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.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.
- 2.Triangle ABC has AB = 10 cm, BC = 7 cm and angle A = 40°. Triangle DEF has DE = 10 cm, EF = 7 cm and angle D = 40°. Sam says the two triangles are congruent by SAS. Which statement about Sam's claim is correct?
- 3.A cylindrical drum and a cuboid box are being wrapped for a display. A shop assistant wraps only the flat faces of each item, not any curved surface. How many flat faces in total does the assistant wrap across both items?
- 4.A sector of a circle has angle 120° and an arc length of 31.4 cm. Using π = 3.14, work out the radius of the circle.
- 5.Work out the exact value of cos 0° − sin 30°.
- 6.A tangent to a circle touches the circle at exactly one point, P. Work out the size of the angle between the tangent and the radius drawn to P.
- 7.In triangle ABC, AB = 13 cm, AC = 9 cm and angle ABC = 35°. Given that angle ACB is obtuse, work out its size. Give your answer to 1 decimal place.
- 8.Work out the exact value of tan 30° + tan 30°.
- 9.A, B and C are points on a circle with centre O. B is on the minor arc AC. Angle ABC = 100°. Work out the size of the non-reflex angle AOC.
- 10.The angle between north and a footpath is 35°, measured clockwise from north. What is the three-figure bearing of the footpath?
- 11.A builder props a straight plank against a vertical wall to reach a window ledge. The foot of the plank is 2.1 m from the base of the wall, and the plank is 3.5 m long. Work out how high up the wall the plank reaches.
- 12.Two similar bottles have heights in the ratio 1 : 2. Write down the ratio of their volumes.
- 13.Triangle ABC has a right angle at B, hypotenuse AC = 13 cm and AB = 5 cm. Triangle DEF has a right angle at E, hypotenuse DF = 13 cm and DE = 5 cm. Which condition proves the two triangles are congruent?
- 14.In triangle ABC the angle at C is 90° and the hypotenuse AB is 12 cm. H is the point on AB for which CH is perpendicular to AB, and AH = 3 cm. Work out the length of AC.
- 15.Two triangles are similar. The sides of the first triangle are 3 cm, 4 cm and 5 cm. The shortest side of the second triangle is 9 cm. Work out the scale factor from the first triangle to the second.
Answer key
- (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.
- (d) Wrong - the given angle is not the included angle — Sides AB and BC meet at vertex B, so the included angle needed for SAS is angle B, not angle A — the information given is SSA. SSA does not prove congruence: with AB = 10 cm, BC = 7 cm and angle A = 40° there are two different triangles that fit, one with angle C ≈ 74.6° and one with angle C ≈ 105.4°, so Sam's triangles need not be the same shape and size at all. Sam is not correct just because two sides and an angle are equal, since the angle must be the one INCLUDED between those two sides. The condition is not ASA either, because ASA needs two angles, and only one angle is given here. It is also not true that nothing matches — the stated lengths and angle DO match between the two triangles; the problem is which angle was given, not whether the values agree.
- (a) 8 — A cylinder has two flat circular faces (the top and the base) and one curved surface, so only its 2 flat faces are wrapped. A cuboid has 6 faces and all of them are flat, so all 6 are wrapped. Adding these: 2 + 6 = 8, so 8 is correct. 9 comes from wrongly counting the cylinder's curved surface as a flat face. 6 comes from counting only the cuboid and forgetting the cylinder's two flat circular faces. 3 comes from counting only the cylinder and including its curved surface in that total.
- (d) 15 cm — Arc length = (angle ÷ 360) × 2 × π × r. Here 120 ÷ 360 = 1/3, and 2 × 3.14 = 6.28, so 31.4 = (1/3) × 6.28 × r. Multiplying both sides by 3 gives 6.28 × r = 94.2, so r = 94.2 ÷ 6.28 = 15 cm. (5 cm comes from forgetting the angle fraction altogether and dividing the arc length by 2 × π alone: 31.4 ÷ 6.28 = 5; 30 cm comes from leaving out the factor of 2, dividing by (1/3) × 3.14 = 1.0467 instead of (1/3) × 6.28: 31.4 ÷ 1.0467 = 30; 7.5 cm comes from correctly finding a radius of 15 cm but then treating that 15 cm as a diameter and halving it.)
- (a) 1/2 — cos 0° = 1 and sin 30° = 1/2, so cos 0° − sin 30° = 1 − 1/2 = 1/2. 1 comes from writing down cos 0° alone and forgetting to subtract sin 30°. 3/2 comes from adding the two values instead of subtracting. −1/2 comes from working out sin 30° − cos 0°, the two terms the wrong way round.
- (c) 90° — A tangent to a circle always meets the radius drawn to the point of contact at a right angle, so the angle between the tangent and the radius at P is 90°. 180° confuses the tangent with the diameter through P, as if the radius continued in a straight line into the tangent. 45° halves the true angle by mistake. 60° comes from confusing this fact with the angle of an equilateral triangle.
