Mathematics • Year 7 • Unit 3 • Lesson 6
Triangle Problem Solving — Mixed Challenge
Bring together everything: angle sum (180°), exterior angle theorem (e = a + b), isosceles base angles, equilateral angles, and algebraic equations in angles. Spot a common Year 7 reasoning slip, then design your own multi-triangle puzzle.
1. Mixed problems — choose the right rule
Each question mixes the rules. Decide WHICH rule applies BEFORE you start writing. Show your working with a reason in brackets after each line. 2 marks each
1.1 A triangle has angles 35°, 75° and x°. Find x.
1.2 An isosceles triangle has apex 26°. Find each base angle.
1.3 The exterior angle at one vertex of a triangle is 130°. The two remote interior angles are equal. Find each of those interior angles.
1.4 A triangle has angles in the ratio 1 : 2 : 3. Find the three angles.
1.5 An equilateral triangle sits next to an isosceles triangle, sharing one side. The isosceles has apex 100°. Find each base angle of the isosceles triangle.
1.6 Triangles ABD and BCD share side BD. A, B, C are collinear. ∠DAB = 40°, ∠ABD = 90°. Find ∠ADB, ∠DBC, and (if ∠BCD = 30°) ∠BDC. Show every step on a new line.
2. Find the mistake
A Year 7 student tried to find the apex angle of an isosceles triangle with one base angle of 40°. Their working is shown below. Exactly one line contains a reasoning error. Spot it, explain why it's wrong, then redo the working correctly. 3 marks
Student's working — isosceles triangle with one base angle 40°:
Line 1: "An isosceles triangle has three equal angles, so the other two are 40° each."
Line 2: Apex + base + base = 180 (∠ sum of △).
Line 3: Apex + 40 + 40 = 180.
Line 4: Apex = 100°.
(a) Which line contains the reasoning error?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the correct final apex.
Stuck? Is it isosceles, or equilateral? Look at Line 1 — three equal angles would mean… ?3. Open-ended challenge — design your own diagram
This question has more than one correct answer. Show your work clearly. 4 marks
3.1 Design a diagram with TWO triangles that share one side and where ONE single given measurement of 50° is enough to find every other angle in both triangles.
Requirements: (i) label both triangles and every vertex; (ii) mark the 50° clearly; (iii) state any extra properties you are giving the triangles (e.g. "left triangle is isosceles with apex marked"); (iv) work out every other angle, with reason in brackets for each line.
Hint: You're allowed to choose properties like "equilateral", "isosceles", "right-angled" for one or both triangles. Choose properties that let one angle unlock everything.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — 35° + 75° + x = 180
x = 180 − 35 − 75 = 70° (∠ sum of △).
1.2 — Isosceles, apex 26°
Each base = (180 − 26) ÷ 2 = 154 ÷ 2 = 77° (base ∠s isos. + ∠ sum of △).
1.3 — Exterior 130°, two equal remote interiors
By the exterior angle theorem, the two remote interiors sum to 130 (ext. ∠ of △). Since they are equal, each = 130 ÷ 2 = 65°.
1.4 — Ratio 1 : 2 : 3
Let angles be x, 2x, 3x. x + 2x + 3x = 180 (∠ sum of △). 6x = 180, x = 30. Angles = 30°, 60°, 90°.
1.5 — Isosceles apex 100°
Each base = (180 − 100) ÷ 2 = 80 ÷ 2 = 40° (base ∠s isos. + ∠ sum of △). The neighbouring equilateral has angles of 60° each — they are independent of the isosceles.
1.6 — Two triangles sharing BD, ∠ABD = 90°
∠ADB = 180 − 40 − 90 = 50° (∠ sum of △ABD).
∠DBC = 180 − 90 = 90° (∠s on str. line at B).
∠BDC = 180 − 90 − 30 = 60° (∠ sum of △BCD).
2 — Find the mistake
(a) The error is on Line 1.
(b) An isosceles triangle has only TWO equal sides and TWO equal base angles — not three. (A triangle with three equal angles is an equilateral triangle, all 60°.) The student confused isosceles with equilateral, and that error then forces a wrong apex calculation that lands on a coincidentally-tidy 100°.
(c) Corrected working — isosceles, one base = 40°:
Line 1 (fixed): Other base = 40° (base ∠s isos.) — only TWO equal angles.
Line 2: Apex + 40 + 40 = 180 (∠ sum of △).
Line 3: Apex = 180 − 80 = 100°.
Final apex 100° is correct, but the reasoning on Line 1 had to be replaced — saying "three equal angles" is wrong even though it accidentally gave the right base angle here.
3 — Open-ended challenge (sample solution)
Sample design. Triangles ABD and BCD share side BD, with A, B, C collinear. Left triangle ABD is equilateral: so ∠BAD = ∠ABD = ∠ADB = 60°. Right triangle BCD is right-angled at B: so ∠DBC = 90°, with one other angle given as 50° at vertex C.
From the equilateral: all three angles = 60° (equilateral).
At B on the straight line: ∠ABD + ∠DBC = 180°, check: 60 + 90 = 150° → that contradicts straight line, so the design needs A, B, C NOT collinear. Adjust: place the equilateral and right-angled triangle simply sharing edge BD without a straight line constraint.
Then in △BCD: ∠DBC = 90°, ∠BCD = 50°, so ∠BDC = 180 − 90 − 50 = 40° (∠ sum of △).
Every angle is now known: 60°, 60°, 60° in the left triangle; 90°, 50°, 40° in the right triangle. The single given measurement (50°) plus the shape labels were enough to unlock everything.
Marking: 1 for a clear labelled diagram; 1 for choosing properties that let the 50° unlock all angles; 2 for correct working with a reason in brackets on every line. Accept any design where the 50° (combined with the labels) determines every other angle.