Printable · GCSE Higher · ages 14-16
Congruence criteria for triangles worksheet — GCSE Higher
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.
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Congruence criteria for triangles worksheet — GCSE Higher
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- 1.Triangle ABC has AB = (2x + 3) cm, BC = 11 cm and angle B = 90°. Triangle DEF has DE = 13 cm, EF = 11 cm and angle E = 90°. Given that the two triangles are congruent by SAS, work out the value of x.
- 2.Triangle ABC has AB = 9 cm, BC = 6 cm and angle A = 35°. Triangle DEF has DE = 9 cm, EF = 6 cm and angle D = 35°. Which of the following correctly identifies the condition shown here?
- 3.Triangle ABC has AB = 6 cm and BC = 9 cm, with angle A = 40°. Triangle DEF has DE = 6 cm and EF = 9 cm, with angle D = 40°. Meera says these facts prove the triangles are congruent by SAS. Is Meera correct?
- 4.A carpenter cuts two triangular metal brackets for a shelf. Bracket P has sides of 12 cm, 16 cm and 20 cm. Bracket Q has sides of 16 cm, 20 cm and 12 cm, listed in a different order. Before fitting them as a matching pair, the carpenter wants to check they are exactly the same shape and size. Which of the following correctly justifies this?
- 5.Triangles PQS and RQS share the common side QS. PQ equals RQ, and PS equals RS. Which condition proves that triangle PQS is congruent to triangle RQS?
- 6.A carpenter measures two triangular braces. Brace 1 has a 55° angle, a 65° angle, and, opposite the 55° angle, a side of 8 cm. Brace 2 has matching 55° and 65° angles and, opposite its 55° angle, a side of 8 cm. The carpenter says that because the equal 8 cm side is not between the two equal angles in either brace, he cannot be sure the two braces are the same size. Is the carpenter correct?
- 7.A sailmaker cuts two triangular sail panels from a pattern. Panel X has a 10 m side, with angles of 50° and 75° at its two ends. Panel Y also has a 10 m side, with angles of 50° and 75° at its two ends, in the same arrangement. The sailmaker wants to check the panels will match exactly, without cutting a third measurement. Which condition proves the two panels are congruent?
- 8.Two triangles are proved congruent using the ASA condition. What can be concluded about the two remaining pairs of corresponding sides that were not part of the original ASA facts?
- 9.Triangle ABC and triangle DEF both contain a right angle, at B and E respectively, and the hypotenuses AC and DF are equal in length. Which extra fact would prove the triangles are congruent by RHS?
- 10.Two metal window frames are checked before installation. Frame A has edges of 40 cm and 65 cm, with a marked angle of 35° at the far end of the 65 cm edge, away from where the two given edges meet. Frame B has matching edges of 40 cm and 65 cm, with a matching 35° angle in the same position. The fitter wants to know if these measurements alone guarantee the two frames are congruent. What should the fitter be told?
- 11.Triangles ABE and CDE share the vertex E, where lines AC and BD cross at E. AE equals CE, and BE equals DE. Angle AEB and angle CED are formed as vertically opposite angles where the lines cross. Which condition proves that triangle ABE is congruent to triangle CDE?
- 12.Triangle ABC has AB = 8 cm, BC = 10 cm and CA = 6 cm. Triangle LMN has LM = 10 cm, MN = 6 cm and NL = 8 cm. The two triangles are congruent by SSS. Which of the following correctly matches the corresponding vertices?
- 13.Triangle ABC has AB = 10 cm, BC = 7 cm and angle A = 40°. Triangle DEF has DE = 10 cm, EF = 7 cm and angle D = 40°. Sam says the two triangles are congruent by SAS. Which statement about Sam's claim is correct?
- 14.Triangle ABC has a right angle at B and hypotenuse AC = 17 cm, with AB = 8 cm. Triangle DEF has a right angle at E and hypotenuse DF = 17 cm. Which extra fact about DEF would complete the RHS condition for proving that ABC ≅ DEF, with the vertices matching in that order?
