Printable · GCSE Foundation · ages 14-16
Congruence criteria for triangles worksheet — GCSE Foundation
Fifteen questions on "congruence criteria for triangles" — DfE statement G5. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Answer key: Congruence criteria for triangles worksheet — GCSE Foundation
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- (d) DE = 8 cm — Method: use the stated correspondence ABC ≅ DEF to work out which side in DEF matches the known side AB in ABC. Working: the correspondence sends A to D, B to E and C to F, so AB corresponds to DE; the right angles at B and E and the equal hypotenuses AC = DF = 17 cm are already given, so DE = 8 cm supplies the third ingredient — Right angle, Hypotenuse, Side. Options: EF = 8 cm matches AB to the wrong side, since EF corresponds to BC, and BC = √(17² − 8²) = 15 cm, not 8 cm; BC = 8 cm states something about triangle ABC rather than the missing fact about DEF, and it is false as well, since BC = 15 cm; angle D = angle A does follow once the triangles are congruent, but RHS is completed by a matching side, not by a matching angle. Answer: DE = 8 cm.
- (c) SSA — Method: recall the four basic congruence criteria for triangles and compare them with each option. Working: the four basic conditions are SSS, SAS, ASA and RHS. SSA (two sides and a non-included angle) is not one of them, because knowing two sides and an angle that is not between them does not always fix a unique triangle. Options: SSS, ASA and RHS are each genuine congruence conditions, so none of them is the correct answer to 'which is NOT'. Answer: SSA.
- (d) SSS - three sides equal — All three pairs of corresponding sides are equal in length (5 cm, 7 cm and 9 cm in both triangles), so the triangles are congruent by SSS. SAS needs an angle to be given as well as two sides, but no angle is given here. ASA needs two angles and the side between them, but no angles are given at all. RHS needs a right angle and the hypotenuse, but no angle is stated to be a right angle.
- (c) ABC ≅ XYZ — Method: match each vertex in ABC to its corresponding vertex in XYZ, using the equal sides and angles given, then write the letters in that matching order. Working: AB matches XY, BC matches YZ, and angle B matches angle Y, so A corresponds to X, B corresponds to Y, and C corresponds to Z, giving ABC ≅ XYZ. Options: 'ABC ≅ ZYX' puts Z in A's position, but A corresponds to X, not Z; 'ABC ≅ YXZ' puts Y in A's position, but A corresponds to X; 'ABC ≅ ZXY' puts Z in A's position and X in B's position, neither of which is correct. Answer: ABC ≅ XYZ.
- (d) No — third angle is also fixed — Since both braces have angles of 55° and 65°, their third angles must both be 60°, because angles in a triangle sum to 180°. All three angles now match, so the braces have the same shape. Both 8 cm sides lie in the same position relative to those angles — opposite the 55° angle in each brace — so one matching pair of corresponding sides fixes the size as well, exactly as ASA or AAS would. The braces are therefore guaranteed to be congruent and the carpenter is incorrect: 'No — third angle is also fixed' is correct. 'Yes — side must be included' is wrong because the side does not have to lie physically between the two named angles; once the third angle is fixed, a corresponding equal side anywhere is enough. 'No — any two angles enough alone' is wrong because two equal angles with no side length at all would only show the triangles are similar, not congruent. 'Yes — third angle may differ' is wrong because the third angle is fixed at 60° by the angle sum and cannot vary.
- (a) SSA - not sufficient to prove congruence — The angle given, angle A, is not the angle between sides AB and BC — it is not the included angle — so this data is SSA (side, side, angle), which is not sufficient to prove congruence on its own; two triangles can share this SSA information without being congruent. SAS is wrong because the given angle is not the one included between the two given sides. ASA is wrong because only one angle (A) is given, not two. AAS is wrong for the same reason — only one angle is given, not two.
- (d) 5 — Method: since the triangles are congruent by SAS, the corresponding sides AB and DE must be equal, because both are the side next to the given right angle that is not BC or EF. Working: AB = DE gives 2x + 3 = 13, so 2x = 10, so x = 5. Options: 10 comes from dropping the coefficient of x and solving x + 3 = 13 instead of 2x + 3 = 13; 4 comes from matching AB to the wrong side, EF, giving 2x + 3 = 11, so 2x = 8, so x = 4; 8 comes from a sign error, solving 2x − 3 = 13 instead of 2x + 3 = 13, giving 2x = 16, so x = 8. Answer: 5.
