1
It all starts with the rectangle.
triangle
A = ½bh
half a rectangle
parallelogram
A = bh
a leaning rectangle
A = ½(12)(5)
the triangle
= 30 square units
half of 60
Slide the top of a parallelogram sideways and the area never changes — the base
and the perpendicular height are all that matter.
The same slide explains a favourite: a triangle drawn on one side of a parallelogram, with its apex anywhere on the opposite side, has the parallelogram’s base and height — so it is always exactly half the parallelogram.
Given the perimeter instead of a side. A rectangle has perimeter 30 and length 9. Then 2(9 + w) = 30, so 9 + w = 15, w = 6 and A = 9 × 6 = 54. With sides as expressions, say w and 3w − 2 and perimeter 44, the same line gives 2(4w − 2) = 44, w = 6, sides 6 and 16, area 96.
A square of equal area. A square with the same area as a 9-by-4 rectangle has s² = 36, so s = √36 = 6 and its perimeter is 4 × 6 = 24. Take the square root of the area to get the side.
2
Height means perpendicular height.
This single point causes more wrong answers than any formula.
wrong
20 × 10 = 200
that is the slanted side
The height may sit outside the shape
For a very slanted triangle the perpendicular from the apex lands beyond the
base. It is still the height, and the formula still works.
3
The trapezoid: average the parallel sides.
A =
a + b2
× h
(8 + 14) ÷ 2 = 11
the average width
A trapezoid is a rectangle whose width changes steadily, so the average width does
the job.
Backwards: a trapezoid with parallel sides 6 and 8 has area 70. The average width is 7, so h = 70 ÷ 7 = 10. Divide the area by the average of the parallel sides.
Only the parallel pair
The two slanted sides never enter the formula. Averaging the wrong pair is the
standard error here.
4
Rhombus and kite: use the diagonals.
A = ½ d₁ d₂
½ × 10 × 24
half the product
30 · 66 · 120
It works for any quadrilateral whose diagonals cross at right angles — rhombus,
kite, and the square, whose diagonals are equal.
5
Three sides and no height: Heron.
For a triangle with sides 13, 14 and 15.
s = (13 + 14 + 15) ÷ 2 = 21
the semi-perimeter
A = √21 · 8 · 7 · 6
s − a, s − b, s − c
= √7056 = 84
square units
| Shape | Area |
|---|
| rectangle / parallelogram | bh |
|---|
| triangle | ½bh |
|---|
| trapezoid | ½(a + b)h |
|---|
| rhombus / kite | ½d₁d₂ |
|---|
| triangle, three sides known | √s(s−a)(s−b)(s−c) |
|---|
6
An awkward shape: take the whole, subtract the corners.
ABCD is a rectangle with AB = 10 and BC = 6. E is on BC with BE = 4, and F is on CD with CF = 5. Find the area of △AEF.
rectangle = 10 × 6 = 60
the whole
△ABE = ½ × 10 × 4 = 20
right angle at B
△ECF = ½ × 2 × 5 = 5
right angle at C, EC = 6 − 4
△FDA = ½ × 5 × 6 = 15
right angle at D, FD = 10 − 5
△AEF = 60 − 20 − 5 − 15 = 20
whole minus corners
Each corner piece is a right triangle, so its two legs are its base and height — no
slanted sides needed. Find the leftover lengths (6 − 4, 10 − 5) before you start. The
same subtraction handles circles cut out of a rectangle: rectangle minus the circles’
areas.
7
Variations: the height outside, a height to another side, mixed units.
In △ABC, ∠B is obtuse and BC = 7. The perpendicular from A meets line CB extended 3 beyond B, and is 4 long. Find the area.
base = BC = 7
the side itself, not 7 + 3
height = 4
it lands outside, still perpendicular
A = ½ × 7 × 4 = 14
the 3 is not used
Different: the foot of the height is off the triangle. The extension only carries the height; it is never part of the base.
A right triangle has legs 6 and 8. Find the altitude to the hypotenuse.
A = ½ × 6 × 8 = 24
the legs are base and height
½ × 10 × h = 24, h = 4.8
same area, other base
Different: one triangle, two base-height pairs, one area. Use the easy pair to get the area, then the other pair to get the unknown height.
A rectangle is 2.5 m by 80 cm. Find its area in square metres.
80 cm = 0.8 m
convert the length first
A = 2.5 × 0.8 = 2 m²
same as 20 000 cm²
Square units convert by the square
1 m = 100 cm, but 1 m² = 100 × 100 = 10 000 cm². Dividing 20 000 cm² by 100 gives 200,
the classic wrong answer. Convert lengths before multiplying and the trap never arises.
8
Equilateral triangles, and what scaling does to area.
A = √34 s²
s = 6: A = (√34) × 36 = 9√3
forwards
A = 25√3: (√34)s² = 25√3
backwards
s² = 100, s = 10
÷ √3, × 4
perimeter = 3 × 10 = 30
answer what was asked
| Change | Area becomes |
|---|
| length × 4, width × ½ | 4 × ½ = 2 times as large |
|---|
| every length × k | k² times as large |
|---|
| base × 3, height unchanged | 3 times as large |
|---|
Scale factors multiply, they do not add
Area is a product, so the factors on length and width multiply. Doubling every side
makes the area 4 times as large, not 2.
On the exam
Area items are usually one step removed: a shaded region (whole minus the pieces), a
missing height or side from a given area, a “what happens if” scaling question, or an
equilateral triangle in surd form. Write the formula, put in what you know and solve
for what you do not. Ten-second check: a triangle is half its rectangle, and a shaded
part is smaller than the whole.
✓
Check: sanity-test each one.
triangle
30 < 12 × 5 = 60
half a rectangle ✓
trapezoid
8×6 = 48 < 66 < 14×6 = 84
between the two rectangles ✓
rhombus
120 = half of 10 × 24
half its bounding box ✓
Remember the flow
Perpendicular height · Average the parallel pair · Half the diagonals
Never use the slanted side as a height.
