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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- (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.
- (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) 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.
- (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.
- (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.
- (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.
- (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) 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) $12\sqrt{3}$ 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.
- (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.
- (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.
- (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.
- (a) 10.1 cm — Method: the area formula 1/2 × a × b × sin C contains the unknown side, so substitute what is known and rearrange. Working: 42 = 1/2 × 9.5 × AC × sin 61°. Multiplying both sides by 2 gives 84 = 9.5 × AC × sin 61°, and 9.5 × sin 61° = 9.5 × 0.87462 = 8.3089, so AC = 84 ÷ 8.3089 = 10.1096. Answer: AC = 10.1 cm to 1 decimal place. The distractors: 5.1 cm comes from forgetting to double the area when clearing the factor 1/2 and working out 42 ÷ 8.3089; 18.2 cm comes from using cos 61° in place of sin 61° in the denominator; 7.7 cm comes from multiplying by sin 61° instead of dividing by it, 84 × sin 61° ÷ 9.5, the standard slip when the unknown is inside a product.
- (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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