Lesson 20 · FINAL ~30 min Unit 3 · Geometry +100 XP

Geometry Synthesis and Review

The grand finale. We sweep through every idea from Unit 3 — angles, triangles, quadrilaterals, parallel lines, polygons, congruence, similarity and constructions — and bring them together in mixed problems. One unit, one big picture.

Today's hook: Twenty lessons, one big geometry toolkit. Time to combine them all and finish like a pro.

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Think First

warm-up

Take 3 minutes. From memory alone, list FIVE different geometric facts you've learned this unit (one for each: angle, triangle, quadrilateral, parallel lines, similarity). Don't peek!

Record your answer in your workbook.

The Big Picture: 20-Lesson Summary

+5 XP

Unit 3 has built a complete toolkit for Stage 4 plane geometry. Below is a quick tour of every lesson:

L1–L2: Points, lines, rays; types of angles (acute, right, obtuse, straight, reflex, revolution).
L3–L4: Angles at a point, on a straight line, vertically opposite.
L5–L6: Parallel lines — alternate, corresponding, co-interior angles.
L7–L9: Quadrilaterals — classifying, angle sum, special properties.
L10–L12: Triangles — classifying, angle sum, exterior angles, isosceles/equilateral properties.
L13–L14: Polygons — sum $(n-2)\times 180^{\circ}$, regular polygons.
L15: Congruence — SSS, SAS, AAS, RHS.
L16–L17: Similarity — scale factor, missing sides, real-world applications.
L18: Multi-step geometric reasoning.
L19: Constructions — perpendicular bisector, angle bisector, perpendicular from a point.
L20: Synthesis — today.

The whole unit is held together by ONE big idea: angles and sides are linked. Equal sides give equal angles; equal angles give equal sides; parallel lines transfer angles between locations; similar shapes preserve angles while scaling sides. Every fact in this unit is a special case of that linkage.

All facts: angles and sides are linked

Always write reasons

Every angle = ... gets a named justification.

Mark the diagram

Known angles pencilled in lead you to the chain.

Sanity-check

Do angles add to expected totals? Are ratios consistent?

What You'll Master

objectives

Know

Every angle fact from the unit (named & written as reasons)
Properties of every triangle and quadrilateral type
The four congruence tests and the similarity test
How to identify which fact applies to a given diagram

Understand

Why congruence is a "special case" of similarity
How constructions PROVE properties (e.g. SSS)
How to pick the most efficient fact for a problem

Can Do

Solve a mixed problem combining 3+ different topics
Justify every step with a written reason
Choose between scale factor and proportion methods

Vocab Megalist

vocabulary

Acute / Obtuse / ReflexAngle types: $< 90^{\circ}$ / $>90^{\circ}$ & $<180^{\circ}$ / $> 180^{\circ}$.

Vertically oppositeEqual angles formed by two crossing lines.

Alternate (Z) / Corresponding (F) / Co-int (C)Parallel-line angle pairs.

Isosceles / Equilateral / ScaleneTriangle types by side count.

Square / Rectangle / Parallelogram / Rhombus / Trapezium / KiteSix special quadrilaterals.

Polygon angle sum$(n - 2) \times 180^{\circ}$.

$\equiv$ CongruentSame shape AND size.

$\sim$ SimilarSame shape, any size; sides in ratio.

Angle Facts Cheatsheet

+5 XP

Every named reason you might use, all in one place:

$\angle$s on a straight line $= 180^{\circ}$
$\angle$s at a point $= 360^{\circ}$
Vertically opposite $\angle$s equal
Alt $\angle$s, $AB \parallel CD$: equal
Corr $\angle$s, $AB \parallel CD$: equal
Co-int $\angle$s, $AB \parallel CD$: add to $180^{\circ}$
$\angle$ sum of $\triangle = 180^{\circ}$
Ext $\angle$ of $\triangle$ = sum of 2 opp int $\angle$s
$\angle$ sum of quad $= 360^{\circ}$
Polygon $\angle$ sum $= (n - 2) \times 180^{\circ}$

If you can match the diagram to ONE of these ten facts, you have a step. Multi-step problems chain 2–4 of these in sequence. Memorise this list — it's literally the menu.

Ten facts — the entire toolbox

Memorise the names

Use the exact wording in your reasons.

Match to diagram

Scan diagram — which of the ten patterns can you see?

Chain them

For multi-step problems, several facts in sequence.

Book notes · Angle facts cheatsheet

Straight line, point, vertically opposite.
Parallel-line trio: alt, corr, co-int.
Triangle sum, quadrilateral sum, polygon sum.

