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.Sam says: '3.2 litres is the same as 320 ml, because you multiply by 100.' Which statement about Sam's claim is correct?
- 2.A quadrilateral has both pairs of opposite sides parallel and equal in length, but none of its angles are right angles. Which statement about this quadrilateral is correct?
- 3.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.)
- 4.In triangle ABC, AB = 9 cm, BC = 6 cm and angle BAC = 35°. This description fits two different triangles. Work out the two possible sizes of angle ACB, each to 1 decimal place.
- 5.A game designer places a coin at (5, 3) on a grid. The game rotates the coin 90° clockwise about the point (2, 3) each time the player presses a button, and this is applied twice in a row. Work out the coordinates of the coin after the button is pressed twice.
- 6.The top of a clock tower is 25 m above level ground. Oliver stands on the ground 25 m from the foot of the tower. Work out the angle of elevation of the top of the tower from the point where Oliver stands.
- 7.A furniture catalogue drawing uses a scale of 1 cm to 1.5 m. A sofa in the drawing is measured as 7 cm long. What is the real length of the sofa, in metres?
- 8.A cuboid has a rectangular base measuring 6 cm by 8 cm, and a height of 4 cm. Work out the angle between the diagonal from a base corner to the opposite top corner, and the base of the cuboid. Give your answer correct to 1 decimal place.
- 9.A triangular prism has a cross-section that is a triangle with a base of 3.5 cm and a perpendicular height of 4 cm. The prism is 10 cm long. Work out the volume of the prism.
- 10.A shop stacks tins in a display block. The front elevation of the block is 4 tins wide and 3 tins high. The side elevation shows the block is 2 tins deep. Every position in the block is filled. How many tins are used in total?
- 11.A raised garden bed is a prism whose cross-section is a triangle with a base of 1.2 m and a perpendicular height of 0.8 m. The bed is 3 m long. Work out the volume of soil needed to fill it, in m³.
- 12.A sector of a circle has radius 6 cm and angle 150°. Using π = 3.14, work out the perimeter of the sector.
- 13.The bearing of a harbour B from a ferry's position A is 260°. What is the bearing of A from B?
- 14.A support strut for a shed roof is in the shape of a right-angled triangle. The two shorter sides of the strut are 5 m and 9 m. Work out the length of the sloping strut (the hypotenuse). Give your answer correct to 1 decimal place.
- 15.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.
Answer key
- (b) Wrong: 1 litre = 1000 ml, so 3.2 l = 3200 ml. — 1 litre = 1000 ml, so 3.2 litres = 3.2 × 1000 = 3200 ml — Sam is wrong because he multiplied by 100 instead of 1000. Saying Sam is correct accepts the wrong multiplier. Saying Sam is wrong only because 3.2 should be rounded first misses the real error, which is the multiplier, not the starting number. Saying '1 litre is 100 ml' misstates the basic fact and blames the wrong part of Sam's working.
- (c) It is a parallelogram but not a rectangle — Method: match the given properties to the definition of each named quadrilateral. Working: having both pairs of opposite sides parallel and equal in length is exactly the definition of a parallelogram; since none of the angles are right angles, it cannot also be a rectangle, which needs four right angles. Options: 'must be a rectangle' wrongly assumes every parallelogram has right angles; 'must be a rhombus' wrongly assumes equal opposite sides means all four sides are equal, but only the opposite pairs are stated as equal here; 'trapezium' is wrong because a trapezium has exactly one pair of parallel sides, while this shape has two pairs, so it is a parallelogram and not a trapezium. Answer: it is a parallelogram but not a rectangle.
- (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).
- (d) 59.4° and 120.6° — Method: two sides and an angle that is not between them can describe two triangles, because an acute angle and its supplement have the same sine. Use the sine rule for the angle, then test whether the supplement also fits inside 180°. Working: AB = 9 cm faces angle ACB and BC = 6 cm faces the 35° angle, so sin ACB = 9 × sin 35° ÷ 6 = 5.1622 ÷ 6 = 0.86036. The inverse sine of 0.86036 is 59.358°, and its supplement is 180° − 59.358° = 120.642°. Both survive the angle-sum test, since 35° + 59.4° = 94.4° and 35° + 120.6° = 155.6°, each less than 180°. Answer: angle ACB is 59.4° or 120.6° to 1 decimal place. The distractors: 22.5° and 157.5° come from putting the sides the wrong way up, 6 × sin 35° ÷ 9 = 0.38238; 30.6° and 149.4° come from pressing the inverse cosine key on 0.86036 instead of the inverse sine key; 59.4° and 85.6° are the two unknown angles of the first triangle, 59.358° and 180° − 35° − 59.358°, given by a candidate who has found one triangle and reported its angles rather than the two possible sizes of the same angle.
