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.State whether the line segment joining A(−4, 3) and B(2, 6) is horizontal, vertical or neither, and give a reason for your answer.
- 2.A triangle has vertices A(1, 1), B(5, 1) and C(5, 4). Work out the lengths of AB, BC and CA, and write down whether triangle ABC is right-angled.
- 3.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.
- 4.A model car is built to a scale of 1 : 24. The real car is 4.32 m long. Work out the length of the model, in centimetres.
- 5.Triangle A has vertices (2, 1), (5, 1) and (2, 4). It is translated by the vector (−3, 2) and the image is then reflected in the x-axis. Work out the coordinates of the image of (5, 1).
- 6.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.
- 7.A solid has 5 faces: one square face and four triangular faces, which all meet at a single point above the square. What is the name of this solid?
- 8.A water tank is a cube with edges of length 2 m. Work out how many cubic centimetres the tank holds when it is full.
- 9.A solid pyramid has a square base of side 6 cm. Its apex is directly above the centre of the base, at a vertical height of 8 cm. Work out the volume of the pyramid.
- 10.In triangle ABC, AB = 8 cm, BC = 11 cm and AC = 6 cm. Work out the size of angle ABC. Give your answer to 1 decimal place.
- 11.Sam says: '3.2 litres is the same as 320 ml, because you multiply by 100.' Which statement about Sam's claim is correct?
- 12.Two bronze statues are geometrically similar and made from the same solid bronze. Their heights are 30 cm and 45 cm. The smaller statue has a mass of 12 kg. Work out the mass of the larger statue, in kg.
- 13.The midpoint of the line segment AB is (3, 5). A is the point (1, 3). Work out the coordinates of B.
- 14.In quadrilateral ABCD the two sides AB and BC are each 6 cm long and meet each other at B. The two sides CD and DA are each 9 cm long and meet each other at D. No side of ABCD is parallel to any other side. Write down the mathematical name of this quadrilateral.
- 15.A rectangular name card has a perimeter of 30 cm. The card is 9 cm long. Work out the width of the card.
Answer key
- (b) Neither, because both coordinates differ — A line segment is horizontal only when both points share the same y-coordinate, and vertical only when both points share the same x-coordinate. Here A has x-coordinate −4 and B has x-coordinate 2, which differ, and A has y-coordinate 3 and B has y-coordinate 6, which also differ, so the segment is neither horizontal nor vertical. Every point has a y-coordinate and an x-coordinate, so simply having one is not a reason for the line to be horizontal or vertical — both of those wrong reasons ignore that the coordinates must match, not just exist. The x-coordinates do increase from A to B, but an increasing x-coordinate on its own describes a slope, not a horizontal line.
- (d) Yes, since 3² + 4² = 5² — AB is horizontal with length 5 − 1 = 4, BC is vertical with length 4 − 1 = 3, and CA = √(4² + 3²) = √25 = 5. Since the two shorter sides satisfy 3² + 4² = 5², the triangle is right-angled, with the right angle at B. "No, since 3 + 4 ≠ 5" wrongly tests Pythagoras' theorem by adding the sides instead of squaring them first. "No, since AB, BC and CA are not all equal" confuses a right-angled triangle with an equilateral one — a triangle does not need equal sides to have a right angle. "Yes, since 4² + 5² = 3²" reaches the correct conclusion but puts the longest side, 5, on the wrong side of the equation, as if it were one of the two shorter sides instead of the hypotenuse.
- (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.
- (b) 18 — Method: convert the real length to centimetres, then divide by the scale factor to shrink it down to the model's size. Working: 4.32 m = 432 cm; 432 cm / 24 = 18 cm. A student who answers 432 has converted the units correctly but forgotten to divide by the scale factor at all. A student who answers 10368 has multiplied by the scale factor instead of dividing (432 x 24). A student who answers 1.8 has converted the metres to centimetres by multiplying by 10 instead of 100, getting 43.2 cm, and then divided by 24. Answer: 18 cm.
- (c) (2, −3) — Method: apply the transformations in the order given, translation first, then reflection, to the point itself. Working: translating (5, 1) by the vector (−3, 2) moves the x-coordinate by −3 and the y-coordinate by 2, giving 5 − 3 = 2 and 1 + 2 = 3, so the image after the translation is (2, 3). Reflecting this point in the x-axis leaves the x-coordinate unchanged and reverses the sign of the y-coordinate, giving (2, −3). Answer: (2, −3). Do the two maps in the stated order and carry the point all the way through both of them: swapping the order, or stopping after the first map, lands you on a different point altogether.
- (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.
- (a) Square-based pyramid — A solid with one square base and four triangular faces meeting at a single apex above the base is a square-based pyramid. A triangular prism has two triangular faces and three rectangular faces, not a square base with four triangles, so that is a different solid. A cube has six square faces, and a cuboid has six rectangular faces — neither has any triangular faces at all. The solid described is a square-based pyramid.
