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.
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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!
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.
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
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.
Book notes · Angle facts cheatsheet
- Straight line, point, vertically opposite.
- Parallel-line trio: alt, corr, co-int.
- Triangle sum, quadrilateral sum, polygon sum.
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.
Book notes · Shape properties
- Triangle types by sides: equilateral, isosceles, scalene.
- Triangle types by angles: acute, right, obtuse.
- Six special quadrilaterals + the family hierarchy.
Every square is also a rectangle, a rhombus AND a parallelogram.
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.
Book notes · What I've Learned
- L-P-S-J: Label, Plan, Solve, Justify.
- Use exact named reasons.
- Chain facts for multi-step problems.
Watch Me Solve It · 3 examples
- 1Find $\angle B$ first$\angle B = 180 - 70 = 110^{\circ}$ (co-int $\angle$s, $AD \parallel BC$).
- 2Triangle $ABC$In $\triangle ABC$: $\angle BAC = 30^{\circ}$, $\angle ABC = 110^{\circ}$, $\angle ACB = ?$
- 3Triangle angle sum$\angle ACB = 180 - 30 - 110 = 40^{\circ}$ ($\angle$ sum of $\triangle$).Two facts chained: co-int + triangle sum.
- 1Find SFSF $= \dfrac{DE}{AB} = \dfrac{20}{8} = 2.5$
- 2Apply SF to $BC$$EF = 12 \times 2.5 = 30$
- 3Check with the third side$DF = 14 \times 2.5 = 35$ (consistent).$EF = 30$, SF $= 2.5$.
- 1Hexagon anglesEach interior angle $= \dfrac{(6 - 2) \times 180}{6} = 120^{\circ}$.
- 2Diagonal cuts the angleThe diagonal from one vertex to the next-but-one vertex splits the $120^{\circ}$ apex into the triangle plus an extra angle.
- 3Isosceles triangleIn the triangle: apex $= 120^{\circ}$, base angles $= \frac{180 - 120}{2} = 30^{\circ}$ each.Polygon formula + isosceles property combined.
Common Pitfalls
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
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.
$(180 - 40)/2$.$70^{\circ}$
Quick Check · 5 mixed questions
Show Your Working · 3 questions
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.
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.
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 $x$.
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].
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.
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.
- 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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