Five formulas, one strategy: identify the shape, find the perpendicular height, substitute — and always write the unit.
50–55 minMS-M13 MC3 SALesson 2 of 22Free
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Think First
You are tiling a bathroom floor. The floor is L-shaped — it is not a simple rectangle. How would you figure out how many tiles you need? What would your strategy be?
Type your initial response below — you will revisit this at the end of the lesson.
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Area Formulas — This Lesson
$A = \ell w$
Rectangle — $\ell$ = length, $w$ = widthFoundation formula — all other area formulas connect back to this
$A = \tfrac{1}{2}bh$
Triangle — $b$ = base, $h$ = perpendicular heightThe $\tfrac{1}{2}$ is always there — a triangle is half its enclosing rectangle
Trapezium — $a$, $b$ = parallel sides, $h$ = perpendicular heightThink: average the two parallel sides, then multiply by height
$A = \pi r^2$
Circle — $r$ = radius = diameter ÷ 2Use the $\pi$ button on your calculator — never substitute 3.14
Know
The area formula for rectangles, triangles, parallelograms, trapeziums and circles
What a composite shape is
The correct units for area answers
Understand
Why different shapes need different formulas — and what each part measures
How to break a complex shape into simpler parts to add or subtract
Why area is always expressed in square units
Can Do
Calculate the area of any standard shape by selecting and applying the correct formula
Find the area of composite shapes, including those requiring subtraction
Write every area answer with the correct unit
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Key Vocabulary
AreaThe amount of flat surface enclosed by a shape — always measured in square units (cm², m², etc.)
Perpendicular heightThe height measured at a right angle to the base — not the slant side of a triangle or parallelogram
Composite shapeA shape made by combining or removing two or more standard shapes
RadiusThe distance from the centre of a circle to its edge — half the diameter
Pi ($\pi$)The ratio of a circle's circumference to its diameter — use the $\pi$ button on your calculator, never 3.14
Misconceptions to Fix
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Wrong: Converting units only requires multiplying by 10.
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Right: Metric conversions use powers of 10, but area conversions use powers of 100 and volume uses powers of 1000.
Core Content
01
Why Area Formulas Work
Before memorising any formula, it helps to understand where it comes from. Every area formula connects back to the most basic one: the rectangle.
Rectangle: $A = \ell w$
This is the foundation. A rectangle with length 5 cm and width 3 cm contains exactly 15 unit squares. Area counts those squares.
Triangle: $A = \tfrac{1}{2}bh$
A triangle is exactly half a rectangle with the same base and height. Draw any triangle, complete the rectangle around it — the triangle fills exactly half. That is why the $\tfrac{1}{2}$ is always there.
Parallelogram: $A = bh$
Slice a triangle off one end of a parallelogram and reattach it to the other end — you get a rectangle with the same base and height. The area equals $bh$, identical to the rectangle formula.
Trapezium: $A = \tfrac{1}{2}(a + b)h$
A trapezium has two parallel sides of different lengths, $a$ and $b$. The formula averages them — $\tfrac{1}{2}(a + b)$ — then multiplies by the height. Think of it as a rectangle with an average width.
Circle: $A = \pi r^2$
The area of a circle depends on the square of its radius. Double the radius and the area quadruples — not doubles. The $r^2$ relationship is what makes circles behave so differently from other shapes.
Pattern check: All five formulas involve multiplying two lengths together in some way — which is why area is always in square units. Length × length = unit². A single length gives a linear unit (cm, m), but two lengths multiplied give a square unit (cm², m²).
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Shape Area Fluency Builder
Show full working for every question: formula → substitution → calculation → answer with unit.
02
The Perpendicular Height Problem
The single most common error across all of these formulas is using the slant side instead of the perpendicular height.
For a triangle or parallelogram, the height $h$ must be measured at 90° to the base. If the shape is drawn as a slanted figure and you are given the slant side, that is not $h$.
