Printable · GCSE Higher · ages 14-16
Area of a triangle: ½ab sin C worksheet — GCSE Higher
Fifteen questions on "area of a triangle: ½ab sin c" — DfE statement G23. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
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Area of a triangle: ½ab sin C worksheet — GCSE Higher
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- 1.A garden is made of two triangular flower beds, ABC and ACD, joined along the edge AC. In triangle ABC, angle ABC = 90°, AB = 5 m and BC = 12 m. In triangle ACD, AD = 9 m and angle CAD = 40°, the angle between AC and AD. Work out the area of flower bed ACD. Give your answer to 1 decimal place.
- 2.A triangular sail for a small boat has two sides of 4.2 m and 3.6 m, with an angle of 115° between them. One litre of waterproofing paint covers 3 m² of sail. Work out the least number of whole tins of paint (1 litre each) needed to cover the sail.
- 3.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.
- 4.A regular hexagon has sides of length 8 cm. By splitting the hexagon into 6 identical triangles that meet at its centre, work out the area of the hexagon. Give your answer to 1 decimal place.
- 5.A triangular solar panel has two edges of length 1.8 m and 1.3 m, with an angle of 72° between them. Manufacturing the panel costs £85 per square metre. Work out the total manufacturing cost. Give your answer to the nearest penny.
- 6.A regular octagonal tabletop has centre O. Each vertex is 30 cm from O. Work out the area of the tabletop. Give your answer to the nearest square centimetre.
- 7.A kite ABCD is symmetrical about the diagonal AC, with AB = AD = 6 cm and CB = CD = 9 cm. Angle ABC = 100°. By splitting the kite into two congruent triangles along AC, work out the area of the whole kite. Give your answer to 1 decimal place.
- 8.Triangle 1 has sides of 10 cm and 13 cm with an angle of 64° between them. Triangle 2 has sides of 11 cm and 12 cm with an angle of 70° between them. Which triangle has the greater area, and by how much? Give your answer to 1 decimal place.
- 9.In triangle ABC, AB = 12 cm, AC = 7 cm and the area of the triangle is 33 cm². Given that angle BAC is obtuse, work out the size of angle BAC. Give your answer to 1 decimal place.
- 10.In triangle PQR, PQ = 8 cm and QR = 11 cm. Angle QPR = 50°. A student calculates the area of the triangle as Area = 1/2 × 8 × 11 × sin 50°. Is the student's method correct?
- 11.In triangle ABC, AB = 15.6 cm, AC = 8.9 cm and angle BAC = 112°, the angle between them. Work out the area of triangle ABC. Give your answer to 1 decimal place.
- 12.A stage designer is building a triangular platform ABC for a school play. AB = 11.5 m, AC = 8.2 m and angle BAC = 72°, the angle between them. The flooring for the platform costs £15.80 per square metre. Work out the total cost of the flooring, to the nearest pound.
- 13.In triangle ABC, AB = 8.6 cm, AC = 11.4 cm and angle BAC = 47°. Work out the area of triangle ABC. Give your answer to 1 decimal place.
- 14.A triangular plot of land ABC is to be covered with turf. AB = 23.5 m, AC = 17.2 m and angle BAC = 108°. Turf costs £4.25 per square metre. Work out the cost of the turf for the plot. Give your answer to the nearest pound.
- 15.A parallelogram-shaped tile has two sides of 6 cm and 9 cm, with an angle of 60° between them. Work out the area of the tile. Give your answer to 1 decimal place.
Answer key
- (c) 37.6 m² — Triangle ABC has a right angle at B, so use Pythagoras' theorem to find AC: AC² = AB² + BC² = 5² + 12² = 25 + 144 = 169, so AC = 13 m. In triangle ACD, use Area = 1/2 × AC × AD × sin(angle CAD) = 1/2 × 13 × 9 × sin 40° = 58.5 × 0.6428 = 37.6 m² (1 d.p.). Adding AB and BC to get AC = 17 m instead of applying Pythagoras gives 1/2 × 17 × 9 × sin 40° = 49.2 m². Using cos 40° instead of sin 40° gives 1/2 × 13 × 9 × cos 40° = 44.8 m². Substituting AB = 5 m directly instead of finding AC first gives 1/2 × 5 × 9 × sin 40° = 14.5 m².
