Mathematics • Year 7 • Unit 3 • Lesson 18
Geometric Reasoning — Mixed Challenge
Combine triangle, quadrilateral, parallel-line and polygon facts in 2–3 step chains. Spot a wrong reason, then design your own multi-step problem.
1. Mixed problems
Each question mixes angle facts. Show your working — every line needs a named reason. 2 marks each
1.1 A triangle has angles 3x, 4x, 5x. Find each angle.
1.2 A regular nonagon (9-sided polygon). Find one interior angle.
1.3 Two parallel lines crossed by a transversal. Angles formed are a (upper) and b (corresponding lower). If a = 4x − 10 and b = 2x + 30, find x and the value of a.
1.4 An isosceles triangle has a base angle of 65°. Find the apex angle.
1.5 A trapezium ABCD has AB ∥ CD, ∠A = 110°, ∠B = 85°. Find ∠C and ∠D.
1.6 A quadrilateral has angles 90°, 80°, x and 2x. Find x.
2. Find the mistake
Another Year 7 student tried to find an angle inside parallel lines. Their working is shown below. Exactly one line contains a mistake (it's not the arithmetic — it's the named reason). Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Problem: Two parallel lines AB and CD are crossed by a transversal. A co-interior pair forms angles labelled p (upper) and q (lower). If p = 105°, find q.
Line 1: p and q are co-interior angles.
Line 2: Co-interior angles are equal (alt ∠s, AB ∥ CD).
Line 3: So q = p = 105°.
Line 4: Answer: q = 105°.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong (think about WHICH fact applies to co-interior angles).
(c) Write out the corrected working in full, with the right reason AND the correct numerical answer.
Stuck? Revisit lesson § Card 5 — Z = alt (equal), F = corr (equal), C = co-int (add to 180°). Co-interior angles are NOT equal.3. Open-ended challenge — design a multi-step problem
This question asks you to invent a problem of your own. Be clear and complete. 4 marks
3.1 Design a multi-step angle problem that uses at least TWO different named angle facts in its solution. Pick from this menu of facts: ∠ sum of △, ∠ sum of quad, alt ∠s on parallel lines, corr ∠s on parallel lines, co-int ∠s on parallel lines, vert opp ∠s, ∠s on str line, ∠ sum of polygon (n − 2) × 180°.
Your problem must include:
(i) A clear description of the figure (or a labelled sketch).
(ii) Given values (at least one angle and any other info).
(iii) A question that requires the chain of TWO facts.
(iv) A FULLY WORKED solution showing every step with its named reason.
Bonus: Make your problem use THREE different facts (e.g. co-int + triangle sum + isosceles base angles).
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Triangle angles 3x, 4x, 5x
3x + 4x + 5x = 180 (∠ sum of △). So 12x = 180, x = 15°.
Angles: 3(15) = 45°, 4(15) = 60°, 5(15) = 75°.
1.2 — Regular nonagon
Sum = (9 − 2) × 180 = 7 × 180 = 1260°. Each angle = 1260 ÷ 9 = 140°.
1.3 — Corresponding angles equation
Corresponding angles on parallel lines are equal, so a = b: 4x − 10 = 2x + 30. 2x = 40, so x = 20.
Then a = 4(20) − 10 = 70°.
1.4 — Isosceles triangle apex
In an isosceles triangle, the two base angles are equal. Both = 65°.
Apex = 180 − 65 − 65 = 50° (∠ sum of △).
1.5 — Trapezium ABCD, AB ∥ CD
∠A and ∠D are co-interior (AB ∥ CD), so ∠D = 180 − 110 = 70°.
∠B and ∠C are co-interior, so ∠C = 180 − 85 = 95°.
Check: 110 + 85 + 95 + 70 = 360 ✓.
1.6 — Quadrilateral with algebra
90 + 80 + x + 2x = 360 (∠ sum of quad). 170 + 3x = 360. 3x = 190. x = 63.3° (to 1 d.p.) or 190/3.
2 — Find the mistake
(a) The mistake is on Line 2.
(b) Co-interior angles on parallel lines are NOT equal — they add to 180° (supplementary). The student named the wrong fact ("alt ∠s") and also wrongly said they were equal. The correct reason for co-interior angles is "co-int ∠s, AB ∥ CD" and they add to 180°.
(c) Corrected working:
Line 1: p and q are co-interior angles. (unchanged)
Line 2 (fixed): Co-interior angles on parallel lines add to 180° (co-int ∠s, AB ∥ CD).
Line 3 (fixed): q = 180 − p = 180 − 105 = 75°.
Line 4 (fixed): Answer: q = 75°.
The fix: Z = alt (equal), F = corr (equal), C = co-int (supplementary, add to 180°).
3 — Open-ended challenge (sample solution)
Sample problem: Lines BC and DE are parallel. Point A lies above BC, with AB extended to E. ∠EAB = 50° (at A, on the upper transversal). Triangle ABC has ∠ABC inside it. ∠BCA = 70°. Find ∠BAC.
Worked solution:
Step 1: ∠ABC = 50° (alt ∠s, BC ∥ DE). The 50° at A on transversal DE matches the angle at B inside the triangle.
Step 2: In △ABC: ∠BAC + ∠ABC + ∠BCA = 180° (∠ sum of △).
Step 3: ∠BAC = 180 − 50 − 70 = 60°.
Two named facts used: alt ∠s + ∠ sum of △.
Bonus (3 facts): add an isosceles condition — "AB = AC, so ∠ABC = ∠ACB (base angles of isosceles △)" before solving. Then one isosceles fact + alt + sum = three.
Marking: 1 mark for clear figure description; 1 mark for given values; 1 mark for question that requires 2 facts; 1 mark for fully worked solution with both named reasons.