection in the x-axis only changes the sign of the y-coordinate, sending (x, y) to (x, −y), leaving the x-coordinate untouched. A reflection in the y-axis only changes the sign of the x-coordinate, sending (x, y) to (−x, y), leaving the y-coordinate untouched. Treating a negative scale factor as though it behaves like a positive one gives no transformation at all, but the minus sign is not decorative — it reverses both coordinates. Both signs flip together, which is exactly what a 180° rotation about the origin does.
- (d) A translation by the vector (8, 0) — Method: two reflections in PARALLEL lines combine into a single translation, perpendicular to the lines, of size twice the distance between them; two reflections in lines that CROSS combine into a rotation instead, never a translation. Working: the lines x = 2 and x = 6 are parallel, a distance of 6 − 2 = 4 apart. Doubling this distance gives 2 × 4 = 8, and the translation runs in the direction from the first line towards the second, so the vector is (8, 0). Answer: a translation by the vector (8, 0). Double the distance between the lines rather than using it directly, keep the direction running from the FIRST line reflected to the SECOND, and remember that two reflections in lines that never meet can only give a translation, never a rotation.
- (c) (3, 2) — For an enlargement centred on the origin, multiply every coordinate by the scale factor: (9 × 1/3, 6 × 1/3) = (3, 2). A pupil who multiplies by 3 instead of by 1/3 gets (27, 18). A pupil who subtracts a third of each coordinate instead of scaling by a third gets (9 − 3, 6 − 2) = (6, 4). A pupil who applies the scale factor to the x-coordinate only gets (3, 6). The correct image is (3, 2).
- (c) Square — Because the cut is parallel to the square base, it produces a shape that is similar to the base — the same shape, scaled down — so it is also a square, with four equal sides, smaller than the base. A candidate who looks at the sloped triangular faces instead of at the horizontal cut answers triangle. A candidate who pictures a slanted cut, which would shorten one pair of sides, answers trapezium, but a cut parallel to the base scales all four sides by the same amount. A candidate who counts an extra side, perhaps including an edge of the base, answers pentagon. The cross-section is a square.
- (b) £2.94 — Method: the price is quoted for each kilogram, so the mass has to be written in kilograms before it is multiplied by the price. Working: 1 kg = 1000 g, so 350 ÷ 1000 = 0.35 and the piece weighs 0.35 kg. The cost is then 8.40 × 0.35 = 2.94. Answer: £2.94. Treating 350 g as 3.5 kg, a division by 100 rather than by 1000, gives 8.40 × 3.5 = 29.40. Multiplying the price by the number of grams gives 8.40 × 350 = 2940. Dividing the price by the mass instead of multiplying gives 8.40 ÷ 0.35 = 24.
- (b) 25 minutes — Method: a rate in litres per minute can only be used on a volume measured in litres, so convert the tank first and then divide. Working: 1 m³ = 1000 litres, so the tank holds 0.45 × 1000 = 450 litres, and the time is 450 ÷ 18 = 25. Answer: 25 minutes. Using 1 m³ = 100 litres gives 45 ÷ 18 = 2.5 minutes. Using 1 m³ = 1 000 000 litres, which is the factor that turns cubic metres into cubic centimetres, gives 450 000 ÷ 18 = 25 000 minutes. Multiplying by the rate instead of dividing by it gives 450 × 18 = 8100.
- (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.
- (a) 035° — A bearing is measured clockwise from north, so an angle of 35° clockwise from north is a bearing of 035° (written with three figures). Choosing 325° measures the angle anticlockwise instead of clockwise (360 − 35 = 325). Choosing 215° adds 180° to the angle, mixing this up with a back-bearing calculation (35 + 180 = 215). Choosing 350° reorders the digits of 035, writing the ones digit before the tens digit by mistake.
- (a) 75° — The three angles in the triangle formed by the two supports and the ground add up to 180°. Two of the angles are 34° and 71°, so the third angle is 180 − 34 − 71 = 75, which is 75°. 105° comes from adding the two given angles together instead of subtracting them from 180° (34 + 71 = 105). 146° comes from computing 180 − 34 = 146 and forgetting to also subtract 71°. 37° comes from subtracting the two given angles from each other instead of using the triangle's angle sum (71 − 34 = 37).
- (d) (3/4)a + (1/4)c — Method: OX = OA + AX, and since AX is a third of XC, AX is 1/4 of the whole of AC, with AC = c − a. Working: OX = a + 1/4(c − a) = a − (1/4)a + (1/4)c = (3/4)a + (1/4)c. Answer: OX = (3/4)a + (1/4)c. Measuring 1/4 of AC from C's end instead of A's swaps the fractions round, giving (1/4)a + (3/4)c; adding (1/4)c onto the whole of a without subtracting a inside the bracket first gives a + (1/4)c; and treating the ratio as though AX and XC were equal gives the midpoint, (1/2)a + (1/2)c. Convert the ratio to a fraction of AC measured from A, subtract before you scale, and then add the result to OA.
- (a) 080° — The back bearing differs from the given bearing by 180°. Because 260° is greater than 180°, subtract 180°: 260 − 180 = 80°, so the bearing of A from B is 080°. Choosing 440° adds 180° instead of subtracting it, even though the result would be more than a full turn (260 + 180 = 440). Choosing 180° assumes the back bearing is always exactly 180°, ignoring the original bearing altogether. Choosing 100° comes from measuring the reflex angle the other way round the circle (360 − 260 = 100) instead of applying the 180° back-bearing rule.
- (c) RHS applies: the right angle, hypotenuse and one matching side are all equal — Both triangles are right-angled, and the hypotenuses (AC and DF) are equal, and one further pair of matching sides (AB and DE) are equal. This matches the Right angle-Hypotenuse-Side condition, which needs only these three facts, not a third side.
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