- (a) 8,000,000 cm³ — Method: change the edge length into centimetres first and then cube it, because 1 m = 100 cm and a volume needs that conversion applied to all three dimensions. Working: 2 m = 2 × 100 = 200 cm, so the volume is 200 × 200 × 200. 200 × 200 = 40,000 and 40,000 × 200 = 8,000,000. Answer: 8,000,000 cm³. The distractors: 8,000 cm³ comes from converting 2 m to 20 cm and cubing that; 80,000 cm³ comes from cubing in metres to get 8 m³ and then multiplying by 10,000, the conversion factor for an area rather than the 1,000,000 a volume needs; 8 cm³ comes from cubing the 2 without converting at all and simply writing cm³ because the question asked for that unit.
- (b) 96 cm³ — Method: the volume of a pyramid is one third of the base area multiplied by the vertical height. Work out the area of the square base, multiply by the height, then divide by 3. Working: the base area is 6 × 6 = 36 cm², then 36 × 8 = 288, and 288 ÷ 3 = 96. Answer: 96 cm³. The distractors: 288 cm³ comes from multiplying the base area by the height and forgetting the one third, which is the volume of a cuboid with the same base and height; 144 cm³ comes from halving that 288 instead of taking a third of it; 16 cm³ comes from using the base edge of 6 cm in place of the base area, (6 × 8) ÷ 3.
- (a) 32.2° — Method: all three sides are known, so rearrange the cosine rule as cos A = (b² + c² − a²) ÷ 2bc, where a is the side facing the angle you want. Working: angle ABC sits between AB = 8 cm and BC = 11 cm and faces AC = 6 cm, so cos ABC = (8² + 11² − 6²) ÷ (2 × 8 × 11) = (64 + 121 − 36) ÷ 176 = 149 ÷ 176 = 0.84659, and the inverse cosine of 0.84659 is 32.157°. Answer: angle ABC = 32.2° to 1 decimal place. The distractors: 102.6° comes from subtracting the square of the longest side, 11, rather than the square of the side the angle actually faces, giving (64 + 36 − 121) ÷ 96; 57.8° comes from taking the inverse sine of 0.84659 instead of the inverse cosine; 147.8° comes from rearranging with the subtraction the wrong way round, (36 − 64 − 121) ÷ 176 = −0.84659, which turns an acute angle into its supplement.
- (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) 40.5 kg — Since the statues are made from the same material, mass is proportional to volume, which scales with the cube of the length scale factor. The length scale factor is 45 ÷ 30 = 1.5, so the volume (and mass) scale factor is 1.5³ = 3.375. The larger statue's mass is 12 × 3.375 = 40.5 kg. 18 kg comes from multiplying by the length factor 1.5 directly. 27 kg comes from using the area scale factor 1.5² = 2.25 instead of the volume scale factor. 54 kg comes from treating 'cubed' as 'multiplied by 3', giving 1.5 × 3 = 4.5 instead of 1.5³.
- (a) (5, 7) — Method: a midpoint is the mean of the two end points, so for each coordinate (start + end) ÷ 2 = midpoint; rearranging that gives end = 2 × midpoint less the start. Working: for x, (1 + x) ÷ 2 = 3, so 1 + x = 6 and x = 5. For y, (3 + y) ÷ 2 = 5, so 3 + y = 10 and y = 7. B is therefore (5, 7). Answer: (5, 7). The distractors: (2, 2) comes from subtracting A from the midpoint, (3 − 1, 5 − 3), which gives the step from A to the midpoint and stops there instead of taking that same step a second time; (4, 8) comes from adding A to the midpoint, (3 + 1, 5 + 3), without doubling the midpoint first; (6, 10) comes from doubling the midpoint, (2 × 3, 2 × 5), and then forgetting to take A off.
- (b) Kite — Method: name a quadrilateral by matching what is given — which sides are equal, whether those equal sides lie next to each other or opposite each other, and whether any sides are parallel — against the definitions of the special quadrilaterals. Working: the two 6 cm sides meet at B and the two 9 cm sides meet at D, so each pair of equal sides is a pair of neighbours rather than a pair of opposites, and the stem rules out any parallel sides. The quadrilateral with two pairs of equal adjacent sides and no parallel sides is a kite. Answer: kite. The distractors: a rhombus is chosen by candidates who see two pairs of equal sides and read that as all four sides being equal, which the two different lengths of 6 cm and 9 cm rule out; a parallelogram is chosen by candidates who remember that a parallelogram has two pairs of equal sides but not that in a parallelogram the equal sides are the opposite ones, and who pass over the statement that nothing is parallel; an isosceles trapezium is chosen by candidates who notice that the shape is symmetrical about the line BD and treat symmetry on its own as the mark of a trapezium, when a trapezium needs a pair of parallel sides.
- (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.
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