Printable · GCSE Higher · ages 14-16
Congruence and similarity: lengths, areas and volumes worksheet — GCSE Higher
Fifteen questions on "congruence and similarity: lengths, areas and volumes" — DfE statement G19. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
part Higher
Congruence and similarity: lengths, areas and volumes worksheet — GCSE Higher
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- 1.Two similar ponds have perimeters in the ratio 4 : 11. The perimeter of the larger pond is 88 m. Work out the perimeter of the smaller pond.
- 2.A drinks company sells a novelty giant bottle that is geometrically similar to its standard 500 ml bottle, scaled up by a length factor of 1.5. Work out the capacity of the giant bottle, in ml.
- 3.A model of a bridge is built to a scale where every length on the model is 1/240 of the matching length on the real bridge. A support beam on the model measures 3.6 cm. Work out the length of the real support beam, giving your answer in metres.
- 4.Two similar signs have lengths in the ratio 9 : 4. The larger sign has a side of length 14.4 cm. Work out the length of the corresponding side of the smaller sign.
- 5.Two similar triangular tiles have lengths in the ratio 3 : 5. The longest side of the smaller tile is 9 cm. Work out the length of the longest side of the larger tile.
- 6.A shop sells a scale model of a van. The model and the real van are similar shapes, with lengths in the ratio 1 : 24. The model is 18.5 cm long. Work out the length of the real van, giving your answer in metres.
- 7.Two similar hexagonal tiles have lengths in the ratio 4 : 7. A side of the smaller tile is 8.4 cm long. Work out the length of the corresponding side of the larger tile.
- 8.Two similar bottles have heights in the ratio 1 : 2. Write down the ratio of their volumes.
- 9.In triangle ABC and triangle DEF, the angle at A and the angle at D are both 40°, and the angle at B and the angle at E are both 70°. No side lengths are given. Write down whether the two triangles must be similar, with the reason.
- 10.A photocopier enlarges a document so that every length is multiplied by the same scale factor. A line on the original document is 3.2 cm long, and the same line measures 11.2 cm on the enlarged copy. A second line on the original document is 2.5 cm long. Work out the length of the second line on the enlarged copy, giving your answer to 2 decimal places.
- 11.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.
- 12.Two similar company logos are printed at different sizes. The area of the larger logo is 2.25 times the area of the smaller logo. Work out the length scale factor from the smaller logo to the larger logo.
- 13.Two spheres are similar solids. Their radii are in the ratio 3 : 4. Write down the ratio of their surface areas.
- 14.Triangle JKL has a right angle at K, hypotenuse JL = 13 cm, and side JK = 5 cm. Triangle MNO has a right angle at N, hypotenuse MO = 13 cm, and side MN = 5 cm. Using only the facts given, and without working out any further lengths, write down the congruence condition that proves triangle JKL is congruent to triangle MNO.
- 15.Two similar shapes have lengths in the ratio 1 : 3. A side of the smaller shape is 4 cm long. Work out the length of the corresponding side of the larger shape.
Answer key
- (a) 32 m — The scale factor from the larger pond to the smaller pond is 4 ÷ 11, so the smaller perimeter is 88 × 4 ÷ 11 = 32 m. The distractor 242 m comes from using the ratio the wrong way round, 88 × 11 ÷ 4 = 242. The distractor 84 m comes from subtracting the smaller ratio number, 88 − 4 = 84, instead of scaling. The distractor 121 m comes from multiplying 11 × 11 = 121, ignoring the given perimeter altogether.
- (d) 1687.5 ml — The length scale factor from the standard bottle to the giant bottle is 1.5, so the volume scale factor is 1.5³ = 3.375. The capacity of the giant bottle is 500 × 3.375 = 1687.5 ml. 750 ml comes from multiplying by the length factor 1.5 directly, without cubing it. 1125 ml comes from using the area scale factor 1.5² = 2.25 instead of the volume scale factor. 2250 ml comes from treating 'cubed' as 'multiplied by 3', giving 1.5 × 3 = 4.5 as the scale factor instead of 1.5³.
- (b) 8.64 m — The real length is 240 times the model length: 3.6 × 240 = 864 cm. Converting to metres, 864 cm ÷ 100 = 8.64 m. The distractor 864 m comes from finding the length correctly in centimetres but forgetting to convert to metres. The distractor 0.0864 m comes from dividing by 100 a second time, converting 864 cm to metres twice over (864 ÷ 10 000) instead of once. The distractor 86.4 m comes from converting centimetres to metres by dividing by 10 instead of 100.
- (d) 6.4 cm — The scale factor from the larger sign to the smaller sign is 4 ÷ 9, so the smaller side is 14.4 × 4 ÷ 9 = 6.4 cm. The distractor 32.4 cm comes from using the ratio the wrong way round, 14.4 × 9 ÷ 4 = 32.4. The distractor 9.4 cm comes from subtracting the difference between the ratio numbers (9 − 4 = 5) from the given length, 14.4 − 5 = 9.4. The distractor 3.6 cm comes from dividing the given length by 4 only, 14.4 ÷ 4 = 3.6, without also using the other ratio number.
