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.
Higher onlyCalculator
Answer key: Area of a triangle: ½ab sin C worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- (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.
- (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) 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.
- (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°.
- (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.
- (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) 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) 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) 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) 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) 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) 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) 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) 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.
Build your own mix at the worksheet builder.