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.
Geometry and measures worksheet — GCSE Higher
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- 1.Triangle PQR has angle P = 90°. The side PQ = 9 cm and the hypotenuse QR = 15 cm. Work out the length of PR.
- 2.A, B and C are points on a circle with centre O. B is on the major arc AC. Angle AOC = 116°. Work out the size of angle ABC.
- 3.A plan of a school hall uses a scale of 1 : 250. A wall is drawn 3.4 cm long on the plan. Sam wants to know the wall's real length in metres. What is it?
- 4.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?
- 5.The area of triangle ABC is 42 cm². AB = 9.5 cm and angle BAC = 61°. Work out the length of AC. Give your answer to 1 decimal place.
- 6.Work out the gradient of the line segment joining A(2, 3) and B(6, 11).
- 7.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.
- 8.A triangular flag has a base of 40 cm and a perpendicular height of 25 cm. Work out its area.
- 9.Lines JK and LM are parallel. A straight line crosses JK at point P and crosses LM at point Q. Angle KPQ = 118°. Angle KPQ and angle MQP are co-interior (allied) angles. Work out angle MQP.
- 10.Triangle ABC has AB = 8 cm, BC = 10 cm and CA = 6 cm. Triangle LMN has LM = 10 cm, MN = 6 cm and NL = 8 cm. The two triangles are congruent by SSS. Which of the following correctly matches the corresponding vertices?
- 11.A square-based pyramid has a square base and four identical triangular sloping faces, with its apex directly above the centre of the base. How many planes of symmetry does it have?
- 12.A tangent to a circle with centre O touches the circle at point P, where OP = 5 cm. Point Q lies on the tangent so that PQ = 12 cm. Using the fact that a tangent is perpendicular to the radius at the point of contact, work out the length OQ.
- 13.A rectangular name card has a perimeter of 30 cm. The card is 9 cm long. Work out the width of the card.
- 14.A stage designer is building a triangular platform ABC for a school play. AB = 11.5 m, AC = 8.2 m and angle BAC = 72°, the angle between them. The flooring for the platform costs £15.80 per square metre. Work out the total cost of the flooring, to the nearest pound.
- 15.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.
Answer key
- (c) 12 cm — By Pythagoras' Theorem, QR² = PQ² + PR², so PR² = QR² − PQ² = 15² − 9² = 225 − 81 = 144. Square root: √144 = 12 cm.
- (b) 58° — Method: the angle at the centre is TWICE the angle at the circumference when both stand on the same arc. Working: angle AOC is the angle at the centre standing on arc AC, and angle ABC is the angle at the circumference standing on the same arc AC, with B on the major arc, so angle AOC = 2 × angle ABC. Rearranging, angle ABC = 116 ÷ 2 = 58 degrees. Answer: 58°. Halve the centre angle, do not double it, and use the angle AOC exactly as given, 116°: you do not need its reflex angle here, because B sits on the major arc, which is precisely the arrangement the theorem is stated for.
- (b) 8.5 m — First apply the scale to convert the plan length to a real length in centimetres: 3.4 × 250 = 850 cm. Then convert centimetres to metres by dividing by 100: 850 ÷ 100 = 8.5, so the wall is 8.5 m long. Choosing 850 m applies the scale correctly but forgets to convert the answer from centimetres into metres. Choosing 0.85 m divides by 1000 instead of 100, confusing the centimetre-to-metre conversion with a metre-to-kilometre one. Choosing 3.4 m ignores the scale factor completely and just restates the plan length as if it were already the real length.
- (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.
- (a) 10.1 cm — Method: the area formula 1/2 × a × b × sin C contains the unknown side, so substitute what is known and rearrange. Working: 42 = 1/2 × 9.5 × AC × sin 61°. Multiplying both sides by 2 gives 84 = 9.5 × AC × sin 61°, and 9.5 × sin 61° = 9.5 × 0.87462 = 8.3089, so AC = 84 ÷ 8.3089 = 10.1096. Answer: AC = 10.1 cm to 1 decimal place. The distractors: 5.1 cm comes from forgetting to double the area when clearing the factor 1/2 and working out 42 ÷ 8.3089; 18.2 cm comes from using cos 61° in place of sin 61° in the denominator; 7.7 cm comes from multiplying by sin 61° instead of dividing by it, 84 × sin 61° ÷ 9.5, the standard slip when the unknown is inside a product.
