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.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?
- 2.A triangular flower bed ABC has AB = 14 m, BC = 9 m and AC = 11 m. Work out the area of the flower bed. Give your answer to 1 decimal place.
- 3.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.
- 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 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.
- 6.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.
- 7.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.
- 8.A rhombus-shaped floor tile has sides of length 12 cm and an interior angle of 55°. Work out the area of the tile. Give your answer to 1 decimal place.
- 9.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.
- 10.In triangle ABC, AB = 15 cm, angle BAC = 38° and the area of the triangle is 61 cm². Work out the length of BC. Give your answer to 1 decimal place.
- 11.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.
- 12.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.
- 13.In triangle ABC, AB = AC = 9 cm and angle ABC = angle ACB = 65°. Work out the area of triangle ABC. Give your answer to 1 decimal place.
- 14.In triangle ABC, angle ABC = 58°, angle ACB = 47° and BC = 14 cm. Work out the area of triangle ABC. Give your answer to 1 decimal place.
- 15.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.
Answer key
- (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) 49.5 m² — Method: the area formula needs two sides and the angle between them, and only sides are given, so find one angle with the cosine rule first and then use the two sides that enclose it. Working: angle ABC lies between AB = 14 m and BC = 9 m, so cos ABC = (14² + 9² − 11²) ÷ (2 × 14 × 9) = (196 + 81 − 121) ÷ 252 = 156 ÷ 252 = 0.61905, giving angle ABC = 51.753° and sin ABC = 0.78535. The area is then 1/2 × 14 × 9 × 0.78535 = 63 × 0.78535 = 49.477. Answer: the area is 49.5 m² to 1 decimal place. The distractors: 60.5 m² comes from using the correct angle at B with the sides 14 m and 11 m, which do not both meet at B, so the angle is no longer the one they enclose; 63.0 m² comes from leaving the sine out and treating the two sides as a base and a perpendicular height; 40.5 m² comes from working out angle BAC = 39.98° instead and using its sine with the sides that meet at B, so that the angle used is not the angle between them.
- (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.
- (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.
- (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.
- (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.
- (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.
- (d) 118.0 cm² — Method: a rhombus is made of two congruent triangles either side of a diagonal, each with area 1/2 × 12 × 12 × sin 55°, so the whole rhombus has area 12 × 12 × sin 55° (side² × sin of the interior angle). Working: area = 12² × sin 55° = 118.0 cm². Stopping at one triangle's area, 1/2 × 12² × sin 55°, and forgetting to double it gives 59.0 cm²; using cos 55° instead of sin 55° gives 82.6 cm²; and multiplying the two sides together with no trig term at all gives 144.0 cm². Splitting the rhombus into its two triangles is the safest way to see why the 1/2 disappears from the whole-shape formula.
- (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.
- (b) 9.3 cm — Method: the area formula gives the second side that encloses the 38° angle, and once two sides and the angle between them are known the cosine rule gives the third side. Working: 61 = 1/2 × 15 × AC × sin 38°, so AC = 2 × 61 ÷ (15 × sin 38°) = 122 ÷ 9.2349 = 13.211 cm. Then BC² = 15² + 13.211² − 2 × 15 × 13.211 × cos 38° = 225 + 174.53 − 312.31 = 87.22, and the square root of 87.22 is 9.339. Answer: BC = 9.3 cm to 1 decimal place. The distractors: 10.6 cm comes from forgetting to double the area when rearranging, so AC is taken as 6.606 cm before the cosine rule is applied; 26.7 cm comes from adding the last term of the cosine rule instead of subtracting it, 399.53 + 312.31; 13.2 cm is the length of AC, written down by a candidate who completes the first step and stops there.
- (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) 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.
- (b) 31.0 cm² — Method: Area = 1/2ab sin C needs the angle BETWEEN the two given sides, so first find angle BAC using the angle sum of a triangle. Working: angle BAC = 180° − 65° − 65° = 50°, the angle between AB and AC; Area = 1/2 × 9 × 9 × sin 50° = 31.0 cm². Using the given base angle 65° in place of the included angle 50° gives 36.7 cm²; leaving out the 1/2 altogether gives 62.0 cm²; and using cos 50° instead of sin 50° gives 26.0 cm². The formula only works with the angle that sits between the two sides being multiplied — here that means finding the missing angle first.
- (c) 62.9 cm² — Method: no two sides are given, so first find AC with the sine rule, then find the area using BC, AC and the angle between them, angle ACB. Working: angle BAC = 180° − 58° − 47° = 75°; by the sine rule, AC = 14 × sin 58° / sin 75° = 12.2915 cm; then area = 1/2 × 14 × 12.2915 × sin 47° = 62.9 cm² — keep the unrounded AC, since rounding it to 12.3 cm shifts the area to 63.0 cm². Pairing 14 with sin 75° and dividing by sin 58° instead (the ratio the wrong way round) gives AC = 15.9 cm and an area of 81.6 cm²; using angle BAC = 75° as the included angle instead of angle ACB gives 83.1 cm²; and assuming the triangle is isosceles with AC = BC = 14 cm, skipping the sine rule step entirely, gives 71.7 cm². The angle used in the area formula must be the one between the two sides being multiplied — here that is angle ACB, between BC and AC.
- (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.
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