Printable · GCSE Foundation · ages 14-16
Congruence and similarity: lengths, areas and volumes worksheet — GCSE Foundation
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 Foundation
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- 1.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.
- 2.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.
- 3.In triangle ABC and triangle DEF, AB = DE, AC = DF, and the angle at A equals the angle at D. Write down the congruence condition that proves the two triangles are congruent.
- 4.In triangle ABC and triangle DEF, the angle at A equals the angle at D, and AB ÷ DE = AC ÷ DF. Write down what this proves about the two triangles.
- 5.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.
- 6.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.
- 7.Two similar triangles have lengths in the ratio 3 : 4. The perimeter of the smaller triangle is 24 cm. Work out the perimeter of the larger triangle.
- 8.Two similar flags have widths in the ratio 2 : 5. The width of the smaller flag is 8 cm. Work out the width of the larger flag.
- 9.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.
- 10.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.
- 11.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.
- 12.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.
- 13.Triangle JKL and triangle MNP are similar, with JK ÷ MN = JL ÷ MP = 2. A student says triangle JKL must be congruent to triangle MNP. Write down why the student is wrong.
- 14.In triangle PQR and triangle STU, PQ = ST, QR = TU and PR = SU. Write down the congruence condition that proves the two triangles are congruent.
- 15.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.
Answer key
- (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.
- (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) SAS — Method: a congruence condition is named by the parts that are given equal and the order in which they sit round the triangle, so count the sides and the angles first. Working: AB = DE and AC = DF are two pairs of equal sides, and the equal angle at A and D lies between AB and AC, so the given parts read side, included angle, side. Answer: SAS. The distractors: SSS needs three pairs of equal sides, and the third pair, BC and EF, is not given — it follows from the proof rather than being part of it; ASA reads the two equal sides as two equal angles, swapping which facts are which; RHS applies only when the triangles contain a right angle and the equal pair includes the hypotenuse, and nothing here says the angle at A is 90°.
- (d) Similar, by SAS — Method: read off whether the given sides are equal or only in the same ratio, and where the given angle sits. Sides in proportion point to similarity; sides equal in length would be needed for congruence. Working: AB ÷ DE = AC ÷ DF gives two pairs of corresponding sides in the same ratio, and the equal angle at A and D lies between AB and AC in the first triangle and between DE and DF in the second, so it is the included angle. Two pairs of sides in proportion with the included angle equal is the SAS condition for similarity. Answer: similar, by SAS. The distractors: 'Similar, by AA' quotes a condition that needs two pairs of equal angles, while only one pair is given here; 'Congruent, by SAS' reads AB ÷ DE = AC ÷ DF as AB = DE and AC = DF, turning a statement about proportion into one about equal lengths; 'Congruent, by SSS' makes that same misreading and adds an assumption that the third pair of sides is equal too, which nothing in the question says.
- (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) 32 cm — Method: a perimeter is lengths added together, so it scales by the length scale factor itself, which is 4 ÷ 3 going from the smaller triangle to the larger one — not by its square. Working: 24 ÷ 3 = 8, and 8 × 4 = 32. Answer: 32 cm. The distractors: 18 cm comes from multiplying by 3 ÷ 4, scaling from the larger triangle down to the smaller one; 25 cm comes from adding the difference between the parts of the ratio, 4 − 3 = 1, to the perimeter; 8 cm comes from dividing by 3 and stopping there, before multiplying by 4.
- (a) 20 cm — The scale factor from the smaller flag to the larger flag is 5 ÷ 2 = 2.5, so the larger width is 8 × 2.5 = 20 cm. The distractor 3.2 cm comes from using the ratio the wrong way round, 8 × 2 ÷ 5 = 3.2. The distractor 11 cm comes from adding the difference between the ratio numbers (5 − 2 = 3) to the given width, 8 + 3 = 11. The distractor 40 cm comes from multiplying by 5 without dividing by 2 first, 8 × 5 = 40.
- (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) 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.
- (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.
- (c) The scale factor is 2, not 1, so the sides are not equal — Congruent shapes must be exactly the same size as well as the same shape, which means a scale factor of 1. Here the scale factor between the triangles is 2, so the sides are different lengths and the triangles cannot be congruent, even though they are similar. 'Similar triangles are never congruent' is too strong — a scale factor of exactly 1 would make them both similar and congruent. 'The angles are not necessarily equal' is wrong, since similar triangles always have equal matching angles. 'Congruent triangles must have a right angle' is an unrelated, false fact about congruence.
- (d) SSS — All three pairs of corresponding sides are equal (PQ = ST, QR = TU, PR = SU), so the triangles are congruent by the SSS (side-side-side) condition. The distractor SAS would apply if two sides and the included angle were given equal, not three sides. The distractor ASA would apply if two angles and the included side were given equal. The distractor RHS would apply only for right-angled triangles with the hypotenuse and one other side equal.
- (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.
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