Printable · GCSE Higher · ages 14-16
Geometry and measures worksheet — GCSE Higher
Fifteen questions across the geometry and measures statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Geometry and measures worksheet — GCSE Higher
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- 1.The angle between north and a footpath is 35°, measured clockwise from north. What is the three-figure bearing of the footpath?
- 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.Two cylindrical storage tins are geometrically similar. The smaller tin has height 4 cm and the larger tin has height 6 cm. Work out the ratio of the surface area of the smaller tin to the surface area of the larger tin, giving your answer in the form a : b in its simplest form.
- 4.In triangle ABC, AB = 10 cm, BC = 7 cm and angle ACB = 35°. Given that angle BAC is acute, work out the length of AC. Give your answer to 1 decimal place.
- 5.A circle has a diameter of 24 cm. A sector of this circle has an angle of 90°. Using π = 3.14, work out the area of the sector.
- 6.A ladder of length 7 m leans against a vertical wall. The foot of the ladder is 3 m from the base of the wall. Work out how high up the wall the ladder reaches. Give your answer correct to 1 decimal place.
- 7.A student is asked to rearrange the cosine rule a² = b² + c² − 2bc cos A to make cos A the subject. They write: cos A = (a² − b² − c²) / (2bc). Is the student's rearrangement correct?
- 8.Loose sweets are sold at £1.20 per 100 g. Work out the cost of 350 g of sweets.
- 9.Two hikers walk from a campsite A to a lookout B, a distance of 14 km on a bearing of 038°. From B they walk to a shelter C, a distance of 20 km on a bearing of 142°. Work out the direct distance from A to C. Give your answer to 1 decimal place.
- 10.The bearing of a harbour B from a ferry's position A is 260°. What is the bearing of A from B?
- 11.Two coastguard stations A and B are 18 km apart, with B due east of A. A boat at C is on a bearing of 062° from A and on a bearing of 315° from B. Work out the distance of the boat from station B. Give your answer to 1 decimal place.
- 12.Two similar hexagonal tiles have lengths in the ratio 4 : 7. A side of the smaller tile is 8.4 cm long. Work out the length of the corresponding side of the larger tile.
- 13.In triangle ABC, angle ABC = 72° and angle ACB = 48°. Side AB = (2x + 1) cm and side AC = (3x − 2) cm. Work out the value of x. Give your answer to 1 decimal place.
- 14.A parallelogram has an area of 136 cm² and a base of 17 cm. Work out the perpendicular height of the parallelogram.
- 15.A sprinkler waters a sector-shaped patch of a garden with radius 8 m and angle 90°. Work out the perimeter of the watered sector, to 1 decimal place. (Use π = 3.14.)
Answer key
- (a) 035° — A bearing is measured clockwise from north, so an angle of 35° clockwise from north is a bearing of 035° (written with three figures). Choosing 325° measures the angle anticlockwise instead of clockwise (360 − 35 = 325). Choosing 215° adds 180° to the angle, mixing this up with a back-bearing calculation (35 + 180 = 215). Choosing 350° reorders the digits of 035, writing the ones digit before the tens digit by mistake.
- (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.
- (b) 4 : 9 — The heights are in the ratio 4 : 6, which simplifies to 2 : 3. For similar solids, area scales with the square of the length ratio, so the surface area ratio is 2² : 3² = 4 : 9. 2 : 3 is just the simplified length ratio, before squaring has been done. 8 : 27 comes from cubing the ratio (2³ : 3³) instead of squaring it — that's the rule for volumes, not areas. 9 : 4 has the correct squared values but in the wrong order, giving the larger tin's area first instead of the smaller.
