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.
Calculator
Geometry and measures worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- 1.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?
- 2.Two straight paths cross at a junction in a park. One of the angles between them, measured with a protractor, is 76°. What is the size of the angle immediately next to it on a straight line?
- 3.In triangle ABC, AB = 8 cm, AC = 7 cm and the area of the triangle is 24 cm². Given that angle BAC is acute, work out the size of angle BAC. Give your answer to 1 decimal place.
- 4.A right-angled triangle has a hypotenuse of 10 cm. One of its other angles is 45°. Work out the exact length of one of the two shorter sides.
- 5.Sam says: '3.2 litres is the same as 320 ml, because you multiply by 100.' Which statement about Sam's claim is correct?
- 6.A cylinder has a volume of 942 cm³ and a height of 12 cm. Using π = 3.14, work out the radius of the cylinder.
- 7.Two identical ladders lean against the same vertical wall from opposite sides, each making an angle of 58.2° with the ground. The two ladders and the ground form a triangle. Work out the size of the angle between the two ladders at the top, where they meet.
- 8.Two bronze statues are geometrically similar and made from the same solid bronze. Their heights are 30 cm and 45 cm. The smaller statue has a mass of 12 kg. Work out the mass of the larger statue, in kg.
- 9.Two hikers walk from a campsite A to a lookout B, a distance of 14 km on a bearing of 038°. From B they walk to a shelter C, a distance of 20 km on a bearing of 142°. Work out the direct distance from A to C. Give your answer to 1 decimal place.
- 10.A rectangle has vertices A(0, 0), B(6, 0), C(6, 4) and D(0, 4). Work out the length of the diagonal AC. Give your answer correct to 2 decimal places.
- 11.In triangle ABC, AB = 9 cm, AC = 6 cm and angle BAC = 110°. Work out the length of BC. Give your answer to 1 decimal place.
- 12.In triangle ABC, AB = 12 cm, AC = 9 cm and angle BAC = 55°. Work out the size of angle ABC. Give your answer to 1 decimal place.
- 13.In triangle ABC, AB = 9 cm, angle BAC = 50° and the area of the triangle is 36 cm². Work out the length of AC. Give your answer to 1 decimal place.
- 14.A cuboid has a rectangular base measuring 6 cm by 8 cm, and a height of 4 cm. Work out the angle between the diagonal from a base corner to the opposite top corner, and the base of the cuboid. Give your answer correct to 1 decimal place.
- 15.In triangle ABC, AB = 7 cm, AC = 6 cm and angle ABC = 40°. Two different lengths of BC are possible. Given that BC is the longer of them, work out the length of BC. Give your answer to 1 decimal place.
Answer key
- (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) 104 — Method: angles on a straight line add up to 180°. Working: 180° − 76° = 104°. A student who answers 76 has mistaken this for the vertically opposite angle, which is equal, rather than the adjacent angle on a straight line. A student who answers 90 has wrongly assumed the two paths must be perpendicular. A student who answers 14 has subtracted 76° from 90° instead of from 180°. Answer: 104°.
- (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) 5√2 cm — Method: the angles of a triangle add to 180°, so the third angle is 45° as well and the two shorter sides are equal. Take one of them as the side opposite a 45° angle and use sin 45° = opposite ÷ hypotenuse. Working: the exact value of sin 45° is √2/2, so the shorter side = 10 × √2 ÷ 2, and half of 10 is 5. Answer: 5√2 cm, which is about 7.07 cm. Remembering sin 45° as √2 rather than as √2 halved gives 10√2 cm, which is longer than the hypotenuse. Halving the hypotenuse because 45° is half of 90° gives 5 cm. Taking the value from the other special triangle, sin 60° = √3/2, gives 5√3 cm.
