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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Geometry and measures worksheet — GCSE Higher
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- 1.ABCD is a convex quadrilateral field with diagonal AC = 20 m. In triangle ABC, AB = 14 m and angle BAC = 35°. In triangle ACD, AD = 16 m and angle DAC = 40°. Work out the total area of the field. Give your answer to 1 decimal place.
- 2.In triangle ABC, AB = 6.4 cm, BC = 9.1 cm and AC = 12.8 cm. Work out the size of the largest angle in the triangle. Give your answer to 1 decimal place.
- 3.A sector of a circle has radius 15 cm and angle 216°. Using π = 3.14, work out the arc length of the sector.
- 4.A trapezium-shaped garden bed has parallel sides of 2.5 m and 4.5 m, and a perpendicular width of 3 m. A gardener wants to know the area of the bed before ordering topsoil. Work out the area of the garden bed.
- 5.A field ABCD is a convex quadrilateral. AB = 40 m, BC = 32 m and angle ABC = 95°. The other two sides are CD = 25 m and DA = 36 m. Work out the area of the field. Give your answer to the nearest square metre.
- 6.In triangle ABC, AB = 11 cm, AC = 9 cm and the area of the triangle is 45 cm². Given that angle BAC is obtuse, work out the size of angle BAC. Give your answer to 1 decimal place.
- 7.A support strut for a shed roof is in the shape of a right-angled triangle. The two shorter sides of the strut are 5 m and 9 m. Work out the length of the sloping strut (the hypotenuse). Give your answer correct to 1 decimal place.
- 8.A cube has edges of length 5 cm. Work out the length of the diagonal that runs from one corner of the cube to the opposite corner. Give your answer as a surd in its simplest form.
- 9.A textbook contains the instruction 'Draw AB' and, on the next line, the statement 'AB = 5 cm'. Which statement about the notation AB is correct?
- 10.A flagpole is 12 m tall. Freya stands 9 m from the base of the flagpole on level ground. Work out the angle of elevation of the top of the flagpole from where Freya stands. Give your answer correct to 1 decimal place.
- 11.In triangle ABC, AB = 7.4 cm, AC = 5.9 cm and angle BAC = 68°. Work out the length of BC. Give your answer to 1 decimal place.
- 12.A cuboid-shaped shipping crate has a length of 12 m, a width of 5 m, and a volume of 360 m³. Work out the height of the crate, in metres.
- 13.A cheese counter sells cheddar at £8.40 per kilogram. Work out the cost of a 350 g piece of cheddar.
- 14.A boat sails from a harbour on a bearing of 090° for 24 km to a buoy, then changes course and sails on a bearing of 000° for 16 km to reach an island. Work out the direct distance from the harbour to the island. Give your answer correct to 1 decimal place.
- 15.A scale drawing shows a room that is 5 cm wide. The real room is 15 m wide. Write the scale of the drawing in the form 1 : n.
Answer key
- (d) 183.1 m² — Method: the diagonal AC splits the field into two triangles; find each triangle's area with 1/2ab sin C using AC as a side in both, then add the two areas. Working: area of triangle ABC = 1/2 × 14 × 20 × sin 35° = 80.3 m²; area of triangle ACD = 1/2 × 16 × 20 × sin 40° = 102.8 m²; total area = 80.3 + 102.8 = 183.1 m². Ignoring the diagonal AC completely and using AB, AD and the combined angle 35° + 40° = 75° as if it were one triangle gives 108.2 m²; averaging the two triangle areas instead of adding them gives 91.6 m²; and reporting only the area of triangle ABC, forgetting triangle ACD entirely, gives 80.3 m². A diagonal that splits a quadrilateral into two triangles means both areas must be added, using the diagonal as a side of each.
- (d) 110.1° — Method: the largest angle in any triangle faces the longest side, so identify that angle first and then use the cosine rule rearranged as cos A = (b² + c² − a²) ÷ 2bc. Working: the longest side is AC = 12.8 cm, which is faced by angle ABC, and the two sides meeting at B are 6.4 cm and 9.1 cm, so cos ABC = (6.4² + 9.1² − 12.8²) ÷ (2 × 6.4 × 9.1) = (40.96 + 82.81 − 163.84) ÷ 116.48 = −40.07 ÷ 116.48 = −0.34401. A negative cosine means an obtuse angle, and the inverse cosine of −0.34401 is 110.12°. Answer: the largest angle is 110.1° to 1 decimal place. The distractors: 28.0° is the angle facing the shortest side, chosen by a candidate who thinks the largest angle sits opposite the smallest side; 69.9° comes from dropping the minus sign and using cos = 0.34401, which turns the obtuse angle into its supplement; 81.4° comes from assuming the angles share out 180° in the same ratio as the sides, 180 × 12.8 ÷ 28.3, which is not how a triangle behaves.