- (d) (−1, 3) — Method: two 90° rotations about the SAME centre, applied one after another, combine into a single 180° rotation about that same centre: use the shortcut (x, y) → (2a − x, 2b − y) for a half-turn about (a, b). Working: with centre (2, 3), doubling each coordinate gives 2 × 2 = 4 and 2 × 3 = 6, so the rule is (x, y) → (4 − x, 6 − y). Applying it to (5, 3) gives 4 − 5 = −1 and 6 − 3 = 3, so the coin ends at (−1, 3). Answer: (−1, 3). Rotate about the centre (2, 3) stated in the game, not about the origin, and remember the button is pressed TWICE: stopping after one press, or rotating about the wrong centre, both leave the coin somewhere else.
- (c) 45° — Method: the tower, the ground and the line of sight form a right-angled triangle in which the 25 m height is opposite the angle of elevation and the 25 m along the ground is adjacent to it, so use tan θ = opposite ÷ adjacent. Working: tan θ = 25 ÷ 25 = 1, so θ = tan⁻¹(1). Answer: 45°. The distractors: 90° comes from using sin θ = 25 ÷ 25 = 1, which treats the 25 m along the ground as the hypotenuse when it is the side next to the angle; 1° comes from writing down the value of tan θ as though it were the angle itself; 50° comes from adding the two given lengths, 25 + 25, instead of comparing them.
- (b) 10.5 — Method: multiply the drawing length by the scale factor to get the real length. Working: 7 cm × 1.5 = 10.5 m. A student who answers 8.5 has added the scale factor to the drawing length instead of multiplying (7 + 1.5). A student who answers 14 has rounded the scale factor up to 2 before multiplying. A student who answers 3.5 has divided the drawing length by 2 instead of multiplying it by 1.5. Answer: 10.5 m.
- (a) 21.8° — The base diagonal has length √(6² + 8²) = √(36 + 64) = √100 = 10 cm, using Pythagoras' theorem on the rectangular base. The angle between the space diagonal and the base lies in the right-angled triangle formed by the height (4 cm, opposite), the base diagonal (10 cm, adjacent) and the space diagonal (hypotenuse), so tan(angle) = 4 ÷ 10 = 0.4, giving angle = 21.8° (1 d.p.). Using the height and one base edge instead of the full base diagonal, tan(angle) = 4 ÷ 8 = 0.5, gives 26.6° instead. Inverting the ratio, tan(angle) = 10 ÷ 4 = 2.5, gives 68.2°, the complement of the angle rather than the angle itself. Using the height and the other base edge, tan(angle) = 4 ÷ 6, gives 33.7°.
- (b) 70 cm³ — Area of the triangular cross-section = base × height ÷ 2. Base × height = 3.5 × 4 = 14 cm², and half of that is 14 ÷ 2 = 7 cm². Volume of the prism = cross-sectional area × length = 7 × 10 = 70 cm³. A pupil who forgets to halve when finding the triangle's area gets 3.5 × 4 × 10 = 140 cm³. A pupil who correctly finds the cross-sectional area but forgets to multiply by the length stops at 7 cm³. A pupil who ignores the height altogether, multiplying base by length, gets 3.5 × 10 = 35 cm³. The correct volume is 70 cm³.
- (a) 24 — The display is 4 tins wide, 3 tins high and 2 tins deep, and every position in that block is filled, so the total number of tins is the product of all three measurements: 4 × 3 × 2 = 24. "12" comes from multiplying only the width and height shown in the front elevation (4 × 3), forgetting the depth entirely. "9" comes from adding the three measurements (4 + 3 + 2) instead of multiplying them. "6" comes from multiplying only the height and depth (3 × 2), forgetting the width shown by the front elevation.
- (c) 1.44 m³ — The cross-section is a triangle, so its area = base × height ÷ 2. Base × height = 1.2 × 0.8 = 0.96 m², and half of that is 0.96 ÷ 2 = 0.48 m². The volume of the prism = cross-sectional area × length = 0.48 × 3 = 1.44 m³. A pupil who forgets to halve when finding the triangle's area gets 1.2 × 0.8 × 3 = 2.88 m³. A pupil who ignores the height altogether, treating the cross-section as base × length, gets 1.2 × 3 = 3.6 m³. A pupil who correctly finds the cross-sectional area but forgets to multiply by the length of the bed stops at 0.48 m³. The correct volume of soil is 1.44 m³.
- (c) 27.7 cm — Arc length = (150 ÷ 360) × 2 × 3.14 × 6 = (5 ÷ 12) × 37.68 = 15.7 cm. The perimeter of a sector also includes the two straight radii, so perimeter = 15.7 + 6 + 6 = 27.7 cm. (15.7 cm comes from stopping after the arc length and forgetting the two straight edges; 21.7 cm comes from adding only one radius instead of two; 31.4 cm comes from doubling the arc length instead of adding the two straight edges.)
- (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) 10.3 m — By Pythagoras' theorem, the hypotenuse = √(5² + 9²) = √(25 + 81) = √106 = 10.29...≈ 10.3 m. "106 m" is the value under the square root sign, correct as far as it goes but with the final square root step left out. "14 m" comes from adding the two shorter sides, 5 + 9, instead of using Pythagoras' theorem at all. "10.2 m" comes from rounding 10.29...m down to 10.2 instead of correctly rounding it up to 10.3.
- (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.)
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