- (a) 3 — Method: find the sail's area with Area = (1/2)ab sin C, then divide by how much one litre covers, and round up to a whole number of tins since paint can only be bought by the tin. Working: Area = 1/2 × 4.2 × 3.6 × sin 115° = 6.85 m² (2 d.p.); 6.85 ÷ 3 = 2.28 tins, which rounds up to 3 tins. Answer: 3. Rounding 2.28 to the nearest whole number instead of rounding up gives 2, which is not enough paint to finish the sail; forgetting the 1/2 in the area formula gives an area of 13.7 m², needing 5 tins; and assuming one litre covers exactly 1 m² of sail, ignoring the coverage rate given, rounds the area itself up to 7 tins. Whenever a quantity can only be bought in whole units, round up even when the decimal part is small — 2.28 tins means you need a third tin.
- (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.
- (b) 166.3 cm² — Method: split the regular hexagon into 6 identical triangles meeting at the centre, each with two sides of 8 cm and a 60° angle between them, and use Area = (1/2)ab sin C on just one of them. Working: one triangle's area = 1/2 × 8 × 8 × sin 60° = 27.7 cm² (1 d.p.); the hexagon is 6 of these, so its area is 6 × 27.7 = 166.3 cm² (1 d.p.). Answer: 166.3 cm². Reporting just one triangle's area, without multiplying by 6, gives 27.7 cm²; treating the angle at the centre as a right angle instead of 60°, using 1/2 × 8 × 8 with no sine factor at all, gives 6 × 32 = 192.0 cm²; and multiplying by 5 instead of 6, miscounting the triangles in the hexagon, gives 5 × 27.7 = 138.6 cm². A regular hexagon always splits into exactly 6 triangles at its centre — count them before you multiply.
- (a) £94.58 — Method: find the area with 1/2ab sin C, then multiply by the cost per square metre. Working: area = 1/2 × 1.8 × 1.3 × sin 72° = 1.11274 m², so cost = 1.11274 × £85 = £94.58. Leaving out the 1/2 gives an area of 2.22547 m² and a cost of £189.17; using cos 72° instead of sin 72° gives a cost of £30.73; and rounding the area to 1 decimal place (1.1 m²) before multiplying by the cost per square metre gives £93.50, which loses accuracy that the final answer needs. Even rounding to 1.113 m² is enough to shift the cost to £94.61 — keep the unrounded area in your calculator until the very last step.
- (d) 2546 cm² — Each of the 8 triangles formed by joining O to the vertices is isosceles, with two sides of 30 cm and an angle at O of 360° ÷ 8 = 45°. The area of one triangle is 1/2 × 30 × 30 × sin 45° = 450 × 0.7071 = 318.2 cm². Multiplying by 8 gives the area of the octagon: 318.2 × 8 = 2545.6 cm², which rounds to 2546 cm². Taking the area of a single triangle as the final answer, without multiplying by 8, gives 318 cm². Multiplying by 6 instead of 8, as for a hexagon, gives 318.2 × 6 = 1909 cm². Leaving out the 1/2 from the triangle area formula gives 30 × 30 × sin 45° × 8 = 5091 cm².
- (d) 53.2 cm² — Method: diagonal AC splits the kite into two congruent triangles, ABC and ADC, each with area 1/2 × AB × CB × sin(ABC), so the whole kite has area 2 × 1/2 × AB × CB × sin(ABC) = AB × CB × sin(ABC). Working: kite area = 6 × 9 × sin 100° = 53.2 cm². Reporting just one triangle's area, 1/2 × 6 × 9 × sin 100°, and forgetting to double it for the whole kite gives 26.6 cm²; multiplying the two sides together without any sine term at all gives 54.0 cm²; and doubling the triangle area twice, as if the kite were made of four congruent triangles instead of two, gives 106.4 cm². A kite split by its axis of symmetry always gives exactly two congruent triangles.