- (b) 56.52 cm — Arc length = (216 ÷ 360) × 2 × 3.14 × 15 = 0.6 × 94.2 = 56.52 cm. (28.26 cm comes from leaving out the factor of 2, using πr instead of 2πr; 94.2 cm comes from finding the full circumference and forgetting the angle fraction; 113.04 cm comes from using the diameter, 30 cm, instead of the radius.)
- (b) 10.5 m² — The area of a trapezium is half of the sum of the parallel sides, multiplied by the width. Add the parallel sides: 2.5 + 4.5 = 7. Multiply by the width: 7 × 3 = 21. Half of 21 is 10.5 m². 21 m² forgets to halve, using the full (2.5+4.5)×3. 3.5 m² averages the two parallel sides, (2.5+4.5)÷2 = 3.5, but forgets to multiply by the width. 6.75 m² treats the bed as a triangle, using only the longer parallel side: half of 4.5 × 3.
- (a) 1023 m² — Method: split the quadrilateral along the diagonal AC into two triangles. Triangle ABC has two sides and the angle between them, so the cosine rule gives AC and the area formula gives its area; triangle ACD then has three known sides, so the cosine rule gives an angle and the area formula gives its area. Working: AC² = 40² + 32² − 2 × 40 × 32 × cos 95° = 1600 + 1024 + 223.12 = 2847.12, so AC = 53.358 m. The area of triangle ABC is 1/2 × 40 × 32 × sin 95° = 640 × 0.99619 = 637.56 m². In triangle ACD, cos ADC = (25² + 36² − 2847.12) ÷ (2 × 25 × 36) = (1921 − 2847.12) ÷ 1800 = −0.51451, so angle ADC = 120.965° and sin ADC = 0.85748, giving an area of 1/2 × 25 × 36 × 0.85748 = 385.87 m². The total is 637.56 + 385.87 = 1023.43. Answer: the field has an area of 1023 m² to the nearest square metre. The distractors: 1071 m² comes from taking cos 95° as positive, so the diagonal is found as 49.00 m instead of 53.358 m and the second triangle comes out too large; 2047 m² comes from leaving the factor 1/2 out of both area calculations; 638 m² is the area of triangle ABC alone, written down by a candidate who finds the diagonal and then forgets that the second triangle is part of the field.
- (d) 114.6° — Method: rearrange Area = (1/2)ab sin C to make sin C the subject, then use the obtuse branch since the question states the angle is obtuse. Working: sin C = 2 × 45 ÷ (11 × 9) = 0.909, so the acute angle is sin⁻¹(0.909) = 65.4°, and the obtuse angle is 180 − 65.4 = 114.6°. Answer: 114.6°. Giving the acute angle straight from the calculator, 65.4°, ignores that the question asks for the obtuse one; forgetting to double the area before dividing gives sin C = 45 ÷ (11 × 9) = 0.4545, whose obtuse angle is 180 − 27.0 = 153.0°; and that same forgotten-doubling error taken on the acute branch instead gives 27.0°. Always double the area first, and then take 180° minus the calculator's answer whenever the question specifically asks for the obtuse angle.
- (d) 10.3 m — By Pythagoras' theorem, the hypotenuse = √(5² + 9²) = √(25 + 81) = √106 = 10.29...≈ 10.3 m. "106 m" is the value under the square root sign, correct as far as it goes but with the final square root step left out. "14 m" comes from adding the two shorter sides, 5 + 9, instead of using Pythagoras' theorem at all. "10.2 m" comes from rounding 10.29...m down to 10.2 instead of correctly rounding it up to 10.3.