- (c) ASA — Method: check which condition matches two angles and the side between them, since that is all the sailmaker has measured. Working: the 10 m side lies between the 50° and 75° angles in both panels, so this is two Angles and the included Side, ASA. Options: SAS would need two sides and the angle between them, but only one side has been measured here; SSS would need three sides, but only one is known; RHS needs a right angle and a hypotenuse, and neither panel has a stated right angle. Answer: ASA.
- (d) SAS, vertically opposite angle included — AE equals CE and BE equals DE give two pairs of equal sides, and angle AEB equals angle CED because they are vertically opposite angles formed where AC and BD cross; vertically opposite angles are always equal without needing to be measured. This included angle sits between the two known sides in each triangle, giving SAS, so 'SAS, vertically opposite angle included' is correct. 'ASA, vertically opposite angle at E' is wrong because ASA needs two pairs of equal angles with the side between them, but only one angle is known in each triangle here, and the two other known facts are sides, not angles. 'SSS, three equal side pairs' is wrong because only two pairs of sides are given; there is no third pair of equal sides. 'Cannot prove — no angle measured' is wrong because vertically opposite angles are always equal automatically when two straight lines cross, so no separate measurement is needed.
- (b) AB = DE — RHS needs a right angle, the hypotenuse, and one OTHER side to be equal; the right angles and hypotenuses are already equal, so a matching pair of the remaining sides, AB = DE, completes RHS. Angle A = angle D is an extra ANGLE fact, not the extra SIDE fact that RHS specifically requires. AC being parallel to DF says nothing about either triangle's side lengths, so it cannot complete a congruence condition. Being drawn the same way up is about orientation on the page, not about any measurement, so it proves nothing about congruence.
- (c) SAS - two sides and the included angle equal — Two sides (AB and BC) and the angle between them (angle B) are equal in both triangles, so this is SAS. SSS would need all three sides given, but only two sides are stated. ASA needs two angles and the side between them, but only one angle is given. AAS needs two angles and a side, but again only one angle is given here.
- (a) Two sides equal, with the angle between them equal — Method: recall what each letter in SAS stands for, and the order they must appear in. Working: SAS stands for Side, Angle, Side — two pairs of equal sides, with the equal angle positioned between them. Options: 'two sides and any one angle' describes SSA, which is not valid, since the angle must specifically be the included one; 'two angles with the side between them' describes ASA, a different condition; 'all three pairs of sides' describes SSS, also different. Answer: two sides equal, with the angle between them equal.
- (b) Right angle, equal hypotenuse, one more side — RHS requires a right angle in each triangle, equal hypotenuses, and one further pair of equal sides, so 'right angle, equal hypotenuse, one more side' is correct. 'Right angle, hypotenuse, one more angle' is wrong because it asks for an extra equal angle instead of an extra equal side, which RHS does not check. 'Two sides and the angle between them' describes SAS, not RHS, so it is wrong. 'Right angle and all three sides equal' describes SSS with an extra right-angle condition, asking for more information than RHS actually needs, so it is wrong.
- (c) A↔N, B↔L, C↔M (ABC≅NLM) — Matching equal side lengths: AB (8 cm) equals NL (8 cm), BC (10 cm) equals LM (10 cm), and CA (6 cm) equals MN (6 cm). This gives the correspondence A with N, B with L, and C with M, so triangle ABC is congruent to triangle NLM, making 'A↔N, B↔L, C↔M (ABC≅NLM)' correct. 'A↔L, B↔M, C↔N (ABC≅LMN)' simply matches the vertices in the order they are written without checking the side lengths: AB (8 cm) would need to equal LM (10 cm), which is false. 'A↔M, B↔N, C↔L (ABC≅MNL)' also fails this check, since AB (8 cm) would need to equal MN (6 cm), which is false. 'A↔N, B↔M, C↔L (ABC≅NML)' gets A correct but swaps B and C, so AB (8 cm) would need to equal NM (6 cm), which is also false.
- (d) RHS — Both triangles have a right angle, their hypotenuses are equal, and one further pair of sides (AB and DE) are equal, exactly the RHS condition, so RHS is correct. SAS needs two sides and the angle between them; here the right angle at B is not between the hypotenuse and AB, so this is not a valid SAS setup, making SAS wrong. SSS needs three pairs of equal sides, but only two sides are given here, so SSS is wrong. ASA needs two angles and the side between them, but only one angle, the right angle, is given, so ASA is wrong.
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