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: Angles, lines and geometry foundations.
L2–L3: Classify triangles by their sides and by their angles.
L4–L6: Triangle angle sum, exterior angles and triangle problem solving.
L7–L10: Quadrilaterals, their special properties and the quadrilateral angle sum.
L11–L12: Parallel lines and transversals; use parallel-line angle properties.
L13: Interior and exterior angle sums in polygons.
L14–L15: Introduce congruent figures, then apply triangle congruence tests.
L16–L17: Introduce similar figures and use scale factors to find missing sides.
L18: Multi-step geometric reasoning.
L19: Constructions: bisect angles and lines.
L20: Synthesis and review, this lesson.

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. B4 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 are $10$ cm, so $OA=OB=5$ cm and $\triangle AOB$ is right-angled at $O$. First find the hypotenuse: $AB=\sqrt{5^2+5^2}=5\sqrt{2}$ cm. The area of $\triangle AOB$ is both $\frac12\times5\times5=12.5$ cm$^2$ and $\frac12\times AB\times OM$. Therefore $OM=25/(5\sqrt2)=5/\sqrt2\approx3.54$ cm. This agrees with $OM=AB/2$, since the altitude from the right angle of an isosceles right triangle also bisects its hypotenuse. The value is approximately $3.54$ 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: Angles, lines and geometry foundations.
L2–L3: Triangle types by sides and by angles.
L4–L6: Triangle angle sum, exterior angles and triangle problem solving.
L7–L10: Quadrilaterals, special quadrilaterals and their angle sum.
L11–L12: Parallel lines, transversals and angle properties.
L13: Polygon angle sums.
L14–L15: Congruent figures and applying congruence.
L16–L17: Similar figures and finding missing sides.
L18: Multi-step geometric reasoning.
L19: Constructions: bisect angles and lines.
L20: Synthesis and review, this lesson!

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