Area — triangle, parallelogram, trapezium, composite shapes: Worked examples

The scenario matrix

Here $a, b$ are the two parallel sides of a trapezium (just defined above), $h$ is a perpendicular height, and $θ$ is an angle at a corner.

Cell	What makes it different	The trap it tests	Example
T1 Right triangle, legs given	base ⟂ height already	none — but must see the legs are perpendicular	Ex 1
T2 Generic (acute) triangle, slant base	height is no side; must build it from two sides + angle	drop a perpendicular / use $h = s sin θ$	Ex 2
T3 Obtuse triangle	height falls outside the base	height still perpendicular, measured to the line of the base	Ex 3
P1 Parallelogram, slant + angle	given slant side, not height	use $h = s sin θ$ , never multiply slant	Ex 4
P2 Degenerate limit	angle $\to 0°$ or $90°$	area collapses to $0$ or maxes at rectangle	Ex 5
Z Zero / degenerate input	a length is $0$ or points collinear	area must come out $0$	Ex 5
Tr1 Trapezium, real-world word problem	must extract $a, b, h$ from a story	pick the two parallel sides only	Ex 6
Tr2 Trapezium limit $a \to b$	parallel sides become equal	formula must reduce to a rectangle	Ex 7
C1 Composite by adding	join two known shapes	line up shared edge, don't double-count	Ex 8
C2 Composite by subtracting (cutout)	hole inside a shape	outer $-$ inner	Ex 9
X Exam-style twist	area known, a length unknown	run the formula backwards	Ex 10

Every cell above gets hit at least once below. Let's build the toolkit picture first.

Figure s01 — the three engines side by side. The chalk sketch shows a triangle, a parallelogram and a trapezium, each with its base(s) in blue and its perpendicular height as a pink dashed line. The one thing to glean: in every shape the pink dashed line goes straight across at $90°$ — it is never the slanted edge. The yellow formula under each shape reminds you what to multiply.

Figure — Area — triangle, parallelogram, trapezium, composite shapes

Ex 1 — Right triangle, legs given (cell T1)

Ex 2 — Generic acute triangle: build the height from two sides + angle (cell T2)

Worked example When no side is the height, construct it

A triangle has two sides $p = 8 cm$ and $q = 6 cm$ meeting at an angle of $θ = 50°$ . There is no right angle and no side that is already a height. Take $p = 8$ as the base. Find the area.

Forecast: We must manufacture a perpendicular height. The vertical drop from the far vertex onto the base comes from the other side $q$ leaning at $50°$ . Guess: is the height bigger or smaller than $6$ ?

Drop a perpendicular from the top vertex to the base. The side $q = 6$ leans off the base at $50°$ ; its vertical part (the height) is $h = q sin θ = 6 sin 50° \approx 6 \times 0.766 = 4.596 cm$ Why this step? No side is perpendicular to the base, so we build the height. $sin θ$ = "opposite over hypotenuse" (from 2.1.4-Trigonometric-ratios) gives exactly the vertical slice of the leaning side $q$ — that vertical slice is the perpendicular gap the formula needs. This is why sine, not cosine: cosine would give the along-the-base part, which is useless for height.
Apply the triangle formula with base $p$ and this built height. $A = \frac{1}{2} p h = \frac{1}{2} \times 8 \times 4.596 \approx 18.38 cm^{2}$ Why this step? Once the perpendicular height exists, it's the ordinary $\frac{1}{2} bh$ .

Verify: The standard trig-area shortcut $A = \frac{1}{2} pq sin θ = \frac{1}{2} \times 8 \times 6 \times sin 50° \approx 18.38 cm^{2}$ — identical, confirming our built height. Units cm². ✓ ( $h \approx 4.6 < 6$ makes sense: a leaning side always has a shorter vertical drop than its full length.)

Figure s02 — building the height. Look at the pink dashed line dropping from the apex straight down onto the base $p$ : that is the manufactured height $h = q sin θ$ . The blue side $q$ is the slanted one at angle $θ$ (yellow arc); notice $h$ is shorter than $q$ — the sine "shrinks" the leaning side to its vertical part.