Choose the correct named reason for: "An angle equals the angle at the same position on a parallel line."

Shape Properties Snapshot

+5 XP

Quick recall of defining properties:

Equilateral $\triangle$: 3 equal sides, 3 equal $60^{\circ}$ angles.
Isosceles $\triangle$: 2 equal sides, 2 equal base angles.
Scalene $\triangle$: All sides and angles different.
Right $\triangle$: One $90^{\circ}$ angle.
Square: 4 equal sides + 4 right angles.
Rectangle: 4 right angles.
Parallelogram: 2 pairs parallel sides.
Rhombus: 4 equal sides.
Trapezium: Exactly 1 pair parallel sides (NSW).
Kite: 2 pairs adjacent equal sides.

For congruence: use SSS, SAS, AAS, or RHS. For similarity: equal angles + sides in ratio. A square is the MOST special quadrilateral — it's also a rectangle, rhombus, parallelogram. An equilateral triangle is the MOST special triangle — it's also isosceles, acute.

Most specific name wins — "square" before "rectangle"

SSS, SAS, AAS, RHS

The four ways to prove triangle congruence.

$\sim$ keeps angles, scales sides

Similarity preserves angles; sides multiply by SF.

Specific names

Go as deep into the family tree as the facts allow.

Book notes · Shape properties

Triangle types by sides: equilateral, isosceles, scalene.
Triangle types by angles: acute, right, obtuse.
Six special quadrilaterals + the family hierarchy.

True or false?

Every square is also a rectangle, a rhombus AND a parallelogram.

What I've Learned

+5 XP

The complete set of skills from Unit 3. Tick them off mentally as you go:

Naming and classifying angles, triangles, quadrilaterals and polygons
Using angle facts to find unknown angles in single and multi-step diagrams
Recognising and applying parallel-line angle relationships
Calculating polygon angle sums with $(n - 2) \times 180^{\circ}$
Identifying congruent triangles (SSS, SAS, AAS, RHS)
Identifying similar figures and using scale factors
Setting up and solving proportions for missing sides
Solving real-world problems involving maps, models and shadows
Writing reasons after every step of working
Performing basic constructions: perpendicular bisector, angle bisector, perpendicular from a point
Combining multiple facts in chains of geometric reasoning

You now have every tool needed to tackle any Stage 4 plane-geometry problem in NSW outcomes. Bring them all together when you face a new diagram: label, plan, solve, justify.

L — P — S — J

Use the routine

L-P-S-J works for every problem.

Practice mixed

Single-topic problems are warm-ups — real exams combine topics.

Trust the toolkit

You've got every fact you need.

Book notes · What I've Learned

L-P-S-J: Label, Plan, Solve, Justify.
Use exact named reasons.
Chain facts for multi-step problems.

A polygon with $n$ sides has interior angle sum equal to $(n - 2) \times$ __________ degrees.

Watch Me Solve It · 3 examples

Watch Me Solve It · Mixed reasoning

+15 XP per step

PROBLEM

A parallelogram $ABCD$ has $\angle A = 70^{\circ}$. The diagonal $AC$ makes an angle of $30^{\circ}$ with side $AB$. Find $\angle ACB$.

1
Find $\angle B$ first
$\angle B = 180 - 70 = 110^{\circ}$ (co-int $\angle$s, $AD \parallel BC$).
2
Triangle $ABC$
In $\triangle ABC$: $\angle BAC = 30^{\circ}$, $\angle ABC = 110^{\circ}$, $\angle ACB = ?$
3
Triangle angle sum
$\angle ACB = 180 - 30 - 110 = 40^{\circ}$ ($\angle$ sum of $\triangle$).
Two facts chained: co-int + triangle sum.

Answer$\angle ACB = 40^{\circ}$.

Watch Me Solve It · Similar triangles in a real problem

+15 XP per step

PROBLEM

Two similar triangles. $\triangle ABC \sim \triangle DEF$. $AB = 8$, $BC = 12$, $AC = 14$. $DE = 20$. Find $EF$ and the scale factor from $ABC$ to $DEF$.

1
Find SF
SF $= \dfrac{DE}{AB} = \dfrac{20}{8} = 2.5$
2
Apply SF to $BC$
$EF = 12 \times 2.5 = 30$
3
Check with the third side
$DF = 14 \times 2.5 = 35$ (consistent).
$EF = 30$, SF $= 2.5$.

Answer$EF = 30$, SF $= 2.5$.

Watch Me Solve It · Polygon + triangle combined

+15 XP per step

PROBLEM

A regular hexagon has one diagonal drawn from a vertex to a non-adjacent vertex, forming an isosceles triangle. Find the apex angle (at the original vertex) and the two base angles of the triangle.

