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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Answer key: Area of a triangle: ½ab sin C worksheet — GCSE Higher
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- (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.
- (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.
- (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) 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°.
- (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) 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) 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.
- (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.
- (c) $12\sqrt{3}$ cm² — Use Area = 1/2 × AB × AC × sin(angle BAC) = 1/2 × 6 × 8 × sin 60°. Since sin 60° = √3⁄2, this is 1/2 × 6 × 8 × √3⁄2 = 24 × √3⁄2 = 12√3 cm². Using AB × AB instead of AB × AC (taking 6 × 6 rather than 6 × 8) gives 1/2 × 6 × 6 × sin 60° = 18 × √3⁄2 = 9√3 cm². Using AC × AC instead of AB × AC (taking 8 × 8 rather than 6 × 8) gives 1/2 × 8 × 8 × sin 60° = 32 × √3⁄2 = 16√3 cm². Leaving out the 1/2 from the area formula gives 6 × 8 × sin 60° = 48 × √3⁄2 = 24√3 cm².
- (c) 152.2 cm² — Method: find the area of one of the 5 isosceles triangles with 1/2ab sin C, then multiply by 5 for the whole pentagon. Working: one triangle has area 1/2 × 8 × 8 × sin 72° = 30.4338 cm², so the pentagon's area = 5 × 30.4338 = 152.2 cm² — keep the unrounded triangle area, since 5 × 30.4 would give 152.0. Reporting the area of a single triangle and forgetting to multiply by 5 gives 30.4 cm²; multiplying by 6 instead of 5, as for a hexagon, gives 182.6 cm²; and leaving out the 1/2 from each triangle before multiplying by 5 gives 304.3 cm². A regular pentagon splits into exactly 5 triangles at its centre, each with a 72° angle, since 360° ÷ 5 = 72°.
- (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.
- (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.
- (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.
- (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.
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