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 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.
- 2.A cylinder has a radius of 3 cm and a height of 10 cm. Using π = 3.14, work out the volume of the cylinder.
- 3.A regular hexagon has sides of length 8 cm. By splitting the hexagon into 6 identical triangles that meet at its centre, work out the area of the hexagon. Give your answer to 1 decimal place.
- 4.A sector of a circle has radius 10 cm and angle 150°. Using π = 3.14, work out the perimeter of the sector, to 1 decimal place.
- 5.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.
- 6.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.
- 7.In triangle ABC, AB = 8.6 cm, AC = 11.4 cm and angle BAC = 47°. Work out the area of triangle ABC. Give your answer to 1 decimal place.
- 8.A footpath is 3750 m long. Work out this distance in kilometres.
- 9.A circle has two straight radii drawn from the centre to the edge, together with the arc between them. Write down the name of the region enclosed by these two radii and the arc.
- 10.Shape S is enlarged by a scale factor of −3 about a fixed centre. Which statement correctly describes the image compared to the original shape?
- 11.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?
- 12.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.
- 13.The bearing of a harbour B from a ferry's position A is 260°. What is the bearing of A from B?
- 14.The angle between north and a cycle path is 40°, but it is measured anticlockwise from north. What is the three-figure bearing of the cycle path?
- 15.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.
Answer key
- (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) 282.6 cm³ — Volume of a cylinder = πr²h = 3.14 × 3² × 10 = 3.14 × 9 × 10 = 282.6 cm³. (94.2 cm³ comes from using πrh and forgetting to square the radius; 90 cm³ comes from using r²h and leaving π out altogether; 1130.4 cm³ comes from using the diameter, 6 cm, in place of the radius.)
- (b) 166.3 cm² — Method: split the regular hexagon into 6 identical triangles meeting at the centre, each with two sides of 8 cm and a 60° angle between them, and use Area = (1/2)ab sin C on just one of them. Working: one triangle's area = 1/2 × 8 × 8 × sin 60° = 27.7 cm² (1 d.p.); the hexagon is 6 of these, so its area is 6 × 27.7 = 166.3 cm² (1 d.p.). Answer: 166.3 cm². Reporting just one triangle's area, without multiplying by 6, gives 27.7 cm²; treating the angle at the centre as a right angle instead of 60°, using 1/2 × 8 × 8 with no sine factor at all, gives 6 × 32 = 192.0 cm²; and multiplying by 5 instead of 6, miscounting the triangles in the hexagon, gives 5 × 27.7 = 138.6 cm². A regular hexagon always splits into exactly 6 triangles at its centre — count them before you multiply.
- (d) 46.2 cm — The perimeter of a sector is the arc length plus its two straight radii. The circumference is 2 × 3.14 × 10 = 62.8 cm, and the arc is 150 ÷ 360 of that: 62.8 × 150 ÷ 360 = 26.2 cm (1 d.p.). Adding the two radii, 26.2 + 10 + 10 = 46.2 cm. Giving just the arc length, without adding the straight edges, gives 26.2 cm. Adding only ONE radius instead of two gives 36.2 cm. Using 150 ÷ 180 instead of 150 ÷ 360 for the fraction gives an arc of 52.3 cm and a perimeter of 72.3 cm.
- (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.)
- (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.
- (a) 35.9 cm² — Method: the area of any triangle is 1/2 × a × b × sin C, where a and b are two sides and C is the angle between them. Working: the 47° angle lies between AB = 8.6 cm and AC = 11.4 cm, so the area is 1/2 × 8.6 × 11.4 × sin 47° = 49.02 × 0.73135 = 35.851. Answer: the area is 35.9 cm² to 1 decimal place. The distractors: 71.7 cm² comes from leaving out the factor 1/2 and working out 8.6 × 11.4 × sin 47°; 33.4 cm² comes from pressing cos instead of sin, 49.02 × cos 47°, which is the same as using the complement 43° in place of 47°; 52.6 cm² comes from pressing tan instead of sin, 49.02 × tan 47°.
- (a) 3.75 km — To convert metres to kilometres, divide by 1000: 3750 ÷ 1000 = 3.75 km. Dividing by 100 instead of 1000 gives 37.5 km. Dividing by 10 instead of 1000 gives 375 km. Dividing by 10 000 instead of 1000 gives 0.375 km.
- (d) sector — The region enclosed by two radii and the arc between them is called a sector. A segment is a different region, enclosed by a CHORD and an arc — it does not touch the centre at all. A chord is a straight line, not a region, so it cannot be the answer to a question asking for a region. A semicircle is only correct when the angle between the two radii is exactly 180°, which is not stated here, so it is too specific to be the general answer.
- (a) 3 times the size, opposite side, rotated 180° — Method: for any enlargement, the MAGNITUDE of the scale factor gives the size ratio between image and object, while the SIGN decides which side of the centre the image falls on; a negative scale factor puts the image on the opposite side, which is equivalent to a 180° rotation about the centre. Working: the scale factor is −3, so the size ratio is the magnitude, which is 3, and the negative sign puts the image on the opposite side of the centre, rotated 180° relative to the original. Answer: 3 times the size, opposite side, rotated 180°. The magnitude of the scale factor controls the SIZE only: do not let the sign leak into it and turn 3 into 1/3. The sign controls the SIDE and orientation, which a positive-only view of enlargement, just 'further away' with the same orientation, misses entirely.
- (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.
- (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) 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.
- (b) 320° — A bearing is always measured clockwise from north. An angle measured anticlockwise must be converted by subtracting it from 360°: 360 − 40 = 320°, so the bearing is 320°. Choosing 040° treats the anticlockwise angle as if it were already a clockwise bearing, without converting it. Choosing 220° adds 180° to the angle, mixing this up with a back-bearing calculation (40 + 180 = 220). Choosing 400° adds the angle to 360° instead of subtracting it (40 + 360 = 400), giving a bearing greater than a full turn.
- (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.
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