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 hiker walks on a bearing of 065°. She then turns clockwise through 90° and continues walking in a straight line. What bearing is she now walking on?
- 2.A tent frame forms a triangle ABC, where AB = 3.5 m is a sloping pole, BC = 2.8 m is the base and AC = 4.2 m is the guy rope. Work out the size of angle ABC, between the pole and the base. Give your answer to 1 decimal place.
- 3.Point P has coordinates (6, 4). P is rotated 90° clockwise about the point (1, 2), and the image is then translated by the vector (−3, 5). Work out the coordinates of the final image of P.
- 4.Two straight roads cross at a junction. The angle between them, measured with a protractor, is 118°. What is the size of the angle vertically opposite to it?
- 5.PT is a tangent to a circle with centre O, touching the circle at T. OT is a radius. Angle OPT = 27°, where P is a point outside the circle. Work out the size of angle POT.
- 6.A round cake has a radius of 10 cm. Icing costs £0.02 per square centimetre. One slice of the cake is a sector with an angle of 90°. Using π = 3.14, work out the cost of icing the top of that slice, to the nearest penny.
- 7.A vertical line passes through the point (5, 2). Write down the equation of this line.
- 8.A ship's radar shows a lighthouse at the point (12, −4) on a grid measured in nautical miles. The ship is at (2, 5). The ship sails along the vector that takes it directly to the lighthouse, then sails along that same vector again. Work out the ship's final position.
- 9.A photograph is 10 cm wide and 15 cm tall. It is enlarged to make a similar poster that is 40 cm wide. The poster costs £0.80 per centimetre of its height to print, based on its full height. Work out the cost of printing the poster.
- 10.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.
- 11.A square-based pyramid has a square base whose diagonal is 14 cm, and each slant edge of the pyramid, from a base vertex to the apex, is 15 cm. Work out the height of the pyramid. Give your answer correct to 1 decimal place.
- 12.In triangle ABC, angle ABC = 47°, AC = 12.5 cm and BC = 9.2 cm. Work out the size of angle BAC. Give your answer to 1 decimal place.
- 13.A shopkeeper builds a display from two cuboid boxes, shown in the diagram. She wants to cover the front of the display with coloured paper. Work out the total area of the front elevation, in square centimetres.
- 14.A parallelogram has an area of 54 cm² and a base of 9 cm. Work out its perpendicular height.
- 15.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.
Answer key
- (c) 155° — Turning clockwise adds to the bearing. Starting on a bearing of 065° and turning clockwise through 90° gives 065° + 90° = 155°. A candidate who instead subtracts, working out 90° − 65° = 25°, has performed the wrong operation, giving 025°. A candidate who turns anticlockwise instead of clockwise works out 065° − 90°, which gives a negative number, and adding 360° to fix this gives 335° — the bearing for turning the other way. A candidate who thinks turning does not change the bearing at all keeps the answer as 065°. The new bearing, turning clockwise, is 155°.
- (d) 82.8° — Method: angle ABC is opposite the given side AC, so rearrange the cosine rule to cos B = (AB² + BC² − AC²) / (2 × AB × BC). Working: cos B = (3.5² + 2.8² − 4.2²) / (2 × 3.5 × 2.8), so angle ABC = 82.8°. Forgetting the negative sign in the rearrangement gives the supplementary angle 97.2°; leaving out the factor of 2 in the denominator gives 75.5°; and using AB and AC (the pair either side of angle A, not angle B) gives 41.4°, which is angle BAC, not angle ABC. Always check which angle sits opposite the side you left out of the pair you are dividing by.
- (b) (0, 2) — Method: to rotate about a point that is not the origin, first subtract the centre's coordinates, apply the rotation rule to the shifted point, then add the centre's coordinates back on; only after that do you apply the translation, in the order the question states them. Working: shifting P relative to the centre gives (6 − 1, 4 − 2) = (5, 2); rotating 90° clockwise sends (x, y) to (y, −x), giving (2, −5); adding the centre back on gives (2 + 1, −5 + 2) = (3, −3); applying the translation (−3, 5) gives (3 − 3, −3 + 5) = (0, 2). Answer: (0, 2). Applying the translation BEFORE the rotation, reversing the order the question gives them in, gives (8, 0); stopping after the rotation and forgetting the translation altogether gives (3, −3); and rotating anticlockwise instead of clockwise, using (x, y) → (−y, x), gives (−4, 12). Always carry out the two transformations in the order stated — rotate about the given centre first, then translate — and check each step before moving to the next.
- (a) 118° — When two straight lines cross, the angles that are vertically opposite each other are always equal. So the angle vertically opposite 118° is also 118°. A candidate who instead finds the angle next to it on the straight line, using 180° − 118° = 62°, has found the adjacent angle, not the vertically opposite one. A candidate who answers 180° has confused the rule with angles on a straight line. A candidate who doubles the angle, giving 236°, has applied no valid angle rule at all. The vertically opposite angle is 118°.