Rule: If you cannot draw a small square (right angle symbol) where the height meets the base, you are not using the correct height. The perpendicular height is sometimes drawn as a dotted line — this is deliberate. Use the dotted line, not the solid slant side.
What $h$ is
Dotted line from apex, perpendicular to base
Distance between the two parallel sides, measured at 90°
Perpendicular distance between the two parallel sides
The width — always perpendicular to the length
What $h$ is NOT
The slant side (hypotenuse or leg)
The slant side connecting the parallel sides
The non-parallel side
N/A — rectangles have no slant sides
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03
Composite Shapes — The 5-Step Strategy
A composite shape is any shape that is not one of the five standard shapes. The strategy is always the same.
Action
Column B
Subtraction example: A square with a circular hole cut from the centre.
Area = area of square − area of circle.
You cannot see both shapes independently — one is missing — but you calculate both and subtract.
Decide before you calculate: Write out "Area = rectangle − circle" (or whatever the combination is) as a plan before touching the numbers. This commits you to the correct operation and earns method marks in the HSC even if you make an arithmetic slip later.
Check Your Understanding
Summarise the method from this section in 2–3 bullet points.
04
Calculator Use for Circle Problems
Using 3.14 instead of the $\pi$ button introduces rounding error early — and that error compounds through the rest of your calculation.
What to do
Use the $\pi$ button, round at the very end
Do not press the $\pi$ button — leave $\pi$ as a symbol
Keep $\pi$ in exact form until the final calculation
Example
$A = \pi \times 7^2 = 153.94$ cm² (2 d.p.)
$A = 25\pi$ cm²
Write $18\pi$ cm², then evaluate at the end: $18\pi = 56.55$ cm²
Never use 3.14: $\pi = 3.14159265...$ — substituting 3.14 gives an answer that is slightly wrong from the very first step. In a multi-part question this error propagates. Always use the $\pi$ button.
Check Your Understanding
Write one sentence summarising the main mathematical idea of this section.
05
Common Mistakes
Mistake 1 — Using the slant side as the height
In a triangle or parallelogram, the slant side is visible and labelled — the perpendicular height is a dotted line and feels less "real." But $h$ must form a right angle with the base. If only the slant side is given and no perpendicular height is marked, you cannot find the area with these formulas alone.
Mistake 2 — Using diameter instead of radius in $A = \pi r^2$
The diameter goes all the way across a circle — it is the more visible measurement. Substituting the diameter directly gives an answer four times too large (because $(2r)^2 = 4r^2$). Every time you see a circle, write $r = d \div 2$ as a separate line before substituting.
Mistake 3 — Adding instead of subtracting in composite shapes
When a region is cut out of a shape, students calculate areas of all visible components and add them. Before calculating anything, decide: joining (add) or removing (subtract)? Label your decision in your working first.
Check Your Understanding
Write one sentence summarising the main mathematical idea of this section.
Worked Examples
06
Worked Example 1
Area of a Triangle
Problem
Find the area of a triangle with base 9 cm and perpendicular height 6 cm.
Step-by-Step Solution
1
Identify shape and formula Triangle: $A = \tfrac{1}{2}bh$
The perpendicular height is given as 6 cm — this is the correct $h$ to use
2
Substitute known values $A = \tfrac{1}{2} \times 9 \times 6$
Replace $b = 9$, $h = 6$ — write the formula first, then substitute
3
Evaluate $A = \tfrac{1}{2} \times 54 = 27$
$9 \times 6 = 54$, then $54 \div 2 = 27$
4
State answer with unit $A = 27\text{ cm}^2$
Area unit = (length unit)² — used cm, so answer is cm². The $\tfrac{1}{2}$ is easily forgotten — marks are lost here.
Check Your Understanding
Write one sentence summarising the main mathematical idea of this section.
07
Worked Example 2
Area of a Circle
Problem
Find the area of a circle with diameter 14 cm. Give your answer correct to 2 decimal places.
Step-by-Step Solution
1
Find the radius $r = 14 \div 2 = 7\text{ cm}$
Diameter is given — always halve it before substituting. Do this as a separate written step.
