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 tap fills a tank at a rate of 18 litres per minute. The tank holds 0.45 m³. Work out how long the tap takes to fill the tank.
- 2.Which of these statements about solving a triangle is correct?
- 3.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.
- 4.A television cable runs in a straight line from the edge of a flat roof 20 m above the ground to a bracket on a wall 5 m above the ground. The horizontal distance between the roof edge and the bracket is 15 m. Work out the length of the cable. Give your answer to 1 decimal place.
- 5.In triangle PQR, PQ = 11 cm, QR = 7 cm and PR = 13 cm. Work out the size of angle PQR. Give your answer to 1 decimal place.
- 6.A warehouse stores identical cube-shaped crates. Its plan view is a 2 by 4 rectangle of crate positions, and every position is filled to a height of 3 crates, except one corner position, which has only 2 crates stacked on it because a delivery was incomplete. How many crates are there in total?
- 7.A bead starts at position (2, −1) on a grid, in centimetres. It is moved by the column vector u, with top number 3 and bottom number 5, and then moved by the column vector v, with top number −7 and bottom number 2. Work out the coordinates of the bead's final position.
- 8.A birthday cake is a cylinder of radius 10 cm and height 8 cm, with a thin ribbon fixed exactly once around the curved side, at half the height, and joined with no overlap. Work out the length of ribbon needed. Use π = 3.14.
- 9.A crane's hook starts at the point (3.4, 9.5) on a construction site plan measured in metres. It moves along the vector to pick up a beam, then along the vector to place it. Work out the single column vector that describes the hook's overall movement from its start position to where it places the beam.
- 10.A textbook contains the instruction 'Draw AB' and, on the next line, the statement 'AB = 5 cm'. Which statement about the notation AB is correct?
- 11.A rhombus-shaped floor tile has sides of length 12 cm and an interior angle of 55°. Work out the area of the tile. Give your answer to 1 decimal place.
- 12.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.
- 13.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.
- 14.A plumber charges a £45 call-out fee, plus £28 for each hour worked. She works on a job for 3 hours 30 minutes, and her hours are billed by rounding up to the next whole hour. Work out the total amount she charges.
- 15.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.
Answer key
- (b) 25 minutes — Method: a rate in litres per minute can only be used on a volume measured in litres, so convert the tank first and then divide. Working: 1 m³ = 1000 litres, so the tank holds 0.45 × 1000 = 450 litres, and the time is 450 ÷ 18 = 25. Answer: 25 minutes. Using 1 m³ = 100 litres gives 45 ÷ 18 = 2.5 minutes. Using 1 m³ = 1 000 000 litres, which is the factor that turns cubic metres into cubic centimetres, gives 450 000 ÷ 18 = 25 000 minutes. Multiplying by the rate instead of dividing by it gives 450 × 18 = 8100.
- (d) Two angles and a side: sine rule finds other sides. — Method: match the data you are given to the rule that needs it. Working: the sine rule a/sin A = b/sin B = c/sin C needs a complete angle-side pair to set up its ratio, so the statement that two angles and a side (AAS or ASA) let the sine rule find the other sides is the correct one — the third angle comes from the angle sum, and each unknown side is then opposite a known angle. The statement that two sides and the angle between them (SAS) call for the sine rule is wrong: no angle-side pair is complete, so the cosine rule is what works there. The statement that the sine rule finds any angle from three sides (SSS) is wrong for the same reason in reverse — no angle is known at all, so the cosine rule must find the first one. The statement that the cosine rule finds a missing angle directly from two sides and a non-included angle (SSA) is wrong: the cosine rule reports the angle enclosed by the two sides it uses, so with SSA it is the sine rule that reaches the missing angle, and the ambiguous case is then settled from the wording of the question.
- (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.
- (d) 21.2 m — Method: the cable is the hypotenuse of a right-angled triangle whose vertical side is the drop from the roof to the bracket and whose horizontal side is 15 m, so use Pythagoras' theorem. Working: the drop is 20 − 5 = 15 m, so c² = 15² + 15² = 225 + 225 = 450 and c = √450 = 21.213…, which is 21.2 m to 1 decimal place. Answer: 21.2 m. The distractors: 25.0 m comes from using the whole 20 m height of the roof as the vertical side and forgetting that the bracket is already 5 m up; 30.0 m comes from adding the two sides of the triangle, 15 + 15, instead of using Pythagoras' theorem; 15.0 m is the horizontal distance on its own, which would be the length of the cable only if it ran level.