1
Hexagon angles
Each interior angle $= \dfrac{(6 - 2) \times 180}{6} = 120^{\circ}$.
2
Diagonal cuts the angle
The diagonal from one vertex to the next-but-one vertex splits the $120^{\circ}$ apex into the triangle plus an extra angle.
3
Isosceles triangle
In the triangle: apex $= 120^{\circ}$, base angles $= \frac{180 - 120}{2} = 30^{\circ}$ each.
Polygon formula + isosceles property combined.

AnswerApex $= 120^{\circ}$, base angles $= 30^{\circ}$ each.

Common Pitfalls

Common Pitfalls (Across the Unit)

heads-up

No reason = no marks

Even a correct numerical answer loses marks if the named reason is missing.

Fix: Every angle line gets a parenthesised reason: e.g. ($\angle$ sum of $\triangle$).

Mixing similar and congruent

$\equiv$ means identical; $\sim$ means same shape any size. Confusing them changes the whole problem.

Fix: Identical (same size) = $\equiv$; scaled = $\sim$.

Wrong vertex order in similarity statement

$\triangle ABC \sim \triangle DEF$ means $A \to D$, etc. Wrong order leads to wrong side ratios.

Fix: Match by equal angles, then write vertices in that order.

Using a protractor in a construction question

Constructions use ONLY straight-edge and compasses — no measuring with degrees.

Fix: Use only arcs and straight lines through identified points.

Copy Into Your Books

Angles

Str line $= 180^{\circ}$
At a point $= 360^{\circ}$
Vert opp equal
Alt, corr equal; co-int $= 180^{\circ}$

Polygons

$\triangle = 180^{\circ}$
Quad $= 360^{\circ}$
$n$-gon $= (n-2) \times 180^{\circ}$
Regular: each $= \frac{(n-2)\times 180}{n}$

$\equiv$ vs $\sim$

$\equiv$ SSS, SAS, AAS, RHS
$\sim$ equal $\angle$s + sides in ratio
SF $=$ new $\div$ old

Constructions

Perp bisector: same-radius arcs
Angle bisector: SSS proves it
Perp from a point: extension of perp bisector

How are you completing this lesson?

Brain Trainer · 4 mixed problems

Brain Trainer · Mixed Review

4 problems

Four mixed problems combining topics across the unit.

1 A regular decagon. Find one interior angle.

Sum $= 8 \times 180 = 1440$. Each $= 1440/10$.$144^{\circ}$
2 Two similar triangles. Scale factor $1.5$. Original sides $6, 8, 10$. New sides?

Multiply each by $1.5$.$9, 12, 15$
3 A trapezium $ABCD$ has $AB \parallel CD$, $\angle A = 65^{\circ}$. Find $\angle D$.

Co-int: $180 - 65$.$115^{\circ}$
4 An isosceles triangle has apex $40^{\circ}$. Find each base angle.

$(180 - 40)/2$.$70^{\circ}$

Complete in your workbook.

Quick Check · 5 mixed questions

A right triangle has one acute angle of $35^{\circ}$. The other acute angle is:

+10 XP

One interior angle of a regular hexagon is:

+10 XP

$\triangle ABC \sim \triangle DEF$. $AB = 6$, $BC = 9$, $DE = 18$. Find $EF$.

+10 XP

Which set of features uniquely defines a square?

+10 XP

Bisecting a right angle creates two angles each of:

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

Apply Easy 3 MARKS

Q6. A parallelogram has one angle of $115^{\circ}$.
(a) Find the angle adjacent to it (with reason).
(b) Find the angle opposite to it (with reason).
(c) Sum-check all four angles.

Answer in your workbook.

Apply Medium 3 MARKS

Q7. A flag pole casts a $9$ m shadow. At the same time, a $1.2$ m tall student casts a $1.5$ m shadow.
(a) Explain why the situation forms similar triangles.
(b) Set up the proportion.
(c) Calculate the height of the flagpole.

Answer in your workbook.

Reason Hard 3 MARKS

Q8. Multi-step: a triangle is inscribed between two parallel lines $\ell_1 \parallel \ell_2$. One side of the triangle makes a $55^{\circ}$ angle with $\ell_1$. Another side makes a $70^{\circ}$ angle with $\ell_2$ on the opposite side of the triangle. The third angle of the triangle is $x$.
(a) Find the angle inside the triangle that's alternate to $55^{\circ}$ (with reason).
(b) Find the angle inside the triangle that's corresponding to $70^{\circ}$ (with reason).
(c) Use $\angle$ sum of $\triangle$ to find $x$.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. C — $55^{\circ}$.