- (a) 2 — Gradient = (change in y) ÷ (change in x) = (11 − 3) ÷ (6 − 2) = 8 ÷ 4 = 2. "0.5" comes from dividing the change in x by the change in y the wrong way round: 4 ÷ 8. "8" is only the change in y, forgetting to divide by the change in x at all. "−2" comes from a sign error, as if the y-coordinate had decreased rather than increased.
- (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.
- (c) 500 cm² — The area of a triangle is half of base × height. First, base × height = 40 × 25 = 1,000. Half of 1,000 is 500 cm². 1,000 cm² forgets to halve and just gives base × height. 65 cm² adds the base and height together instead of multiplying them. 2,000 cm² doubles base × height instead of halving it.
- (a) 62° — Angle KPQ and angle MQP are co-interior (allied) angles between parallel lines, so they add up to 180°: angle MQP = 180° − 118° = 62°. 118° comes from treating the angles as equal, as if this were a corresponding or alternate angle pair, instead of using the co-interior rule. 90° comes from wrongly assuming the crossing line meets JK and LM at right angles. 59° comes from halving 118° instead of subtracting it from 180°.
- (c) A↔N, B↔L, C↔M (ABC≅NLM) — Matching equal side lengths: AB (8 cm) equals NL (8 cm), BC (10 cm) equals LM (10 cm), and CA (6 cm) equals MN (6 cm). This gives the correspondence A with N, B with L, and C with M, so triangle ABC is congruent to triangle NLM, making 'A↔N, B↔L, C↔M (ABC≅NLM)' correct. 'A↔L, B↔M, C↔N (ABC≅LMN)' simply matches the vertices in the order they are written without checking the side lengths: AB (8 cm) would need to equal LM (10 cm), which is false. 'A↔M, B↔N, C↔L (ABC≅MNL)' also fails this check, since AB (8 cm) would need to equal MN (6 cm), which is false. 'A↔N, B↔M, C↔L (ABC≅NML)' gets A correct but swaps B and C, so AB (8 cm) would need to equal NM (6 cm), which is also false.
- (d) 4 — Method: a plane of symmetry must pass through the apex and cut the base along one of the base's own lines of symmetry. Working: a square has 4 lines of symmetry (2 through opposite edge midpoints, 2 through opposite corners), and each of these, combined with the apex, gives one plane of symmetry of the pyramid. A student who answers 2 has only found the planes through the edge midpoints, or only the ones through the corners, and missed the other pair. A student who answers 8 has doubled the correct count, perhaps confusing it with a different solid. A student who answers 1 has only spotted the one obvious front-to-back plane. Answer: 4.
- (d) 13 — Method: OP and PQ meet at a right angle because of the tangent–radius fact, so triangle OPQ is right-angled at P; use Pythagoras' theorem. Working: OQ² = OP² + PQ² = 5² + 12² = 25 + 144 = 169; OQ = √169 = 13. A student who answers 17 has simply added the two given lengths (5 + 12) instead of using Pythagoras' theorem. A student who answers 7 has subtracted the two given lengths (12 − 5) instead of using Pythagoras' theorem. A student who answers 144 has correctly squared 12 but stopped there, forgetting to add 5² and take the square root. Answer: 13 cm.
- (a) 6 cm — Method: a rectangle is made of two lengths and two widths, so the perimeter is 2 × (length + width). Half the perimeter is therefore one length plus one width, and the width is what is left of that half once the length is taken away. Working: 30 ÷ 2 = 15 cm is one length added to one width, and 15 − 9 = 6. Answer: 6 cm. The distractors: 21 cm comes from taking the length off the whole perimeter, 30 − 9, so three sides are still counted in the figure that is written down; 12 cm comes from removing both lengths correctly, 30 − 2 × 9 = 12, and then recording that remainder as the width instead of halving it, because it is in fact the two widths together; 7.5 cm comes from dividing the perimeter by four, 30 ÷ 4, which is the calculation for the side of a square and throws away the length that was given.
- (b) £709 — Method: first find the area of the triangular platform with Area = (1/2)ab sin C, then multiply by the cost per square metre. Working: Area = 1/2 × 11.5 × 8.2 × sin 72° = 44.8 m² (1 d.p.); cost = area × £15.80, which rounds to £709 to the nearest pound. Answer: £709. Leaving out the 1/2 in the area formula doubles the area, giving a cost of £1417; using cos 72° instead of sin 72° gives a much smaller area and a cost of £230; and squaring the 11.5 m side instead of multiplying the two different given sides together gives a cost of £994. Find the exact area first — don't round it early — then multiply by the cost per square metre and round only the final answer.
- (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.
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