- (a) 89.6° — Angle PQR is at vertex Q, so the side opposite it is PR = 13 cm. Using the cosine rule rearranged for an angle: cos(PQR) = (121 + 49 − 169) ÷ (2 × 11 × 7) = 1 ÷ 154 = 0.0065. Taking the inverse cosine, angle PQR = 89.6° (1 d.p.). Using PR as one of the enclosing sides instead of as the opposite side, finding angle P instead of angle Q, gives (121 + 169 − 49) ÷ (2 × 11 × 13) = 241 ÷ 286 = 0.8427, and angle = 32.6°. Swapping the sign inside the bracket, computing 169 − 121 − 49 instead, gives −1 ÷ 154 = −0.0065, and angle = 90.4°. Using PR in the denominator instead of QR gives (121 + 49 − 169) ÷ (2 × 11 × 13) = 1 ÷ 286 = 0.0035, and angle = 89.8°. Keep PR as the opposite side, QR and PQ as the enclosing sides, and angle PQR comes to 89.6°.
- (b) 23 — If every position were filled to the full height of 3, the total would be 2 × 4 × 3 = 24 crates. One corner position has only 2 crates instead of 3, one crate short of full height there, so the actual total is 24 − 1 = 23. "24" comes from using the full height everywhere and forgetting the one incomplete corner. "22" comes from removing 2 crates for the incomplete corner instead of the 1 that is actually missing (3 − 2 = 1, not 2). "21" comes from removing all 3 crates at that corner, as though the position were completely empty rather than 2 crates short.
- (d) (−2, 6) — Method: add the top numbers of both vectors to the starting x-coordinate, and the bottom numbers of both vectors to the starting y-coordinate. Working: x-coordinate 2 + 3 + (−7) = −2; y-coordinate −1 + 5 + 2 = 6. Answer: (−2, 6). A candidate who only applies vector u and forgets v gets (5, 4). A candidate who only applies vector v and forgets u gets (−5, 1). A candidate who works out the combined vector u + v but forgets to add it to the starting point gets (−4, 7).
- (a) 62.8 cm — The ribbon goes once around the circular cross-section, so its length equals the circumference: 2πr = 2 × 3.14 × 10 = 62.8 cm.
- (b) $\binom{-0.4}{-7}$ — The overall movement is the sum of the two vectors: (−1.2 + 0.8, −4.5 + (−2.5)) = (−0.4, −7). '$\binom{-2}{-2}$' comes from subtracting the second vector from the first instead of adding them. '$\binom{-0.4}{7}$' gets the top number right but drops the negative sign on the bottom number. '$\binom{2}{-7}$' comes from treating −1.2 + 0.8 as if the signs did not matter, giving +2 instead of −0.4.
- (c) AB means the segment or its length, shown by context — Method: recall the standard convention for a two-letter label such as AB. Working: AB names the segment AND its length; the sentence around it shows which is meant — 'draw AB' means the segment, 'AB = 5 cm' means the length. Options: 'a different symbol is needed' and 'must be written as |AB|' both describe stricter rules than the convention GCSE actually uses; 'AB can only mean the segment' ignores that the same label is also used for the length. Answer: AB means the segment or its length, shown by context.
- (d) 118.0 cm² — Method: a rhombus is made of two congruent triangles either side of a diagonal, each with area 1/2 × 12 × 12 × sin 55°, so the whole rhombus has area 12 × 12 × sin 55° (side² × sin of the interior angle). Working: area = 12² × sin 55° = 118.0 cm². Stopping at one triangle's area, 1/2 × 12² × sin 55°, and forgetting to double it gives 59.0 cm²; using cos 55° instead of sin 55° gives 82.6 cm²; and multiplying the two sides together with no trig term at all gives 144.0 cm². Splitting the rhombus into its two triangles is the safest way to see why the 1/2 disappears from the whole-shape formula.
- (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) 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.
- (b) £157 — Method: round the hours worked up to the next whole hour, multiply by the hourly rate, then add the call-out fee. Working: 3 hours 30 minutes rounds up to 4 hours; 28 × 4 = 112; 112 + 45 = 157. Answer: £157. A candidate who uses the unrounded time of 3.5 hours, working out 28 × 3.5 = 98 and adding 45, gets £143. A candidate who forgets the call-out fee, giving only 28 × 4, gets £112. A candidate who rounds down to 3 hours instead of up, working out 28 × 3 = 84 and adding 45, gets £129.
- (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.
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