- (b) Wrong: 1 litre = 1000 ml, so 3.2 l = 3200 ml. — 1 litre = 1000 ml, so 3.2 litres = 3.2 × 1000 = 3200 ml — Sam is wrong because he multiplied by 100 instead of 1000. Saying Sam is correct accepts the wrong multiplier. Saying Sam is wrong only because 3.2 should be rounded first misses the real error, which is the multiplier, not the starting number. Saying '1 litre is 100 ml' misstates the basic fact and blames the wrong part of Sam's working.
- (c) 5 cm — Volume of a cylinder = πr²h, so r² = V ÷ (πh) = 942 ÷ (3.14 × 12) = 942 ÷ 37.68 = 25, and r = √25 = 5 cm. A pupil who finds r² = 25 but forgets to take the square root gives 25 cm. A pupil who forgets to divide by π, using r² = 942 ÷ 12 = 78.5, gets r = √78.5 ≈ 8.9 cm. A pupil who forgets to divide by the height, using r² = 942 ÷ 3.14 = 300, gets r = √300 ≈ 17.3 cm. The correct radius is 5 cm.
- (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) 40.5 kg — Since the statues are made from the same material, mass is proportional to volume, which scales with the cube of the length scale factor. The length scale factor is 45 ÷ 30 = 1.5, so the volume (and mass) scale factor is 1.5³ = 3.375. The larger statue's mass is 12 × 3.375 = 40.5 kg. 18 kg comes from multiplying by the length factor 1.5 directly. 27 kg comes from using the area scale factor 1.5² = 2.25 instead of the volume scale factor. 54 kg comes from treating 'cubed' as 'multiplied by 3', giving 1.5 × 3 = 4.5 instead of 1.5³.
- (b) 21.5 km — Method: find angle ABC from the two bearings, then use the cosine rule. Working: the bearing of A from B is 038° + 180° = 218°, so angle ABC = 218° − 142° = 76°. Then AC² = 14² + 20² − 2 × 14 × 20 × cos 76°, so AC = 21.5 km. Leaving out the factor of 2 in the cosine rule gives AC = 23.0 km; using 142° − 38° = 104° as the angle instead of the correct 76° gives AC = 27.0 km; and simply adding the two distances as if the path were a straight line gives 34.0 km. The angle between the two legs must come from the bearings, not from subtracting them directly.
- (d) 7.21 units — Method: the diagonal AC is the hypotenuse of the right-angled triangle ABC, whose shorter sides are AB and BC, so Pythagoras' theorem gives its length. Working: AB runs from (0, 0) to (6, 0), so AB = 6; BC runs from (6, 0) to (6, 4), so BC = 4. Then AC² = 6² + 4² = 36 + 16 = 52, so AC = √52 = 7.2111…, which is 7.21 correct to 2 decimal places. Answer: 7.21 units. The distractors: 10.00 units comes from adding the two sides, 6 + 4, instead of adding their squares and taking the root; 4.47 units comes from subtracting the squares, √(36 − 16), which is the form of Pythagoras used to find a shorter side rather than the hypotenuse; 26.00 units comes from halving 52 in place of taking its square root.
- (c) 12.4 cm — Using the cosine rule, BC² = AB² + AC² − 2 × AB × AC × cos(A) = 9² + 6² − 2 × 9 × 6 × cos(110°) = 81 + 36 − 108 × cos(110°). Since cos(110°) ≈ −0.34202, 108 × cos(110°) ≈ −36.94, so BC² ≈ 117 + 36.94 = 153.94. Taking the square root, BC ≈ 12.4072, which rounds to 12.4 cm. 8.9 cm comes from treating cos(110°) as if it were positive (using +0.342 instead of −0.342), which wrongly subtracts instead of adds and gives BC² ≈ 80.06. 153.9 cm is BC² itself, rounded, with the square root never taken. 11.6 cm comes from leaving out the factor of 2 in the formula, computing BC² = 81 + 36 − 9 × 6 × cos(110°) ≈ 135.47 instead.
