Mathematics • Year 7 • Unit 3 • Lesson 15

Applying Congruence — Real World

Tents, kites, washing lines, brick walls, and surveying triangles. Use proved congruent triangles to read off lengths or angles that you cannot measure directly.

Apply · Real-World Maths

1. Word problems

For each: state the congruence (which test) THEN deduce the unknown side or angle, citing "(matching … of ≡ △s)".

1.1 — Symmetric tent. A pyramid tent has a square base ABCD and apex P directly above the centre of the base. The four sloping sides PA, PB, PC and PD are all equal (3 m each). The four base edges are all 2 m. (a) Prove △PAB ≡ △PCB (use SSS — note PB is shared). (b) Deduce that the two sloping triangular panels meet at angles ∠PAB = ∠PCB.    3 marks

Stuck? In △PAB and △PCB: PA = PC (3 m), AB = CB (2 m), PB = PB (common). SSS.

1.2 — Kite. A kite ABCD has AB = AD and CB = CD. The diagonal AC is drawn. (a) Prove △ABC ≡ △ADC. (b) Deduce that AC bisects ∠BAD (i.e. ∠BAC = ∠DAC).    3 marks

Stuck? AB = AD (given), CB = CD (given), AC = AC (common). SSS.

1.3 — Washing line. Two clothes posts AB and CD are vertical (both perpendicular to the ground), the same height (1.8 m), and 4 m apart. A washing line is strung from top of post A to top of post C. The line and the two posts plus the ground form a rectangle. The diagonal AC is drawn. Prove △ABC ≡ △CDA (use the diagonal as the common side). Then deduce AB = CD (which we already know) using your congruence.    3 marks

Stuck? In the rectangle ABCD, AB ∥ DC and AD ∥ BC. Use alternate angles + common diagonal AC → AAS.

1.4 — Surveying across a river. A surveyor wants to know the distance AB across a river without crossing it. She picks a point P on her side of the river, then walks to another point Q so that PQ = AP and angle APB = angle QPC, where C is a point she can see on her side that lines up correctly. After construction she gets two congruent triangles △APB ≡ △CPQ with CQ measurable on her side. She measures CQ = 18 m. What is AB?    2 marks

Stuck? Matching sides of ≡ △s: AB matches CQ. So AB = 18 m.

1.5 — Brick wall pattern. A bricklayer cuts a rectangular brick exactly along its diagonal into two right triangles. The two right triangles have legs 6 cm and 12 cm. (a) Prove the two triangles are congruent. (b) Deduce that the two diagonal cuts (which are the hypotenuses) are equal in length.    2 marks

Stuck? Two sides + included 90° = SAS. Or: both right-angled with equal short sides + equal hypotenuses → RHS once you note the hypotenuse equals √(6² + 12²) in both.

2. Explain your thinking

Full sentences. 4 marks

2.1 An engineer designs a triangular brace for a bridge. She wants to mass-produce the brace so every copy is exactly the same shape and size. She tells the factory: "Just make sure every brace has the same three side lengths: 30 cm, 40 cm and 50 cm." In a short paragraph: (i) Explain why specifying the three side lengths IS enough to guarantee every brace is congruent. Name the congruence test used. (ii) Compare this to specifying only the three ANGLES — would that be enough? Why or why not? (iii) Cite the lesson rule for "once congruent, every matching part is equal".

Stuck? Three sides given = SSS, which is enough. Three angles only proves similar (same shape, possibly different size).

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — Symmetric tent

(a) In △PAB and △PCB:
PA = PC (= 3 m, given)
AB = CB (= 2 m, base edges of square)
PB = PB (common)
Three sides equal → △PAB ≡ △PCB (SSS).
(b) ∴ ∠PAB = ∠PCB (matching ∠s of ≡ △s).

1.2 — Kite

(a) In △ABC and △ADC:
AB = AD (given)
CB = CD (given)
AC = AC (common)
Three sides equal → △ABC ≡ △ADC (SSS).
(b) ∴ ∠BAC = ∠DAC (matching ∠s of ≡ △s). So AC bisects ∠BAD.

1.3 — Washing line / rectangle ABCD

In △ABC and △CDA:
∠BAC = ∠DCA (alt. ∠s, AB ∥ DC)
∠BCA = ∠DAC (alt. ∠s, AD ∥ BC)
AC = CA (common)
Two angles + common side → △ABC ≡ △CDA (AAS).
∴ AB = CD (matching sides of ≡ △s) — confirms both posts are equal height (1.8 m).

1.4 — Surveying across the river

From △APB ≡ △CPQ, AB matches CQ. Matching sides of ≡ △s are equal, so AB = CQ = 18 m.

1.5 — Brick cut diagonally

(a) Each right triangle has legs 6 cm and 12 cm with the 90° angle BETWEEN them → two sides + included angle → SAS. ∴ the two triangles are congruent (SAS).
(b) The two hypotenuses match (each is the diagonal of the rectangle). By matching sides of ≡ △s, the two diagonal cuts are equal in length. (Both equal √(6² + 12²) = √180 ≈ 13.42 cm.)

2.1 — SSS guarantees congruence; AAA does not (sample response)

Specifying three side lengths (30 cm, 40 cm, 50 cm) IS enough because three pairs of equal corresponding sides match the SSS congruence test from Lesson 14 — once SSS holds, the two triangles are guaranteed to be identical in both shape AND size. Specifying only the three angles is NOT enough: three pairs of equal angles only proves the triangles are SIMILAR (same shape) but they could be different sizes — a tiny brace and a huge brace can both have angles 37°, 53°, 90° (the angles of any 3-4-5 triangle). To force a specific size, you must give at least one side. Lesson 15 also tells us that "once two triangles are proved congruent, EVERY pair of corresponding sides and angles is automatically equal" — so SSS on the three sides also forces all three pairs of angles to match, guaranteeing the braces are interchangeable.

Marking: 1 for "SSS is enough"; 1 for the test name; 1 for "AAA only proves similar"; 1 for citing the "every matching part is equal" rule.