- (b) 63° — Method: a tangent meets the radius drawn to the point of contact at a right angle, so triangle OPT has a 90° angle at T; the three angles of the triangle then sum to 180°. Working: angle OTP = 90°, angle OPT = 27°, so angle POT = 180 − 90 − 27 = 63 degrees. Answer: 63°. The tangent-radius angle is a fixed 90°, not something to assume equal to another angle in the triangle, and the three angles of ANY triangle sum to 180°, never 360°: that total belongs to a quadrilateral, not a triangle.
- (b) £1.57 — Slice area = (90 ÷ 360) × 3.14 × 10² = 0.25 × 314 = 78.5 cm². Cost = 78.5 × £0.02 = £1.57. (£0.16 comes from forgetting to square the radius, using 0.25 × 3.14 × 10 for the area; £6.28 comes from finding the area of the whole cake and forgetting the angle fraction; £0.31 comes from using the arc length formula instead of the sector area formula.)
- (b) x = 5 — Every point on a vertical line has the same x-coordinate, so the equation of a vertical line through (5, 2) is x = 5. "y = 5" mixes up the coordinates, using the x-value of 5 to write a y-equation. "y = 2" is the equation of the horizontal line through (5, 2), not the vertical one. "x = 2" uses the correct letter but the wrong coordinate, the y-value of 2 instead of the x-value of 5.
- (b) (22, −13) — The vector from the ship to the lighthouse is (12, −4) − (2, 5) = (10, −9). Sailing along this vector twice from the start gives (2, 5) + 2 × (10, −9) = (2 + 20, 5 − 18) = (22, −13). '(12, −4)' stops after the ship reaches the lighthouse and ignores the second identical leg. '(−18, 23)' comes from finding the vector the wrong way round, as (2, 5) − (12, −4) = (−10, 9), and then doubling that. '(2, 5)' comes from adding the vector and then subtracting it again, wrongly cancelling the two legs instead of adding them.
- (d) £48 — Scale factor = new width ÷ original width = 40 ÷ 10 = 4. Poster height = 15 × 4 = 60 cm. Cost = 60 × £0.80 = £48. (£12 comes from forgetting to scale the height at all, and pricing the original 15 cm height; £15.20 comes from adding the scale factor 4 to the height instead of multiplying by it; £3 comes from dividing the height by the scale factor instead of multiplying by it.)
- (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) 13.3 cm — The apex is directly above the centre of the square base, so the height, half the base diagonal, and a slant edge form a right-angled triangle with the slant edge as the hypotenuse. Half the base diagonal is 14 ÷ 2 = 7 cm. Using Pythagoras' theorem, height = √(15² − 7²) = √(225 − 49) = √176 = 13.3 cm (1 d.p.). Using the slant edge itself as the height, without applying Pythagoras' theorem at all, gives 15 cm. Using the full base diagonal (14 cm) instead of half of it gives √(15² − 14²) = √(225 − 196) = √29 = 5.4 cm (1 d.p.), far too short for a pyramid this size. Adding the two squares instead of subtracting them, √(15² + 7²) = √(225 + 49) = √274 = 16.6 cm (1 d.p.), gives a length longer than the slant edge itself, which cannot be the height.
- (c) 32.6° — Method: an angle is wanted from two sides and the angle facing one of them, so use the sine rule in the form sin A/a = sin B/b. Working: BC = 9.2 cm faces angle BAC and AC = 12.5 cm faces the 47° angle, so sin BAC = 9.2 × sin 47° ÷ 12.5 = 6.7285 ÷ 12.5 = 0.5383, and the inverse sine of 0.5383 is 32.566°. Because BC is shorter than AC, angle BAC must be smaller than 47°, so the acute value is the only one that fits. Answer: angle BAC = 32.6° to 1 decimal place. The distractors: 147.4° comes from taking 180° − 32.566°, the obtuse angle with the same sine, without checking it — 147.4° and 47° already add to more than 180°, so no such triangle exists; 83.6° comes from putting the sides the wrong way up, 12.5 × sin 47° ÷ 9.2, which gives a sine of 0.9937; 57.4° comes from pressing the inverse cosine key on 0.5383 instead of the inverse sine key.
- (a) 4700 cm² — Method: split the T-shaped outline into the two rectangles it is made from, find the area of each, then add them together. Working: the wide base gives a rectangle 90 cm × 30 cm = 2700 cm²; the narrower block on top gives a rectangle 40 cm × 50 cm = 2000 cm²; adding these, 2700 + 2000 = 4700 cm². Answer: 4700 cm². The distractors: 2700 cm² comes from finding only the area of the base rectangle and forgetting to add the block on top. 2000 cm² comes from finding only the area of the top block and forgetting the base. 7200 cm² comes from treating the whole outline as one large rectangle, 90 cm wide by (30 + 50) = 80 cm tall, instead of splitting it into the two separate rectangles that actually make up the shape.
- (c) 6 cm — For a parallelogram, area = base × height, so height = area ÷ base = 54 ÷ 9 = 6 cm. 12 cm comes from using the triangle's reverse formula, height = 2 × area ÷ base, which does not apply to a parallelogram. 45 cm comes from subtracting the base from the area, 54 − 9, instead of dividing. 486 cm comes from multiplying the area by the base, 54 × 9, instead of dividing.
- (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.
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