- (d) 14.9 cm — By the sine rule, sin(BAC) ÷ BC = sin(ACB) ÷ AB, so sin(BAC) = 7 × sin35° ÷ 10 = 0.4015. This gives angle BAC = 23.7° or angle BAC = 156.3°; since angle BAC is acute, angle BAC = 23.7°. Angle ABC = 180° − 35° − 23.7° = 121.3°. By the sine rule again, AC ÷ sin(ABC) = AB ÷ sin(ACB), so AC = 10 × sin121.3° ÷ sin35° = 14.9 cm. Using angle BAC in the final step instead of angle ABC gives AC = 10 × sin23.7° ÷ sin35° = 7.0 cm — this just recomputes the given side BC. Using angle ACB again in the final step, instead of the newly found angle ABC, gives AC = 10 × sin35° ÷ sin35° = 10.0 cm — this just returns AB unchanged. Making a sign slip when finding angle ABC, computing 180° − 35° + 23.7° = 168.7° instead of subtracting it, and carrying that through gives AC = 10 × sin168.7° ÷ sin35° = 3.4 cm. Find angle BAC, then angle ABC, then apply the sine rule once more, and AC comes to 14.9 cm.
- (b) 113.04 cm² — Sector area is (angle ÷ 360) × π × radius², and radius means the RADIUS, not the diameter: here the diameter is 24 cm, so the radius is 12 cm. The fraction is 90 ÷ 360 = 1/4, and 12² = 144, so the area is 0.25 × 3.14 × 144 = 113.04 cm². Using the diameter itself as if it were the radius gives 0.25 × 3.14 × 576 = 452.16 cm². Using the arc-length formula, 2 × π × radius, instead of the area formula gives 0.25 × 2 × 3.14 × 12 = 18.84 cm². Using the radius instead of its square gives 0.25 × 3.14 × 12 = 9.42 cm².
- (c) 6.3 m — By Pythagoras' theorem, the height = √(7² − 3²) = √(49 − 9) = √40 = 6.32...≈ 6.3 m. "6.4 m" rounds 6.32...m up to 6.4 instead of correctly rounding it down to 6.3. "4.0 m" comes from subtracting the two given lengths directly, 7 − 3 = 4, instead of subtracting their squares. "10.0 m" comes from adding the two given lengths, 7 + 3 = 10, instead of using Pythagoras' theorem at all.
- (b) No — should divide by −2bc, not +2bc; sign is wrong. — Method: rearrange a² = b² + c² − 2bc cos A step by step and compare with the student's version. Working: subtracting b² + c² from both sides gives a² − b² − c² = −2bc cos A, then dividing both sides by −2bc gives cos A = (a² − b² − c²) / (−2bc), which is the same as cos A = (b² + c² − a²) / (2bc). The student divided by +2bc instead of −2bc, so their expression is the negative of the correct one — the verdict is No. Saying the algebra is fine because 'either sign order gives a valid result' ignores that only one of the two signed expressions matches the original equation. Saying the denominator should be bc rather than 2bc is a different, unrelated error — the coefficient 2bc in the original formula is correct and must stay.
- (a) £4.20 — 350 g is 3.5 lots of 100 g, since 350 ÷ 100 = 3.5, so the cost is 3.5 × £1.20 = £4.20. Multiplying the mass in grams directly by the price, without dividing by 100 first, gives £420.00. Working out 100 ÷ 350 instead of 350 ÷ 100 inverts the ratio and gives about £0.34. Rounding 350 g down to 300 g gives 3 × £1.20 = £3.60.
- (b) 21.5 km — Method: find angle ABC from the two bearings, then use the cosine rule. Working: the bearing of A from B is 038° + 180° = 218°, so angle ABC = 218° − 142° = 76°. Then AC² = 14² + 20² − 2 × 14 × 20 × cos 76°, so AC = 21.5 km. Leaving out the factor of 2 in the cosine rule gives AC = 23.0 km; using 142° − 38° = 104° as the angle instead of the correct 76° gives AC = 27.0 km; and simply adding the two distances as if the path were a straight line gives 34.0 km. The angle between the two legs must come from the bearings, not from subtracting them directly.