2
Write formula and substitute $A = \pi r^2 = \pi \times 7^2 = \pi \times 49$
$r^2 = 7^2 = 49$. Keep $\pi$ exact at this stage — do not evaluate yet.
3
Evaluate using $\pi$ button $A = 153.9380...$
Use the $\pi$ button on your calculator — never 3.14. Round only at this final step.
4
Round and state answer $A = 153.94\text{ cm}^2$
Round to 2 decimal places as instructed. If you substituted 14 instead of 7: $\pi \times 196 = 615.75$ cm² — four times too large.
Check Your Understanding
Write one sentence summarising the main mathematical idea of this section.
08
Worked Example 3
Area of a Trapezium
Problem
A trapezium has parallel sides of 5 m and 11 m, and a perpendicular height of 4 m. Find its area.
Step-by-Step Solution
1
Identify and label $a = 5,\ b = 11,\ h = 4$
Label the parallel sides $a$ and $b$ before substituting — avoids confusion with base and height
2
Write formula and substitute $A = \tfrac{1}{2}(a + b)h = \tfrac{1}{2}(5 + 11) \times 4$
Write the full formula before substituting — earns the method mark even if arithmetic goes wrong later
3
Evaluate brackets first $= \tfrac{1}{2} \times 16 \times 4$
Rectangle: $80 \times 40 = 3200$ m²; Two semicircles = one circle, $r = 20$ m: $\pi \times 400 = 1256.6$ m²; Total = 4456.6 m²
Revisit Your Initial Thinking
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Multiple Choice
1 A circle has diameter 10 cm. Its area, correct to 2 decimal places, is:
A 31.42 cm²
B 78.54 cm²
C 314.16 cm²
D $100\pi$ cm²
B — $r = 10 \div 2 = 5$ cm. $A = \pi \times 5^2 = 25\pi = 78.54$ cm². Option C results from using diameter 10 directly: $\pi \times 100 = 314.16$ — four times too large.
2 A triangle has base 12 m, a slant side of 8 m, and perpendicular height 6 m. The area of the triangle is:
A 96 m²
B 48 m²
C 36 m²
D 24 m²
C — $A = \tfrac{1}{2}bh = \tfrac{1}{2} \times 12 \times 6 = 36$ m². The slant side of 8 m is a deliberate distractor — it plays no role in the area formula.
3 A composite shape is formed by removing a semicircle of radius 4 cm from a rectangle measuring 14 cm × 10 cm. The area of the remaining shape, to 2 decimal places, is:
A 114.87 cm²
B 140 cm²
C 164.87 cm²
D 25.13 cm²
A — Rectangle: $14 \times 10 = 140$ cm². Semicircle: $\tfrac{1}{2} \times \pi \times 16 = 8\pi = 25.13$ cm². Remaining: $140 - 25.13 = 114.87$ cm². Option C (164.87) results from adding instead of subtracting.
Short Answer
10
SA 42 marks
Find the area of a trapezium with parallel sides 7 cm and 13 cm and a perpendicular height of 8 cm.
Syllabus: MS11-3 | 1 mark correct substitution, 1 mark correct answer with unit
Working space in book
Saved
11
SA 53 marks
A logo is made from a square of side 6 cm with a circle of diameter 6 cm inscribed inside it (fitting exactly within the square). Find the area of the square that is not covered by the circle, correct to 2 decimal places.
Syllabus: MS11-3 | 1 mark radius, 1 mark both areas, 1 mark correct subtraction and rounding
Working space in book
Saved
12
SA 64 marks
A garden is in the shape of a rectangle 12 m × 8 m. A triangular garden bed with base 4 m and perpendicular height 3 m is cut from one corner, and a semicircular garden bed of diameter 8 m is added along one long edge.
(a) Find the area of the triangle removed. (1 mark)
(b) Find the area of the semicircle added, correct to 2 decimal places. (1 mark)
(c) Find the total area of the garden, correct to 2 decimal places. (2 marks)