- (b) 124.1° — Method: use the sine rule AB/sin(ACB) = AC/sin(ABC), since AB is opposite angle ACB and AC is opposite angle ABC. Working: sin(ACB) = AB × sin(ABC) / AC = 13 × sin 35° / 9 = 0.8285, which gives angle ACB = 55.9° or its supplement 180 − 55.9 = 124.1°; both of these give a valid triangle, and since angle ACB is obtuse the answer is 124.1°. Reporting the acute solution instead, without checking the word 'obtuse' in the question, gives 55.9°; putting the sides the wrong way round in the ratio, sin(ACB) = 9 × sin 35° / 13, gives 23.4°; and subtracting the acute solution and the given 35° from 180° as if angle ACB were the remaining triangle angle gives 89.1°. A given SSA fact always has two possible angle solutions unless the stem states which one is meant.
- (a) 2√3/3 — tan 30° = √3/3, so tan 30° + tan 30° = 2 × √3/3 = 2√3/3. √3 comes from wrongly treating tan 30° + tan 30° as tan(30° + 30°) = tan 60° = √3 — adding angles is not the same as adding ratios. √3/3 comes from forgetting to double the value and just writing down tan 30° on its own. 2√3 comes from doubling the numerator of √3/3 but forgetting to keep the denominator of 3.
- (b) 160° — B is on the minor arc AC, so the angle at the centre theorem applies to the reflex angle AOC: reflex angle AOC = 2 × 100° = 200°. The non-reflex angle AOC is the rest of the full turn: 360° − 200° = 160°. Reporting the reflex angle itself, without subtracting it from 360°, gives 200°. Subtracting angle ABC from 180° instead, as if this were a cyclic quadrilateral, gives 80°. Halving angle ABC instead of doubling it gives 50°. Double first, then take the angle away from a full turn, and 160° is what's left.
- (a) 035° — A bearing is measured clockwise from north, so an angle of 35° clockwise from north is a bearing of 035° (written with three figures). Choosing 325° measures the angle anticlockwise instead of clockwise (360 − 35 = 325). Choosing 215° adds 180° to the angle, mixing this up with a back-bearing calculation (35 + 180 = 215). Choosing 350° reorders the digits of 035, writing the ones digit before the tens digit by mistake.
- (b) 2.8 m — Use Pythagoras' Theorem: the plank is the hypotenuse (3.5 m) of a right-angled triangle formed with the wall and the ground (2.1 m). height² = 3.5² − 2.1² = 12.25 − 4.41 = 7.84. height = √7.84 = 2.8 m. A student who subtracts the two given lengths directly instead of using Pythagoras gets 3.5 − 2.1 = 1.4 m. A student who doubles the distance from the wall by mistake gets 2.1 × 2 = 4.2 m.
- (a) 1 : 8 — Method: in similar solids the ratio of the volumes is the cube of the ratio of corresponding lengths. Working: the heights are in the ratio 1 : 2, so the volumes are in the ratio 1³ : 2³ = 1 : 8. Answer: 1 : 8. The distractors: 1 : 2 leaves the height ratio unchanged, as though a volume scaled in the same way as a length; 1 : 4 comes from squaring, 1² : 2², which is the rule for surface areas and not for volumes; 1 : 6 comes from multiplying each part by 3 instead of raising each part to the power 3.
- (b) RHS - right angle, hypotenuse and one side equal — A right angle, the hypotenuse (13 cm) and one other side (5 cm) are equal in both triangles, so this is RHS. SAS would need the equal angle to be the one INCLUDED between the two equal sides, but the right angle at B is not between AB and the hypotenuse AC — it is opposite the hypotenuse instead, so SAS does not apply directly here. SSS needs all three sides, but only two sides are stated. ASA needs two angles, but only one angle (the right angle) is given.
- (d) 6 cm — Method: CH cuts the triangle into two smaller right-angled triangles, so Pythagoras' theorem can be written in each of the three right-angled triangles and the results combined. Working: HB = 12 − 3 = 9 cm. In triangle ACH, CH² = AC² − 3² = AC² − 9; in triangle CHB, CH² = CB² − 9² = CB² − 81. Setting those equal gives AC² − 9 = CB² − 81. In triangle ABC, AC² + CB² = 12² = 144, so CB² = 144 − AC². Substituting gives AC² − 9 = 144 − AC² − 81, so 2 × AC² = 72 and AC² = 36, giving AC = 6. Answer: 6 cm. The distractors: 36 cm comes from stopping at AC² = 36 and never taking the square root; 9 cm is HB, the other part of the hypotenuse, written down in place of AC; 4 cm comes from working out 12 ÷ 3, treating AH as a scale factor between the two triangles rather than as a length.
- (b) 3 — Method: the scale factor is a length of the second triangle divided by the corresponding length of the first, and corresponding means shortest paired with shortest. Working: the shortest side of the first triangle is 3 cm and the shortest side of the second is 9 cm, so the scale factor is 9 ÷ 3 = 3. Answer: 3. The distractors: 1/3 comes from dividing 3 by 9, which is the scale factor from the second triangle back to the first; 6 comes from subtracting, 9 − 3, treating the scale factor as a difference rather than a multiplier; 1.8 comes from dividing 9 by 5, pairing the second triangle's shortest side with the first triangle's longest side.
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