- 15.Triangle ABC has AB = 6 cm, BC = 8 cm and angle B = 90°. Triangle XYZ has XY = 6 cm, YZ = 8 cm and angle Y = 90°. Which congruence statement correctly shows the matching vertices?
Answer key
- (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.
- (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.
- (b) No — the angle given is not the included angle — Method: check whether the given angle sits between the two given sides, since SAS requires the included angle. Working: sides AB and BC meet at vertex B, so the angle between them is angle B — but the angle given is angle A, which is not between the two given sides, and the same mismatch happens in triangle DEF. Options: 'two sides and one angle match' restates SAS's ingredients without checking their positions, which is exactly Meera's mistake; 'SSS needs three equal sides' is a true fact about a different condition, but it is not the reason Meera is wrong here; 'SAS allows any equal angle' states a rule that is not how SAS works, since the angle must be the included one. Answer: no, the angle given is not the included angle.
- (a) Yes - SSS, the three side lengths all match — Bracket P's sides (12 cm, 16 cm, 20 cm) can each be matched to one of Bracket Q's sides (16 cm, 20 cm, 12 cm) — the same three lengths, just listed differently — so the brackets are congruent by SSS. SAS is wrong here because no angle is stated for either bracket, only three sides. The order the sides are listed in does not matter for SSS — only whether the SET of three lengths matches, and it does, so 'listed in a different order' is not a reason to say no. Nothing extra is needed: SSS proves congruence from side lengths alone, without any angles, so it can be determined.
- (d) SSS, using shared side QS — PQ equals RQ and PS equals RS are two given pairs of equal sides, and QS is common to both triangles, so QS equals itself and gives a third pair of equal sides. Three pairs of equal sides is exactly the SSS condition, so 'SSS, using shared side QS' is correct. 'SAS, using the angle at Q' is wrong because no angle is given anywhere in this question; angle PQS and angle RQS are not stated to be equal, and assuming they are would be assuming the very thing being proved. 'Only two pairs of sides — not enough' is wrong because it forgets that the shared side QS is itself a third pair of equal sides. 'Cannot prove — no angle given' is wrong because SSS is one of the four basic congruence conditions and specifically requires no angle at all.
- (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.
- (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.
- (c) They must also be equal — Once two triangles are proved congruent by any condition, including ASA, they are identical in every respect: every pair of corresponding sides and every pair of corresponding angles must be equal, not just the ones originally used to prove the congruence. So the two remaining pairs of corresponding sides must also be equal, making 'they must also be equal' correct. 'They might be equal or not' and 'not enough information to say' both wrongly suggest that congruence only guarantees the specific facts used to prove it, when congruence actually guarantees the triangles are identical overall. 'They must be different' is backwards: the triangles being identical is the entire point of proving congruence, not a reason for a side to differ.
- (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.
- (a) No — the angle is not between the two sides — Method: check whether the given angle sits between the two given sides. Working: the 35° angle is marked away from the corner where the 40 cm and 65 cm edges meet, so it is not the included angle — this is SSA, which is not one of the four basic congruence conditions, so it does not guarantee congruence. Options: 'Yes — SAS' wrongly treats any two sides plus any angle as SAS, without checking the angle's position; 'Yes — SSS' wrongly counts the angle as if it were a third side; 'No — only two sides measured' is not the real reason, since SAS itself only needs two sides, so this reasoning is beside the point. Answer: no, the angle is not between the two sides.
- (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.
- (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) Wrong - the given angle is not the included angle — Sides AB and BC meet at vertex B, so the included angle needed for SAS is angle B, not angle A — the information given is SSA. SSA does not prove congruence: with AB = 10 cm, BC = 7 cm and angle A = 40° there are two different triangles that fit, one with angle C ≈ 74.6° and one with angle C ≈ 105.4°, so Sam's triangles need not be the same shape and size at all. Sam is not correct just because two sides and an angle are equal, since the angle must be the one INCLUDED between those two sides. The condition is not ASA either, because ASA needs two angles, and only one angle is given here. It is also not true that nothing matches — the stated lengths and angle DO match between the two triangles; the problem is which angle was given, not whether the values agree.
- (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) 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.
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