- (c) 5√3 cm — The space diagonal of a cube with edge a satisfies d² = a² + a² + a² = 3a², using Pythagoras' theorem in three dimensions. With a = 5, d² = 3 × 5² = 3 × 25 = 75, so d = √75 = √(25 × 3) = 5√3 cm. Finding the diagonal of one face instead, using only two of the three edges, gives d = √(5² + 5²) = √50 = 5√2 cm, which leaves out the third dimension. Adding the three edges directly, 5 + 5 + 5 = 15 cm, ignores that Pythagoras' theorem works with squares of lengths, not the lengths themselves. Squaring the edge and multiplying by 3 correctly, 3 × 5² = 75, but then forgetting to take the square root, leaves 75 cm — the squared length, not the diagonal itself.
- (c) AB means the segment or its length, shown by context — Method: recall the standard convention for a two-letter label such as AB. Working: AB names the segment AND its length; the sentence around it shows which is meant — 'draw AB' means the segment, 'AB = 5 cm' means the length. Options: 'a different symbol is needed' and 'must be written as |AB|' both describe stricter rules than the convention GCSE actually uses; 'AB can only mean the segment' ignores that the same label is also used for the length. Answer: AB means the segment or its length, shown by context.
- (b) 53.1° — The angle of elevation is opposite the height of the flagpole, 12 m, and adjacent to the distance from its base, 9 m, so tan θ = 12/9 = 1.333..., giving θ = tan⁻¹(1.333...) = 53.13...° ≈ 53.1°. "36.9°" finds the OTHER acute angle of the triangle, 90° − 53.1°, the angle at the top of the flagpole rather than the angle of elevation at Freya's position. "48.6°" comes from wrongly treating 9/12 as a sine ratio and finding sin⁻¹(0.75) = 48.6°, when neither side here is the hypotenuse. "41.4°" comes from wrongly treating 9/12 as a cosine ratio and finding cos⁻¹(0.75) = 41.4°, again without a hypotenuse in the ratio at all.
- (a) 7.5 cm — Method: two sides and the angle between them are given, so use the cosine rule a² = b² + c² − 2bc cos A, with BC as the side facing the 68° angle. Working: BC² = 7.4² + 5.9² − 2 × 7.4 × 5.9 × cos 68° = 54.76 + 34.81 − 87.32 × 0.3746 = 89.57 − 32.71 = 56.86, and the square root of 56.86 is 7.5405. Answer: BC = 7.5 cm to 1 decimal place. The distractors: 11.1 cm comes from adding the last term instead of subtracting it, 89.57 + 32.71 = 122.28, the commonest sign slip on the cosine rule; 56.9 cm is the value of BC² written down as though it were BC, stopping one step before the square root; 2.9 cm comes from pressing sin instead of cos, using 87.32 × sin 68° = 80.96 as the term to subtract.
- (b) 6 m — Volume of a cuboid = length × width × height, so height = volume ÷ (length × width) = 360 ÷ (12 × 5) = 360 ÷ 60 = 6 m. A pupil who divides the volume by the length only gets 360 ÷ 12 = 30 m. A pupil who divides the volume by the width only gets 360 ÷ 5 = 72 m. A pupil who subtracts length × width from the volume instead of dividing gets 360 − 60 = 300 m. The correct height is 6 m.
- (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) 28.8 km — A bearing of 090° is due east and a bearing of 000° is due north, so the two legs of the journey are at right angles to each other, meeting at the buoy. Pythagoras' theorem therefore applies directly, with the direct distance from the harbour to the island as the hypotenuse: distance² = 24² + 16² = 576 + 256 = 832. Taking the square root, distance = √832 = 28.8 km (1 d.p.). Adding the two legs of the journey directly, 24 + 16 = 40 km, treats the route as if it were a straight line, ignoring that the boat actually turns through a right angle partway. Using only the first leg of the journey, 24 km, ignores the second leg entirely. Subtracting the two legs instead of combining them with Pythagoras' theorem, √(24² − 16²) = √(576 − 256) = √320 = 17.9 km (1 d.p.), also gives the wrong distance.
- (c) 1 : 300 — First convert both lengths to the same unit: 15 m = 1500 cm. The scale compares 5 cm on the drawing to 1500 cm in real life, so dividing both parts by 5 gives a scale of 1 : 300. A candidate who compares 5 to 15 without converting units gets 1 : 3. A candidate who converts 15 m to 150 cm, using the wrong conversion factor, gets 1 : 30. A candidate who converts 15 m to 15000 cm, again using the wrong conversion factor, gets 1 : 3000. The scale of the drawing is 1 : 300.
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