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) 27.7 cm — Arc length = (150 ÷ 360) × 2 × 3.14 × 6 = (5 ÷ 12) × 37.68 = 15.7 cm. The perimeter of a sector also includes the two straight radii, so perimeter = 15.7 + 6 + 6 = 27.7 cm. (15.7 cm comes from stopping after the arc length and forgetting the two straight edges; 21.7 cm comes from adding only one radius instead of two; 31.4 cm comes from doubling the arc length instead of adding the two straight edges.)
- (b) £26.25 — First multiply the base and height: 2.4 × 1.75 = 4.2. The area of the triangular sail is half of that: half of 4.2 is 2.1 m². Then multiply by the cost per m²: 2.1 × £12.50 = £26.25. £52.50 forgets to halve in the area formula, giving an area of 4.2 m² and doubling the true cost. £30.00 multiplies the base length by the cost per m² (2.4 × £12.50) without ever finding the area. £25.00 rounds the area to 2 m² before multiplying by the cost, losing accuracy.
- (b) 18 — Method: convert the real length to centimetres, then divide by the scale factor to shrink it down to the model's size. Working: 4.32 m = 432 cm; 432 cm / 24 = 18 cm. A student who answers 432 has converted the units correctly but forgotten to divide by the scale factor at all. A student who answers 10368 has multiplied by the scale factor instead of dividing (432 x 24). A student who answers 1.8 has converted the metres to centimetres by multiplying by 10 instead of 100, getting 43.2 cm, and then divided by 24. Answer: 18 cm.
- (a) 114.6° — Using the sine rule, sin(ACB) = AB × sin(ABC) ÷ AC = 6 × sin(40°) ÷ 9. Since sin(40°) ≈ 0.64279, this gives sin(ACB) ≈ 3.8567 ÷ 9 ≈ 0.42852, so angle ACB ≈ 25.4° or its supplement, 154.6°. Testing the obtuse candidate: 40° + 154.6° = 194.6°, which already exceeds 180°, so angle BAC would have to be negative — impossible, so 154.6° is rejected. With angle ACB ≈ 25.4°, angle BAC = 180° − 40° − 25.4° = 114.6°. 25.4° is angle ACB, not angle BAC that the question asks for. 14.6° comes from using the invalid 154.6° candidate anyway and then wrongly turning the resulting negative angle sum (−14.6°) positive instead of rejecting it. 65.4° comes from inverting the sine rule ratio — dividing AC × sin(ABC) by AB instead of AB × sin(ABC) by AC — which gives a different, incorrect candidate for angle ACB entirely.
- (d) 4.44 m — The real van's length is 18.5 × 24 = 444 cm, which converts to 4.44 m. '0.77 m' comes from dividing by 24 instead of multiplying, using the ratio the wrong way round (18.5 ÷ 24 = 0.77 cm), and then writing that figure down as metres. '44.4 m' comes from converting 444 cm to metres with the decimal point in the wrong place. '444 m' comes from working out 444 cm correctly but forgetting to convert it into metres at all.
- (a) 17 cm — Convert the real distance to centimetres first: 3.4 km = 340 000 cm. Then divide by the scale factor, 20 000, since the map is 20 000 times smaller than real life: 340 000 ÷ 20 000 = 17, so the map distance is 17 cm. Choosing 1700 cm divides by 200 instead of 20 000, losing two zeros from the scale factor (340 000 ÷ 200 = 1700). Choosing 170 cm divides by 2 000 instead of 20 000, losing one zero from the scale factor (340 000 ÷ 2 000 = 170). Choosing 0.17 cm divides by 2 000 000 instead of 20 000, adding two extra zeros to the scale factor (340 000 ÷ 2 000 000 = 0.17).
- (b) 128° — Angle AOC and angle ABC stand on the same arc AC (the minor arc, which does not contain B), so by the angle at the centre theorem angle ABC = 104° ÷ 2 = 52°. ABCD is a cyclic quadrilateral, so its opposite angles ABC and ADC sum to 180°: angle ADC = 180° − 52° = 128°. A candidate who finds angle ABC = 52° correctly but then treats opposite angles as equal, as in a parallelogram, instead of supplementary, writes down 52° and stops there. Skipping the halving step and using 104° as angle ABC gives 180° − 104° = 76°. Reading off the given centre angle itself as the final answer, without applying either theorem, gives 104°. Work through both theorems in order and you land on 128°.
- (d) 10 cm — For a triangle to exist, any two sides must add up to more than the third side. 9 + 10 = 19 > 15, and 15 − 9 = 6 < 10, so 10 cm satisfies the triangle inequality. The other lengths fail: 6 cm gives 9 + 6 = 15, which is not more than 15; 24 cm and 26 cm are each at least as large as 9 + 15 = 24.
- (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.
- (d) Opposite angles in a cyclic quadrilateral sum to 180° — ABCD is a cyclic quadrilateral, and angle ABC and angle ADC are a pair of opposite angles in it — they don't sit next to each other around the quadrilateral. The theorem that fixes the sum of a cyclic quadrilateral's opposite angles at 180° is exactly the one needed here. The other three theorems don't apply to this figure at all: the angle-in-a-semicircle theorem gives a 90° angle only when one side is a diameter, and no diameter is mentioned; the angle-at-the-centre theorem needs a centre point and an inscribed angle on the same arc, and no centre is given here; the tangent–radius theorem gives a 90° angle only where a tangent line touches the circle, and there is no tangent in this figure. Only the cyclic-quadrilateral theorem fits what is actually drawn.
- (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.
- (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.
- (a) 10 cm — Sector area is (angle ÷ 360) × π × radius². Here 90 ÷ 360 = 1/4, and 1/4 × 3.14 = 0.785, so radius² is 78.5 ÷ 0.785 = 100, and the radius is √100 = 10 cm. Stopping after finding radius² and not taking the square root gives 100 cm. Treating 78.5 as the area of the WHOLE circle, ignoring the 90° fraction, gives radius² = 78.5 ÷ 3.14 = 25, so a radius of 5 cm. Correctly finding a radius of 10 cm but then doubling it, mistaking the question for asking the diameter, gives 20 cm.
- (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.
- (a) 7.7 m — AB is parallel to DC, so those two sides are each parallel to another side. BC and AD are stated to be not parallel to each other, so neither one is parallel to any other side — these are the two sides that need edging. Adding these: 3.2 + 4.5 = 7.7 m, so 7.7 m is correct. 7.6 m comes from an arithmetic slip when adding 3.2 and 4.5. 15.4 m comes from doubling the correct total, mistakenly assuming edging strip is needed along both faces of each side. 4.5 m comes from using only the longer of the two non-parallel sides and forgetting to add the shorter one.
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