Mathematics • Year 10 • Unit 3 • Lesson 8

Congruence in Everyday Shapes

Apply the four congruence tests (SSS, SAS, AAS, RHS) from Lesson 8 to real geometric shapes — parallelograms, kites, rhombuses, isosceles triangles and roof trusses. For each, identify the test, write the formal proof, and use CPCTC (corresponding parts of congruent triangles are congruent) to deduce unknown sides or angles.

Apply · Real-World Maths

1. Word problems

For each problem: name the two triangles, list matching parts with reasons, state the test, then use CPCTC where asked. A bare answer with no reasons only earns half marks.

1.1 — Roof truss. A triangular roof truss has two equal sloping rafters AB and AC meeting at the apex A, and a horizontal tie BC. A vertical strut AM is built from A down to the midpoint M of BC.

(a) Write a formal proof that △ABM ≡ △ACM.
(b) Hence, show that AM is perpendicular to BC. 3 marks

Stuck? AB = AC (given), BM = CM (M is midpoint), AM = AM (common side). What test fits? Then ∠AMB = ∠AMC by CPCTC, and they sum to 180° on a straight line.

1.2 — Rhombus diagonals. ABCD is a rhombus (a parallelogram with all four sides equal). Diagonal BD is drawn.

(a) Write a formal proof that △ABD ≡ △CBD.
(b) Hence, explain why diagonal BD bisects ∠ABC and ∠ADC. 3 marks

Stuck? AB = CB (all sides of rhombus equal), AD = CD (same), BD = BD (common). SSS. Then ∠ABD = ∠CBD by CPCTC.

1.3 — Isosceles triangle base angles. In △PQR, PQ = PR. Drop a perpendicular from P to the midpoint M of QR.

(a) Write a formal proof that △PQM ≡ △PRM.
(b) Use CPCTC to deduce that ∠Q = ∠R (the famous "base angles of an isosceles triangle are equal" result). 3 marks

Stuck? PQ = PR (given), QM = RM (M is midpoint), PM = PM (common). SSS. Then ∠Q = ∠R by CPCTC.

1.4 — Plot of land. A surveyor checks a rectangular plot ABCD by measuring both diagonals AC and BD.

(a) Using opposite sides of a rectangle being equal, write a formal proof that △ABC ≡ △BAD.
(b) Hence explain why the two diagonals AC and BD must be equal in length. 3 marks

Stuck? AB = BA (common), BC = AD (opposite sides of rectangle), ∠ABC = ∠BAD = 90° (corners of rectangle). What test? Then AC = BD by CPCTC.

1.5 — Garden gate hinge. A garden gate is hinged so that the two triangular bracing panels on either side share a common vertical post. Each panel has the same upper horizontal beam (1.2 m), the same lower horizontal beam (1.2 m), and the same diagonal brace (1.5 m).

(a) Which congruence test proves the two bracing panels are identical?
(b) Write the formal proof. 3 marks

Stuck? Each panel is a triangle with sides 1.2 m, 1.2 m and 1.5 m. Three sides equal → SSS.

2. Explain your thinking

This question is about reasoning, not just numbers. Use full sentences. 4 marks

2.1 A friend says: "If two triangles have the same three angles, they must be congruent." Using Lesson 8's tests, explain (i) what is true about the friend's claim, (ii) what is fundamentally missing, (iii) name the geometric concept that two-equal-angles-only describes (without congruence), and (iv) state which extra piece of information would convert "same three angles" into a valid congruence proof, and what the proof would be called. Use the word "AAS" somewhere in your answer.

Stuck? Three equal angles give similarity, not congruence. To upgrade to congruence you need one corresponding side as well (then it's AAS).

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — Roof truss

(a) In triangles ABM and ACM:
AB = AC (given — equal rafters)
BM = CM (M is midpoint of BC)
AM = AM (common side)
∴ △ABM ≡ △ACM (SSS).
(b) By CPCTC, ∠AMB = ∠AMC. But ∠AMB + ∠AMC = 180° (angles on a straight line BC). So 2∠AMB = 180°, giving ∠AMB = 90°. Therefore AM ⊥ BC.

1.2 — Rhombus diagonal

(a) In triangles ABD and CBD:
AB = CB (all sides of rhombus equal)
AD = CD (all sides of rhombus equal)
BD = BD (common side)
∴ △ABD ≡ △CBD (SSS).
(b) By CPCTC, ∠ABD = ∠CBD, so BD bisects ∠ABC. Also ∠ADB = ∠CDB, so BD bisects ∠ADC.

1.3 — Isosceles base angles

(a) In triangles PQM and PRM:
PQ = PR (given)
QM = RM (M is midpoint of QR)
PM = PM (common side)
∴ △PQM ≡ △PRM (SSS).
(b) By CPCTC, ∠Q = ∠R. This proves the base angles of any isosceles triangle are equal.

1.4 — Rectangle diagonals

(a) In triangles ABC and BAD:
AB = BA (common side)
BC = AD (opposite sides of rectangle)
∠ABC = ∠BAD = 90° (corners of rectangle)
∴ △ABC ≡ △BAD (SAS — two sides and the included right angle).
(b) By CPCTC, AC = BD. So the two diagonals of a rectangle are equal.

1.5 — Garden gate bracing

(a) SSS — all three sides are equal.
(b) Calling the left panel △ABC and the right panel △DEF, with the shared post as one of the sides:
AB = DE = 1.2 m (upper beams)
BC = EF = 1.2 m (lower beams)
AC = DF = 1.5 m (diagonal braces)
∴ △ABC ≡ △DEF (SSS).

2.1 — Explain your thinking (sample response)

(i) The friend is correct that two triangles with the same three angles have the same shape. (ii) What is missing is any information about size — you cannot tell whether one triangle is a tiny version of the other or a giant version. The four valid congruence tests (SSS, SAS, AAS, RHS) all include at least one side length precisely because shape alone does not fix size. (iii) Same-three-angles describes similar triangles, not congruent ones. (iv) To upgrade to a valid congruence proof we need one pair of corresponding sides to be equal. Once we add that, two angles plus a corresponding side is the AAS test.

Marking: 1 for noting shape is fixed (the true part), 1 for identifying size as missing, 1 for naming "similar triangles", 1 for adding a side and naming AAS.