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.A line passes through the points (1, 2) and (3, 6). The line is reflected in the x-axis, and the image is then translated by the vector (2, 0). Work out the gradient of the image line after both transformations.
- 2.Loose sweets are sold at £1.20 per 100 g. Work out the cost of 350 g of sweets.
- 3.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.
- 4.In triangle ABC, AB = 6 cm, angle ABC = 40° and AC = 9 cm. Using the sine rule, sin(ACB) = AB × sin(ABC) ÷ AC, which gives two possible values for angle ACB: 25.4° or 154.6° (each to 1 decimal place). By considering the angle sum of the triangle, decide which of these two values angle ACB can actually take, and hence work out angle BAC. Give your answer to 1 decimal place.
- 5.A cylindrical water tank has a solid hemispherical dome on top, both with radius 2 m. Work out the volume of the hemispherical dome alone. Use π = 3.14 and give your answer correct to 3 decimal places.
- 6.A running track has a straight section and a semicircular bend. The bend is a semicircle with radius 32 m. Work out the length of the curved part of the bend, to 1 decimal place. (Use π = 3.14.)
- 7.A sector of a circle has radius 10 cm and angle 150°. Using π = 3.14, work out the perimeter of the sector, to 1 decimal place.
- 8.A ship leaves port P and sails 24 km on a bearing of 040° to a buoy Q. It then sails 31 km on a bearing of 115° to a lighthouse R. Work out the direct distance from P to R. Give your answer to 1 decimal place.
- 9.A cuboid has length 3 cm, width 4 cm and height 12 cm. Work out the length of the diagonal that runs from one corner of the cuboid to the opposite corner.
- 10.A window is made from a rectangle with a semicircle on top. The rectangle is 40 cm wide and 60 cm tall. The semicircle has the same width as the rectangle. Work out the total area of the window. Use π = 3.14. Give your answer to the nearest whole number.
- 11.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.
- 12.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.
- 13.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?
- 14.In a phone game, a character starts at the point (−5, 2) on a grid. It moves by the vector to collect a coin. The player now wants the character's next single move to finish at the point (3, 0). Write down the column vector of that second move.
- 15.Two traffic cones are geometrically similar. The smaller cone has height 6 m and the larger cone has height 9 m. Work out the ratio of the volume of the smaller cone to the volume of the larger cone, giving your answer in the form a : b in its simplest form.
Answer key
- (b) −2 — The gradient of the original line is (6 − 2) ÷ (3 − 1) = 4 ÷ 2 = 2. Reflecting in the x-axis sends every y-coordinate to its negative, which flips the sign of the gradient: the image line has gradient −2. Translating by (2, 0) is a horizontal shift, which does not change the line's steepness or direction at all, so the gradient stays at −2. Assuming the gradient is unaffected by the reflection gives 2, the original gradient carried straight through. Thinking a reflection in the x-axis turns a gradient into its positive reciprocal gives 1/2. Combining that same wrong idea with the sign flip from the reflection gives −1/2. Only the sign flips, from the reflection, and translating never changes a gradient at all, so the answer is −2.
- (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) 114.6° — Using the sine rule, sin(ACB) = AB × sin(ABC) ÷ AC = 6 × sin(40°) ÷ 9. Since sin(40°) ≈ 0.64279, this gives sin(ACB) ≈ 3.8567 ÷ 9 ≈ 0.42852, so angle ACB ≈ 25.4° or its supplement, 154.6°. Testing the obtuse candidate: 40° + 154.6° = 194.6°, which already exceeds 180°, so angle BAC would have to be negative — impossible, so 154.6° is rejected. With angle ACB ≈ 25.4°, angle BAC = 180° − 40° − 25.4° = 114.6°. 25.4° is angle ACB, not angle BAC that the question asks for. 14.6° comes from using the invalid 154.6° candidate anyway and then wrongly turning the resulting negative angle sum (−14.6°) positive instead of rejecting it. 65.4° comes from inverting the sine rule ratio — dividing AC × sin(ABC) by AB instead of AB × sin(ABC) by AC — which gives a different, incorrect candidate for angle ACB entirely.
- (b) 16.747 m³ — Volume of a sphere = (4/3)πr³, so a hemisphere is half that: (2/3)πr³. Substitute r = 2: (2/3) × 3.14 × 2³ = (2/3) × 3.14 × 8 = (2/3) × 25.12 = 16.7467 m³, which rounds to 16.747 m³.
