Mathematics • Year 7 • Unit 3 • Lesson 18

Geometric Reasoning — Multi-step Problems

Build fluency with multi-step angle problems. Each step needs value + named reason: e.g. (∠ sum of △) or (alt ∠s, AB ∥ CD). No reason = no mark.

Build · I Do / We Do / You Do

1. I do — fully worked example

Read every line. Each step has a NAMED reason in brackets — that's what the marker wants.

Problem. Two parallel lines are crossed by a transversal that makes an angle of 65° on top. A triangle drops below, sharing one side with the transversal. Its other base angle is 80°. Find x, the third angle of the triangle.

Step 1 — Label and plan.

Known: 65° on top line, 80° base angle of triangle. Unknown: x (top angle of triangle).

Plan: use alternate angles to bring 65° INSIDE the triangle, then use triangle angle sum.

Step 2 — Use alternate angles (Z-shape).

Angle inside the triangle = 65° (alt ∠s, lines parallel)

Reason: alternate angles on parallel lines are equal. The 65° on top equals the angle at the same Z-position inside the triangle.

Step 3 — Apply triangle angle sum.

x + 65 + 80 = 180 (∠ sum of △)

Reason: interior angles of any triangle add to 180°.

Step 4 — Solve.

x = 180 − 65 − 80 = 35°

Sanity-check: 35 + 65 + 80 = 180 ✓.

Answer: x = 35°. Two reasons chained: alternate angles + triangle sum.

Stuck? Revisit lesson § Card 5 "Parallel Lines + Triangles" — Z = alt, F = corr, C = co-int.

2. We do — fill in the missing steps

Same structure as Section 1, with the working faded. Fill in each blank. 4 marks

Problem. In quadrilateral ABCD, ∠A = 95°, ∠B = 110°, ∠C = 75°. Find ∠D. Then in △ABD: ∠A = 95°, ∠ABD = 50°. Find ∠ADB.

Step 1 — Use quadrilateral angle sum to find ∠D:

∠A + ∠B + ∠C + ∠D = 360° (∠ sum of quad)

95 + 110 + 75 + ∠D = _______

∠D = 360 − 95 − 110 − 75 = _______ °

Step 2 — Switch to triangle ABD. State the angle sum:

∠A + ∠ABD + ∠ADB = _______ ° (∠ sum of △)

Step 3 — Substitute the known values:

_______ + _______ + ∠ADB = 180°

Step 4 — Solve for ∠ADB:

∠ADB = 180 − 95 − 50 = _______ °

Stuck? Revisit lesson § Watch Me Solve It · Q2 — same problem, fully worked.

3. You do — independent practice

Show your working under each problem. Each step must have a written reason. The first four are foundation, the middle two are standard, and the last two are extension.

Foundation — one fact, one step

3.1 A triangle has angles 60°, 50° and x. Find x, with reason. 1 mark

3.2 Co-interior angles between two parallel lines: one is 70°. Find the other. 1 mark

3.3 A pentagon has interior angle sum equal to ___ degrees. Show how you got it. 1 mark

3.4 A quadrilateral has angles 100°, 90°, 75° and x. Find x. 1 mark

Standard — two facts chained

3.5 A triangle has angles 45°, x, 2x. Find x. 2 marks

3.6 A regular hexagon. Find one interior angle, showing all working. 2 marks

Extension — push your thinking

3.7 Two parallel lines are crossed by a transversal. One angle made with the upper line is 50°. A triangle drops below the lower line, sharing one side with the transversal. Its other base angle is 70°. Find the third angle of the triangle. Show TWO reasons in your working (alt + triangle sum). 3 marks

3.8 A regular pentagon has one of its diagonals drawn. The triangle formed is isosceles (two equal sides, because the pentagon is regular). The apex angle of the triangle is one interior angle of the pentagon. Find the apex angle and each base angle of the isosceles triangle. 3 marks

Stuck on 3.8? Pentagon interior angle = 540 ÷ 5 = 108°. Then in the isosceles triangle: 180 − 108 = 72 split between two equal base angles.

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

Section 2 — We do (quadrilateral ABCD then △ABD)

Step 1: 95 + 110 + 75 + ∠D = 360. So ∠D = 80°.
Step 2: ∠A + ∠ABD + ∠ADB = 180° (∠ sum of △).
Step 3: 95 + 50 + ∠ADB = 180.
Step 4: ∠ADB = 35°.

3.1 — Triangle angle

x = 180 − 60 − 50 = 70°   (∠ sum of △).

3.2 — Co-interior partner

Co-interior angles add to 180°: other = 180 − 70 = 110°   (co-int ∠s, lines parallel).

3.3 — Pentagon angle sum

Sum = (n − 2) × 180 = (5 − 2) × 180 = 3 × 180 = 540°.

3.4 — Quadrilateral missing angle

x = 360 − 100 − 90 − 75 = 95°   (∠ sum of quad).

3.5 — Algebra in triangle

45 + x + 2x = 180 (∠ sum of △).
3x = 135, so x = 45°. The angles are 45°, 45°, 90°.

3.6 — Regular hexagon interior angle

Sum = (6 − 2) × 180 = 4 × 180 = 720°.
Each angle = 720 ÷ 6 = 120°.

3.7 — Parallel + triangle

Step 1: angle inside triangle = 50° (alt ∠s, lines parallel).
Step 2: third angle = 180 − 50 − 70 = 60° (∠ sum of △).

3.8 — Pentagon diagonal triangle

Pentagon interior angle = (5 − 2) × 180 ÷ 5 = 540 ÷ 5 = 108°.
Apex of isosceles triangle = 108°. Base angles together = 180 − 108 = 72° (∠ sum of △).
Each base angle = 72 ÷ 2 = 36°.