- (d) Triangle 2, by 3.6 cm² — Method: find both areas with 1/2ab sin C, then compare them. Working: area of triangle 1 = 1/2 × 10 × 13 × sin 64° = 58.4 cm²; area of triangle 2 = 1/2 × 11 × 12 × sin 70° = 62.0 cm²; triangle 2 is larger, by 62.0 − 58.4 = 3.6 cm². Getting the right difference but naming triangle 1 as the larger one, the subtraction done the wrong way round, gives 'Triangle 1, by 3.6 cm²'; leaving out the 1/2 when finding triangle 1's area (giving 116.8 cm² instead of 58.4 cm²) and then subtracting gives 'Triangle 1, by 54.8 cm²'; and using cos 64° instead of sin 64° for triangle 1 (giving 28.5 cm² instead of 58.4 cm²) gives 'Triangle 2, by 33.5 cm²'. Work out both areas fully and correctly before comparing which is bigger.
- (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°.
- (c) No — 50° is angle P, not between PQ and QR. — Method: 1/2ab sin C only gives the area when C is the angle between the two named sides. Working: PQ and QR meet at vertex Q, so the angle between them is angle PQR, not angle QPR (which is the angle at vertex P, between sides PQ and PR). Since angle QPR is not included between PQ and QR, the student's calculation 1/2 × 8 × 11 × sin 50° does not give the area of the triangle — the verdict is No. Claiming that any angle works with any two sides ignores that the formula is only true for the INCLUDED angle. Claiming the formula needs all three sides is a different, incorrect statement about the formula itself — Area = 1/2ab sin C is valid with exactly two sides and their included angle, when that angle is actually known.
- (c) 64.4 cm² — Method: with two sides and the angle between them, use Area = (1/2)ab sin C. Working: Area = 1/2 × 15.6 × 8.9 × sin 112° = 64.4 cm² (1 d.p.). Answer: 64.4 cm². Leaving out the 1/2 gives 128.7 cm²; using cos 112° instead of sin 112° gives a negative value, which a candidate who drops the minus sign reads as 26.0 cm²; and squaring one side instead of multiplying the two different given sides together gives 112.8 cm². Sin C is never negative for an angle between 0° and 180°, so a negative area is always a sign that cos was used by mistake — check you used sin before you trust your answer.
- (b) £709 — Method: first find the area of the triangular platform with Area = (1/2)ab sin C, then multiply by the cost per square metre. Working: Area = 1/2 × 11.5 × 8.2 × sin 72° = 44.8 m² (1 d.p.); cost = area × £15.80, which rounds to £709 to the nearest pound. Answer: £709. Leaving out the 1/2 in the area formula doubles the area, giving a cost of £1417; using cos 72° instead of sin 72° gives a much smaller area and a cost of £230; and squaring the 11.5 m side instead of multiplying the two different given sides together gives a cost of £994. Find the exact area first — don't round it early — then multiply by the cost per square metre and round only the final answer.
- (a) 35.9 cm² — Method: the area of any triangle is 1/2 × a × b × sin C, where a and b are two sides and C is the angle between them. Working: the 47° angle lies between AB = 8.6 cm and AC = 11.4 cm, so the area is 1/2 × 8.6 × 11.4 × sin 47° = 49.02 × 0.73135 = 35.851. Answer: the area is 35.9 cm² to 1 decimal place. The distractors: 71.7 cm² comes from leaving out the factor 1/2 and working out 8.6 × 11.4 × sin 47°; 33.4 cm² comes from pressing cos instead of sin, 49.02 × cos 47°, which is the same as using the complement 43° in place of 47°; 52.6 cm² comes from pressing tan instead of sin, 49.02 × tan 47°.
- (d) £817 — Method: find the area of the plot with 1/2 × a × b × sin C, then multiply the area by the cost of a square metre. Working: the 108° angle is between AB and AC, so the area is 1/2 × 23.5 × 17.2 × sin 108° = 202.1 × 0.95106 = 192.21 m². The cost is 192.21 × 4.25 = 816.89. Answer: the turf costs £817 to the nearest pound. The distractors: £1634 comes from leaving out the factor 1/2, so the area is taken as 384.42 m²; £859 comes from leaving the sine out and using 202.1 m² as the area, which treats the two sides as a base and a perpendicular height; £192 is the area of the plot written down as though it were the cost, stopping one step short of the question.
- (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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