- (c) 100.5 m — The curved part of a semicircular bend is half of a full circle's circumference. The full circumference would be 2 × 3.14 × 32 = 200.96 m, and half of that is 200.96 ÷ 2 = 100.48 m, which rounds to 100.5 m. Choosing 201.0 m uses the FULL circumference, forgetting to halve it for a semicircle. Choosing 50.2 m halves the radius as well as taking a semicircle, using 3.14 × 16 = 50.24 m instead of the correct radius of 32 m. Choosing 64.0 m uses the diameter, 2 × 32 = 64, as if it were the curved length, ignoring π and the semicircle shape entirely.
- (d) 46.2 cm — The perimeter of a sector is the arc length plus its two straight radii. The circumference is 2 × 3.14 × 10 = 62.8 cm, and the arc is 150 ÷ 360 of that: 62.8 × 150 ÷ 360 = 26.2 cm (1 d.p.). Adding the two radii, 26.2 + 10 + 10 = 46.2 cm. Giving just the arc length, without adding the straight edges, gives 26.2 cm. Adding only ONE radius instead of two gives 36.2 cm. Using 150 ÷ 180 instead of 150 ÷ 360 for the fraction gives an arc of 52.3 cm and a perimeter of 72.3 cm.
- (b) 43.8 km — Method: find the angle of the triangle at Q from the two bearings, then use the cosine rule on the two known sides and that included angle. Working: the bearing of P from Q is the back bearing of 040°, which is 040° + 180° = 220°, and the bearing of R from Q is 115°, so angle PQR = 220° − 115° = 105°. Then PR² = 24² + 31² − 2 × 24 × 31 × cos 105° = 576 + 961 − 1488 × (−0.25882) = 1537 + 385.12 = 1922.12, and the square root of 1922.12 is 43.842. Answer: PR = 43.8 km to 1 decimal place. The distractors: 33.9 km comes from subtracting the bearings, 115° − 040° = 75°, and using 75° as the angle at Q, which is the angle the ship turns through, not the angle inside the triangle; 19.9 km comes from using the first bearing itself, 040°, as the angle at Q; 55.0 km comes from adding the two legs, 24 + 31, which would be the distance only if the ship had sailed in a straight line.
- (d) 13 cm — Use Pythagoras' theorem in three dimensions: for a cuboid with edges a, b and c the space diagonal d satisfies d² = a² + b² + c². Substitute a = 3, b = 4, c = 12: d² = 3² + 4² + 12² = 9 + 16 + 144 = 169. Take the square root: d = √169 = 13 cm. Adding the three edges directly, 3 + 4 + 12 = 19 cm, ignores that Pythagoras' theorem is about squares, not lengths, and gives 19 cm. Stopping after squaring and adding, without taking the square root, leaves 169 cm — the squared length, not the length itself. Using only the 4 cm and 12 cm edges finds the diagonal of one face, √(4² + 12²) = √160 = 12.6 cm (1 d.p.), and leaves out the third dimension entirely.
- (c) 3028 cm² — Split the window into a rectangle and a semicircle. Rectangle area = 40 × 60 = 2400 cm². The semicircle has radius 40 ÷ 2 = 20 cm, so its area = 0.5 × 3.14 × 20² = 628 cm². Total area = 2400 + 628 = 3028 cm². A student who uses a full circle instead of a semicircle on top of the rectangle gets 3.14 × 20² = 1256 cm² for the circle, plus 2400 cm² for the rectangle, totalling 3656 cm². A student who uses the rectangle's full width (40 cm) as the radius instead of halving it gets 0.5 × 3.14 × 40² = 2512 cm² for the semicircle, plus 2400 cm² for the rectangle, totalling 4912 cm².
- (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) 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.
- (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.
- (c) $\binom{2}{7}$ — First find the character's position after the first move: (−5 + 6, 2 + (−9)) = (1, −7). The second move takes it from (1, −7) to (3, 0), so subtract the current position from the target: (3 − 1, 0 − (−7)) = $\binom{2}{7}$. $\binom{8}{−2}$ works from the starting point (−5, 2) and ignores the first move — (3 − (−5), 0 − 2) = $\binom{8}{−2}$. $\binom{−2}{−7}$ subtracts the wrong way round, current position minus target, instead of target minus current position — (1 − 3, −7 − 0) = $\binom{−2}{−7}$. $\binom{4}{−7}$ adds the target's coordinates to the current position instead of subtracting — (1 + 3, −7 + 0) = $\binom{4}{−7}$.
- (b) 8 : 27 — The heights are in the ratio 6 : 9, which simplifies to 2 : 3. For similar solids, volume scales with the cube of the length ratio, so the volume ratio is 2³ : 3³ = 8 : 27. 2 : 3 is only the simplified length ratio, without cubing. 4 : 9 comes from squaring instead of cubing — that's the rule for areas, not volumes. 27 : 8 has the correct cubed values but in the wrong order, giving the larger cone's volume first instead of the smaller.
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