Mathematics • Year 7 • Unit 3 • Lesson 20
Geometry Synthesis and Review
Build fluency across every Unit 3 topic: angles at a point/line, parallel-line angles, triangles, quadrilaterals, polygons, congruence, similarity and constructions. Use the L–P–S–J routine: Label, Plan, Solve, Justify.
1. I do — fully worked example
Read every line. This is a typical Unit 3 synthesis problem combining a parallelogram and a triangle.
Problem. A parallelogram ABCD has ∠A = 70°. The diagonal AC makes an angle of 30° with side AB. Find ∠ACB.
Step 1 (Label) — Identify what's given and what's asked.
Given: parallelogram ABCD, ∠A = 70°, ∠BAC = 30°. Find: ∠ACB.
Step 2 (Plan) — Decide which facts to use.
Plan: (i) Use co-interior angles on parallel sides AD ∥ BC to find ∠B. (ii) Use ∠ sum of △ABC to find ∠ACB.
Step 3 (Solve) — Find ∠B.
∠A + ∠B = 180° (co-int ∠s, AD ∥ BC)
∠B = 180 − 70 = 110°
Step 4 (Solve) — Apply triangle angle sum in △ABC.
∠BAC + ∠ABC + ∠ACB = 180° (∠ sum of △)
30 + 110 + ∠ACB = 180
∠ACB = 180 − 30 − 110 = 40°
Step 5 (Justify) — Reasons used:
co-interior angles (parallel sides) + triangle angle sum.
Answer: ∠ACB = 40°.
2. We do — fill in the missing steps
Same structure as Section 1, with the working faded. Fill in each blank. 4 marks
Problem. A regular hexagon has one of its diagonals drawn from one vertex to a non-adjacent vertex. The diagonal forms an isosceles triangle whose apex angle is one interior angle of the hexagon. Find the apex angle and each base angle of the isosceles triangle.
Step 1 — Find the polygon angle sum:
Sum = (n − 2) × 180° where n = _______ , so sum = (___ − 2) × 180 = _______ °
Step 2 — Find each interior angle of the regular hexagon:
Each angle = _______ ÷ 6 = _______ °
Step 3 — Apex of the isosceles triangle = one interior angle of the hexagon:
Apex = _______ °
Step 4 — Use triangle angle sum to find the two equal base angles:
Base angles together = 180 − _______ = _______ °. Each base angle = _______ ÷ 2 = _______ °.
3. You do — independent practice
Show your working under each problem. Every step gets a named reason. The first four are foundation, the middle two are standard, and the last two are extension.
Foundation — single topic
3.1 A right triangle has one acute angle of 35°. Find the other acute angle (with reason). 1 mark
3.2 A regular decagon has interior angle sum equal to ___° and each interior angle equal to ___° . 1 mark
3.3 Two similar triangles. SF = 1.5. Original sides 6, 8, 10. Find the new side lengths. 1 mark
3.4 An isosceles triangle has an apex of 40°. Find each base angle. 1 mark
Standard — combine two topics
3.5 A trapezium ABCD has AB ∥ CD and ∠A = 65°. Find ∠D, with reason. 2 marks
3.6 △ABC ∼ △DEF. AB = 6, BC = 9, AC = 12. DE = 9. Find EF, DF, and the scale factor from ABC to DEF. 2 marks
Extension — push your thinking
3.7 In a square ABCD, draw the diagonal AC. Identify the triangle △ABC: what type is it? Find each of its interior angles. 2 marks
3.8 A regular pentagon has all its diagonals drawn from ONE vertex. This splits the pentagon into THREE triangles. Confirm that the three triangle angle sums add to the same total as the pentagon's interior angle sum, and explain WHY this works in general (any n-gon split into n − 2 triangles from one vertex). 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (regular hexagon diagonal)
Step 1: n = 6, sum = (6 − 2) × 180 = 720°.
Step 2: Each angle = 720 ÷ 6 = 120°.
Step 3: Apex = 120°.
Step 4: Base angles together = 180 − 120 = 60°. Each base = 60 ÷ 2 = 30°.
3.1 — Right triangle acute angle
Other acute = 180 − 90 − 35 = 55° (∠ sum of △).
3.2 — Regular decagon
Sum = (10 − 2) × 180 = 1440°. Each interior angle = 1440 ÷ 10 = 144°.
3.3 — Similar triangle with SF = 1.5
6 × 1.5 = 9; 8 × 1.5 = 12; 10 × 1.5 = 15. New sides: 9, 12, 15.
3.4 — Isosceles base angles
Each base = (180 − 40) ÷ 2 = 140 ÷ 2 = 70°.
3.5 — Trapezium co-interior
∠A and ∠D are co-interior on AB ∥ CD, so they add to 180°.
∠D = 180 − 65 = 115° (co-int ∠s, AB ∥ CD).
3.6 — Similar triangles
SF (ABC → DEF) = DE ÷ AB = 9 ÷ 6 = 1.5.
EF = 9 × 1.5 = 13.5. DF = 12 × 1.5 = 18.
3.7 — Diagonal of a square
The diagonal AC splits the square into two right-angled isosceles triangles. △ABC has a right angle at B (∠ABC = 90°) and the two sides AB = BC (sides of the square).
So △ABC is a right-angled isosceles triangle.
Angles: ∠ABC = 90°; the other two angles are equal (isosceles base angles) and sum to 180 − 90 = 90°, so each = 45°.
Final: ∠ABC = 90°, ∠BAC = 45°, ∠BCA = 45°.
3.8 — Pentagon split into triangles
From one vertex of a pentagon you can draw diagonals to the 2 non-adjacent vertices. This creates 3 triangles.
Sum of all triangle angles: 3 × 180 = 540°.
Pentagon interior angle sum: (5 − 2) × 180 = 540°. They match ✓.
Why it works: the triangles together cover the WHOLE inside of the pentagon. Every interior angle of the pentagon is made up of one or more triangle angles at that vertex, so the total of all triangle angles = total of all pentagon interior angles. For an n-gon, you can split it into (n − 2) triangles from one vertex, so the sum is (n − 2) × 180° — which is exactly the polygon angle-sum formula.