- (b) 47.2° — Method: two sides and the angle between them are given, so find the third side with the cosine rule and then use the sine rule for the angle. Working: BC² = 12² + 9² − 2 × 12 × 9 × cos 55° = 144 + 81 − 216 × 0.57358 = 225 − 123.89 = 101.11, so BC = 10.0552 cm. Angle ABC faces AC = 9 cm and the 55° angle faces BC = 10.0552 cm, so sin ABC = 9 × sin 55° ÷ 10.0552 = 7.3724 ÷ 10.0552 = 0.73318, and the inverse sine of 0.73318 is 47.154°. Since AC is not the longest side, angle ABC is not the largest angle and the acute value is the one that fits. Answer: angle ABC = 47.2° to 1 decimal place. The distractors: 77.8° comes from pairing the angle at B with AB = 12 cm, the side beside it, instead of AC = 9 cm, the side it faces, which actually produces angle ACB; 23.2° comes from a sign slip in the cosine rule step, 225 + 123.89 = 348.89, giving BC = 18.68 cm before the sine rule is applied; 37.9° comes from dividing by AB = 12 cm rather than by the side facing the 55° angle.
- (c) 10.4 cm — Method: rearrange Area = (1/2)ab sin C to make the unknown side the subject: b = 2 × Area ÷ (a × sin C). Working: b = 2 × 36 ÷ (9 × sin 50°) = 10.4 cm (1 d.p.). Answer: 10.4 cm. Forgetting to double the area before dividing gives b = 36 ÷ (9 × sin 50°) = 5.2 cm; using cos 50° instead of sin 50° gives b = 2 × 36 ÷ (9 × cos 50°) = 12.4 cm; and multiplying by sin 50° instead of dividing by it — inverting the rearrangement — gives b = 2 × 36 × sin 50° ÷ 9 = 6.1 cm. Always double the area before dividing, and check whether the unknown should be multiplied or divided by sin C once you've rearranged.
- (a) 21.8° — The base diagonal has length √(6² + 8²) = √(36 + 64) = √100 = 10 cm, using Pythagoras' theorem on the rectangular base. The angle between the space diagonal and the base lies in the right-angled triangle formed by the height (4 cm, opposite), the base diagonal (10 cm, adjacent) and the space diagonal (hypotenuse), so tan(angle) = 4 ÷ 10 = 0.4, giving angle = 21.8° (1 d.p.). Using the height and one base edge instead of the full base diagonal, tan(angle) = 4 ÷ 8 = 0.5, gives 26.6° instead. Inverting the ratio, tan(angle) = 10 ÷ 4 = 2.5, gives 68.2°, the complement of the angle rather than the angle itself. Using the height and the other base edge, tan(angle) = 4 ÷ 6, gives 33.7°.
- (a) 9.3 cm — Method: the 40° angle is not between the two known sides, so call the unknown side x and put it into the cosine rule, which turns into a quadratic equation with two positive roots. Working: AC faces angle ABC, so 6² = 7² + x² − 2 × 7 × x × cos 40°, that is 36 = 49 + x² − 10.7246x, which rearranges to x² − 10.7246x + 13 = 0. The discriminant is 10.7246² − 4 × 13 = 115.02 − 52 = 63.02, whose square root is 7.9384, so x = (10.7246 + 7.9384) ÷ 2 = 9.3315 or x = (10.7246 − 7.9384) ÷ 2 = 1.3931. The longer of the two is wanted. Answer: BC = 9.3 cm to 1 decimal place. The distractors: 1.4 cm is the shorter root, taken by a candidate who solves the quadratic correctly but does not read which of the two lengths is wanted; 4.5 cm comes from treating the 40° as the angle between the two given sides and working out 7² + 6² − 2 × 7 × 6 × cos 40° directly, when 40° lies at B and faces AC; 3.6 cm comes from assuming the triangle is right-angled with AB as the hypotenuse and using the square root of 7² − 6².
Build your own mix at the worksheet builder.