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.

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Answer key: Congruence criteria for triangles worksheet — GCSE Foundation

Maths · GCSE Foundation · G5 · Congruence criteria for triangles · 15 questions

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(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.
(b) SAS — Method: check which condition matches two sides and the angle between them, since that is what has been measured for each bed. Working: the 60° angle is marked at the corner where the 5 m and 7 m edges meet, so it is the included angle — this is two Sides and the included Angle, SAS. Options: SSS would need a third side measured, but only two edges are known; ASA would need two angles and the side between them, but only one angle is measured; RHS needs a right angle, and 60° is not a right angle. Answer: SAS.
(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) RHS, using AM as common side — Triangle ABM and triangle ACM both have a right angle at M, since AM is perpendicular to BC. AB and AC are the hypotenuses of the two triangles and are equal, and AM is a side common to both triangles, giving a right angle, equal hypotenuses and one further equal side, exactly RHS, so 'RHS, using AM as common side' is correct. 'SAS, right angle as included angle' wrongly treats the right angle at M as included between AB and AM, but AB is the hypotenuse, not one of the two sides forming that right angle. 'SSS, using BM = CM as a fact' wrongly assumes BM equals CM as a given fact, when this is only true because of the RHS congruence, not before it, so it cannot be used to prove that congruence. 'ASA, AB as the included side' again wrongly labels a side as if it could sit between two angles when only one angle, the right angle, is actually known.
(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) RHS - right angle, hypotenuse and one side equal — A right angle, the hypotenuse (13 cm) and one other side (5 cm) are equal in both triangles, so this is RHS. SAS would need the equal angle to be the one INCLUDED between the two equal sides, but the right angle at B is not between AB and the hypotenuse AC — it is opposite the hypotenuse instead, so SAS does not apply directly here. SSS needs all three sides, but only two sides are stated. ASA needs two angles, but only one angle (the right angle) is given.
(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.
(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.
(c) RHS — Method: check which basic congruence condition matches the facts given — a right angle, the hypotenuse, and one other side, in both triangles. Working: both triangles have a right angle (at M and Q), the hypotenuse is given for both (LN = PR = 15 cm), and one other side is given for both (LM = PQ = 9 cm) — this is exactly Right angle, Hypotenuse, Side. Options: SAS would need the given angle to sit between the two given sides, but the right angle at M is not between LM and LN, since LN is the hypotenuse, opposite the right angle; SSS would need three sides given in each triangle, but only two sides are known here; ASA would need two angles and the side between them, but only one angle is given. Answer: RHS.
(a) Similar, but AAA alone does not prove congruence — Three equal corresponding angles (AAA) show that the two triangles are similar — the same shape — but says nothing about their size, so it does not prove congruence. They could be congruent, or one could simply be an enlargement of the other; without matching side lengths, congruence is not established, so 'congruent because AAA proves congruence' is wrong. Equal angles do not force equal sides — a triangle can be enlarged to any size while keeping the same angles, so that option is also wrong. Something CAN be said here — that the triangles are similar — so 'no relationship can be determined' is wrong too.
(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 - 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.
(d) All angles equal and all sides equal — Congruent triangles are identical in both shape and size, so every corresponding angle is equal and every corresponding side is equal, so 'all angles equal and all sides equal' is correct. 'All angles equal, sides may differ' describes 'similar' triangles, which have equal angles but sides in the same ratio rather than necessarily equal, not congruent ones. 'All sides equal, angles may differ' is not geometrically possible for triangles, since equal sides throughout force the angles to match too. 'Same area, angles and sides may differ' is wrong because equal area alone does not guarantee congruence; two triangles can share an area with completely different shapes.
(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) ASA - two angles and the included side equal — Two angles (A and B) are given, and AB is the side between them, so this is ASA. SAS needs two sides and the angle between them, but only one side is given. AAS also uses two angles and a side, but the side must NOT be between the two angles — here AB is between angle A and angle B, so it is ASA, not AAS. RHS needs a right angle, and neither 40° nor 65° is 90°.

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