- (a) 080° — The back bearing differs from the given bearing by 180°. Because 260° is greater than 180°, subtract 180°: 260 − 180 = 80°, so the bearing of A from B is 080°. Choosing 440° adds 180° instead of subtracting it, even though the result would be more than a full turn (260 + 180 = 440). Choosing 180° assumes the back bearing is always exactly 180°, ignoring the original bearing altogether. Choosing 100° comes from measuring the reflex angle the other way round the circle (360 − 260 = 100) instead of applying the 180° back-bearing rule.
- (d) 8.8 km — Method: turn each bearing into an angle of triangle ABC, find the third angle from the angle sum, then use the sine rule. Working: B is due east of A, so AB itself lies on a bearing of 090°, and the angle at A between AB and AC is 090° − 062° = 28°. From B, station A lies due west on a bearing of 270°, and C lies on 315°, so the angle at B is 315° − 270° = 45°. The third angle is 180° − 28° − 45° = 107°. The side BC faces the 28° angle and AB = 18 km faces the 107° angle, so BC = 18 × sin 28° ÷ sin 107° = 8.4505 ÷ 0.95630 = 8.8366. Answer: the boat is 8.8 km from B, to 1 decimal place. The distractors: 13.3 km comes from pairing BC with the 45° angle at B instead of the 28° angle it faces, which gives the distance AC; 12.0 km comes from never working out the third angle and dividing by sin 45° instead of sin 107°; 16.6 km comes from using the bearing 062° itself as the angle at A, instead of the 28° between AC and AB.
- (c) 14.7 cm — The scale factor from the smaller tile to the larger tile is 7 ÷ 4 = 1.75, so the larger side is 8.4 × 1.75 = 14.7 cm. '4.8 cm' comes from scaling by 4 ÷ 7 instead, using the ratio the wrong way round. '11.4 cm' comes from adding the difference between the ratio numbers, 7 − 4 = 3, onto 8.4, instead of scaling. '58.8 cm' comes from multiplying by 7 on its own, using a ratio number as the scale factor instead of working out 7 ÷ 4 = 1.75 first.
- (b) 7.4 — Method: AB is opposite angle ACB and AC is opposite angle ABC, so the sine rule gives AB/sin(ACB) = AC/sin(ABC). Working: (2x + 1)/sin 48° = (3x − 2)/sin 72°; cross-multiplying and collecting the x terms gives x = 7.4. Using angle BAC = 180° − 72° − 48° = 60° in place of angle ABC in the ratio gives x = 4.7; ignoring the sine rule altogether and solving 2x + 1 = 3x − 2 as if the two sides were simply equal gives x = 3.0; and pairing each side with the wrong angle — AB with sin 72° and AC with sin 48° — gives x = 1.9. Each side in the sine rule must be paired with the sine of the angle directly opposite it.
- (b) 8 cm — Area = base × height, so height = area ÷ base = 136 ÷ 17 = 8 cm. (16 cm comes from using the triangle formula, thinking area = 1/2 × base × height and so height = 2 × area ÷ base; 2312 cm comes from multiplying the area by the base instead of dividing; 119 cm comes from subtracting the base from the area instead of dividing the area by the base.)
- (a) 28.6 m — The perimeter of a sector is the two straight radii plus the curved arc. This sector's 90° angle is one quarter of a full turn, so its arc length is one quarter of the full circle's circumference. The full circumference is 2 × 3.14 × 8 = 50.24 m, and one quarter of that is 50.24 ÷ 4 = 12.56 m. Add the two 8 m radii: 12.56 + 8 + 8 = 28.56 m, which rounds to 28.6 m. Choosing 12.6 m gives the arc length alone (rounded), forgetting the two straight edges of the sector. Choosing 20.6 m adds only one radius to the arc length instead of two, missing one of the two straight sides. Choosing 41.1 m comes from using the diameter, 16 m, as if it were the radius when working out the arc length (2 × 3.14 × 16 = 100.48, one quarter of which is 25.12), then adding the two correct 8 m radii (25.12 + 8 + 8 = 41.12).
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