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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 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.
- 2.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.
- 3.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.
- 4.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.
- 5.In triangle ABC, AB = 9.4 cm, AC = 7.2 cm and angle BAC = 63°, the angle between them. Work out the area of triangle ABC. Give your answer to 1 decimal place.
- 6.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.
- 7.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.
- 8.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.
- 9.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.
- 10.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.
- 11.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.
- 12.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.
- 13.Which of these statements about the area formula Area = 1/2ab sin C is correct?
- 14.A circle has centre O and radius 10 cm. Points A and B lie on the circle such that angle AOB = 130°. Work out the area of the minor segment cut off by the chord AB. Give your answer to 1 decimal place.
- 15.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.
Answer key
- (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) 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) 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.
- (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.
- (d) 30.2 cm² — Method: with two sides and the angle between them, use Area = (1/2)ab sin C. Working: Area = 1/2 × 9.4 × 7.2 × sin 63° = 30.2 cm² (1 d.p.). Answer: 30.2 cm². Leaving out the 1/2 altogether gives 60.3 cm²; using cos 63° instead of sin 63° gives 15.4 cm²; and squaring one side instead of multiplying the two different given sides together gives 39.4 cm². Always check you are using sin, not cos, and that the 1/2 is there before you multiply the two given sides together.
- (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.
- (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.
- (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.
- (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) 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.
- (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.
- (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.
- (c) AB, AC and angle BAC: Area = 1/2 × AB × AC × sin(BAC) — Method: Area = 1/2ab sin C only works when the angle used is the one included between the two sides being multiplied. Working: AB and AC meet at A, and angle BAC is the angle at A between them, so the statement pairing AB, AC and angle BAC is the correct one. The statement that three sides with no angle can still go into 1/2 AB × AC × sin(BAC) is wrong: with no angle known, sin(BAC) cannot be evaluated, so a different method must find an angle first. The statement pairing AB and BC with sin(BAC) is wrong: AB and BC meet at B, so the angle between them is angle ABC, not angle BAC — it names the wrong angle for the sides it uses. The statement pairing AB and AC with sin(ABC) is wrong for the same reason: AB and AC meet at A, so their included angle is angle BAC, and angle ABC is not between them at all.
- (b) 75.1 cm² — Method: a segment is a sector with its triangle cut away, so find the sector area and the triangle area (using Area = (1/2)r² sin C on the two radii) and subtract. Working: sector area = (130 ÷ 360) × π × 10² = 113.4 cm² (1 d.p.); triangle area = 1/2 × 10 × 10 × sin 130° = 38.3 cm² (1 d.p.); segment area = 113.4 − 38.3 = 75.1 cm². Answer: 75.1 cm². Reporting the sector area on its own, without subtracting the triangle, gives 113.4 cm²; reporting the triangle area on its own gives 38.3 cm²; and using the reflex angle, 360° − 130° = 230°, in the sector but still subtracting the triangle gives 162.4 cm², which is neither segment — the major segment would be the 230° sector PLUS the triangle, 239.0 cm². The minor segment is the smaller piece, cut off by the shorter arc, so use the angle actually given, 130°, and subtract the triangle from that sector.
- (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.
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