- (c) 15 cm — Method: going from the smaller tile to the larger one, every length is multiplied by the scale factor 5 ÷ 3. Working: 9 ÷ 3 = 3, and 3 × 5 = 15. Answer: 15 cm. The distractors: 11 cm comes from adding the difference between the parts of the ratio, 5 − 3 = 2, to the 9 cm side, treating a ratio as a gap rather than a multiplier; 45 cm comes from multiplying by 5 and forgetting to divide by 3; 27 cm comes from multiplying by 3, which is the part of the ratio belonging to the smaller tile.
- (d) 4.44 m — The real van's length is 18.5 × 24 = 444 cm, which converts to 4.44 m. '0.77 m' comes from dividing by 24 instead of multiplying, using the ratio the wrong way round (18.5 ÷ 24 = 0.77 cm), and then writing that figure down as metres. '44.4 m' comes from converting 444 cm to metres with the decimal point in the wrong place. '444 m' comes from working out 444 cm correctly but forgetting to convert it into metres at all.
- (c) 14.7 cm — The scale factor from the smaller tile to the larger tile is 7 ÷ 4 = 1.75, so the larger side is 8.4 × 1.75 = 14.7 cm. '4.8 cm' comes from scaling by 4 ÷ 7 instead, using the ratio the wrong way round. '11.4 cm' comes from adding the difference between the ratio numbers, 7 − 4 = 3, onto 8.4, instead of scaling. '58.8 cm' comes from multiplying by 7 on its own, using a ratio number as the scale factor instead of working out 7 ÷ 4 = 1.75 first.
- (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.
- (d) Yes, by the AA condition — Method: similarity is decided by the angles, and because the three angles of a triangle add up to 180°, two matching pairs force the third pair to match as well. Working: the angles at A and D are both 40° and the angles at B and E are both 70°, so the third angles are both 180° − 40° − 70° = 70° and all three pairs are equal. Two pairs were enough, and that is the AA condition. Answer: yes, by the AA condition. The distractors: 'Yes, by the SSS condition' names a condition about three pairs of sides in proportion, and the question gives no side lengths at all; 'No, because no side lengths are given' treats side information as necessary, which it is for congruence but not for similarity; 'No, because the triangles may be different sizes' turns the definition of similarity into an objection, since similar figures are allowed to differ in size and only their shape must match.
- (d) 8.75 cm — The scale factor of the enlargement is 11.2 ÷ 3.2 = 3.5. Applying this to the second line, 2.5 × 3.5 = 8.75 cm. The distractor 10.50 cm comes from adding the difference between the first line's two lengths (11.2 − 3.2 = 8) to the second line's original length, 2.5 + 8 = 10.5. The distractor 0.71 cm comes from using the scale factor the wrong way round, 2.5 × (3.2 ÷ 11.2) = 0.71 (to 2 d.p.). The distractor 8.70 cm comes from directly subtracting 11.2 − 2.5 = 8.7, muddling the two different lines instead of scaling the second one.
- (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.
- (c) 1.5 — The area scale factor is the length scale factor squared, so if n is the length factor, n² = 2.25. Taking the positive square root gives n = 1.5 (check: 1.5² = 2.25). 2.25 is just the area factor restated, with no root taken. 1.125 comes from halving 2.25 instead of taking its square root. 5.0625 comes from squaring 2.25 instead of rooting it.
- (c) 9 : 16 — Method: in similar solids the ratio of the surface areas is the square of the ratio of corresponding lengths. Working: the radii are in the ratio 3 : 4, so the surface areas are in the ratio 3² : 4² = 9 : 16. Answer: 9 : 16. The distractors: 27 : 64 is 3³ : 4³, the ratio of the volumes, which cubes the length ratio instead of squaring it; 3 : 4 leaves the length ratio untouched, as though surface area scaled in the same way as a length; 81 : 256 comes from squaring a second time, applying the rule to the ratio 9 : 16 rather than to the radii.
- (c) RHS — Both triangles are right-angled, and have equal hypotenuses (13 cm) and one equal corresponding side (5 cm), so they are congruent by the RHS (right angle, hypotenuse, side) condition — the three facts given are exactly a right angle, a hypotenuse and one other side. The distractor SAS would apply if two sides and the angle between those two sides were matched, but the right angle here lies between the 5 cm side and the third side, not between the 5 cm side and the hypotenuse. The distractor SSS needs all three pairs of sides matched, and only two sides of each triangle are given — the third side would first have to be calculated by Pythagoras, so SSS is not the condition the given facts supply. The distractor AAS would need two angles and a non-included side matched, but only one angle in each triangle is given.
- (b) 12 cm — Method: in similar shapes every length is multiplied by the same scale factor, and a ratio of 1 : 3 means that factor is 3 going from the smaller shape to the larger one. Working: 4 × 3 = 12. Answer: 12 cm. The distractors: 7 cm comes from adding 3 to the side instead of multiplying by it, which is what happens when a ratio is read as a difference; 36 cm comes from multiplying by 3² = 9, the factor that scales areas, and applying it to a length; 4 cm comes from treating the two shapes as congruent, so that corresponding sides stay equal — similar shapes have equal angles, but their sides are in proportion.
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