2. A — Regular hexagon = $120^{\circ}$.

3. D — SF = 3, $EF = 27$.

4. B — 4 equal sides AND 4 right angles.

5. A — $45^{\circ}$.

Show Your Working Model Answers

Q6 (3 marks): (a) $180 - 115 = 65^{\circ}$ (co-int $\angle$s, opp sides parallel) [1]. (b) Opposite angle $= 115^{\circ}$ (opp $\angle$s of parallelogram equal) [1]. (c) $115 + 65 + 115 + 65 = 360^{\circ}$ ($\angle$ sum of quad) ✓ [1].

Q7 (3 marks): (a) Same sun angle means the two object+shadow triangles are similar (AAA equivalent) [1]. (b) $\frac{h}{1.2} = \frac{9}{1.5}$ [1]. (c) $h = 1.2 \times 6 = 7.2$ m [1].

Q8 (3 marks): (a) Inside angle $= 55^{\circ}$ (alt $\angle$s, $\ell_1 \parallel \ell_2$) [1]. (b) Inside angle $= 70^{\circ}$ (corr $\angle$s, $\ell_1 \parallel \ell_2$) [1]. (c) $x = 180 - 55 - 70 = 55^{\circ}$ ($\angle$ sum of $\triangle$) [1].

Stretch Challenge · +30 XP, +15 coins

Boss Battle: Mixed Geometry

A square $ABCD$ is divided by both diagonals, meeting at $O$. From $O$, you draw a line perpendicular to $AB$ meeting $AB$ at $M$. (a) Name the type of triangle $AOB$ and justify with two facts. (b) Find $\angle OAB$ and $\angle OBA$. (c) What is the relationship between $OM$ and $AB$? Explain using the perpendicular bisector concept. (d) If the diagonals of the square have length $10$ cm, find the length of $OM$.

Reveal solution

(a) $\triangle AOB$ is isosceles right-triangle: diagonals of a square are equal and bisect each other, so $OA = OB$ (isosceles); they cross at right angles in a square, so $\angle AOB = 90^{\circ}$. (b) Each base angle $= (180 - 90)/2 = 45^{\circ}$. (c) $OM$ is the perpendicular from $O$ to $AB$. Because $OA = OB$, point $O$ lies on the perpendicular bisector of $AB$, so $OM$ passes through the midpoint of $AB$. (d) Diagonals length $10$ cm, so $OA = OB = 5$ cm. $\triangle AOB$ is isosceles right-angled with legs $5$ — $AB$ is the hypotenuse and $OM$ is the perpendicular height from the right-angle to the hypotenuse. By similarity, $OM = 2.5$ cm (half the diagonal). Alternative: since the square has diagonal $10$, its side $= \frac{10}{\sqrt{2}} \approx 7.07$ cm, and $OM = \frac{1}{2} \times \frac{10}{\sqrt{2}} \approx 3.54$ cm.

Final Recap · The Whole Unit

Angles

Str line $=180^{\circ}$, point $=360^{\circ}$, vert opp equal.

Parallel lines

Alt $=$, corr $=$, co-int add to $180^{\circ}$.

Triangles

Sum $=180^{\circ}$; equilateral/isosceles/scalene.

Quadrilaterals

Sum $=360^{\circ}$; six special shapes.

Polygons

Sum $=(n-2)\times 180^{\circ}$.

Congruence

SSS, SAS, AAS, RHS — identical shapes.

Similarity

Same shape, scale factor; missing-side problems.

Reasoning

Label, plan, solve, justify — written reasons every step.

Constructions

Perp bisector, angle bisector, perp from a point.

What I've Learned — Master Checklist

unit complete

L1–L2: Points, lines, rays, angle types.
L3–L4: Angles on a straight line, at a point, vertically opposite.
L5–L6: Parallel lines: alternate, corresponding, co-interior.
L7: Introducing quadrilaterals; angle sum $= 360^{\circ}$.
L8–L9: Parallelograms, rectangles, rhombuses, kites and trapeziums.
L10: Triangle types (sides and angles).
L11: Triangle angle sum.
L12: Exterior angle of a triangle.
L13: Polygon angle sum.
L14: Regular polygons.
L15: Congruent triangles (SSS, SAS, AAS, RHS).
L16: Introduction to similar figures.
L17: Finding missing sides in similar figures (maps, models, shadows).
L18: Multi-step geometric reasoning.
L19: Constructions: bisecting angles and lines.
L20: Synthesis and review — this lesson!

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