Ex 3 — Obtuse triangle, height outside the base (cell T3)

Worked example The perpendicular can land off the edge

A triangle has base $B C = 8 cm$ along the ground. Its apex $A$ sits to the left, so the foot of the perpendicular from $A$ lands outside segment $B C$ , at distance $2 cm$ left of $B$ . The perpendicular height is $5 cm$ . Find the area.

Forecast: Does the " $2 cm$ overhang" get added into the area? (Careful — the formula only wants the perpendicular gap, not where its foot lands.)

Identify base and height correctly. Base $= B C = 8 cm$ ; height $=$ perpendicular distance from $A$ to the line through $B C = 5 cm$ . Why this step? The height is measured to the infinite line containing the base, even when the foot falls outside the segment. The $2 cm$ overhang is not a length used by the formula.
Apply the formula. $A = \frac{1}{2} \times 8 \times 5 = 20 cm^{2}$ Why this step? Standard triangle area — the overhang never enters.

Verify (subtraction picture): Big right triangle (base $2 + 8 = 10$ , height $5$ ) minus the left sliver (base $2$ , height $5$ ): $\frac{1}{2} (10) (5) - \frac{1}{2} (2) (5) = 25 - 5 = 20 cm^{2}$ . ✓ Same answer — proof the overhang cancels.

Figure s03 — the foot lands off the base. The base line $B C$ is extended as a dotted chalk line to the left; the pink dashed height meets it at a yellow dot that sits outside $B C$ . Glean this: the height is still measured to that extended line, and the labelled "overhang 2" is decoration — it is not fed into $\frac{1}{2} bh$ .

Ex 4 — Parallelogram from slant + angle (cell P1)

Worked example Never multiply the slant side

A parallelogram has base $b = 10 m$ and a slant side $s = 7 m$ ; the angle between them is $θ = 60°$ . Find the area.

Forecast: The lazy answer is $10 \times 7 = 70$ . Is the real area more or less than that? (The shape leans, so it must be less.)

Find the true perpendicular height. The slant side leans at $60°$ , so its vertical part is $h = s sin θ = 7 sin 60° = 7 \times \frac{3}{2} \approx 6.062 m$ Why this step? $sin θ$ (from 2.1.4-Trigonometric-ratios) is "opposite over hypotenuse" = the vertical fraction of the slant. That vertical drop is the perpendicular gap the area formula wants — this is exactly why we reach for sine and not cosine.
Apply the parallelogram formula. $A = bh = 10 \times 6.062 \approx 60.62 m^{2}$ Why this step? Area $=$ base × perpendicular height.

Verify: $60.62 < 70$ as forecast — the lean loses area. As a bound, $10 \times 7 = 70$ is the maximum (reached only at $θ = 90°$ ). ✓

Figure s04 — slant vs. height. The blue slant side leans at the yellow $60°$ arc; the pink dashed line beside it is the true height $h = 7 sin 60° \approx 6.06$ . See how the pink line is shorter than the blue slant — that shortfall is exactly the "lost" area the $70$ -trap ignores.

Ex 5 — Limits & degenerate inputs (cells P2, Z)

Worked example What happens at the extremes?

Using the same parallelogram $b = 10, s = 7$ , examine three special angles: $θ = 90°$ , $θ \to 0°$ , and the case $s = 0$ .

Forecast: Which of these gives the largest area, and which gives zero?

Case $θ = 90°$ (upright — it's a rectangle). $A = 10 \times 7 sin 90° = 10 \times 7 \times 1 = 70 m^{2}$ Why this step? At $90°$ the slant is vertical, so height $=$ slant. This is the maximum area.
Case $θ \to 0°$ (flattened flat). $A = 10 \times 7 sin 0° = 10 \times 7 \times 0 = 0 m^{2}$ Why this step? As the shape squashes flat, top and bottom edges overlap — no space enclosed. Area must vanish, and $sin 0° = 0$ delivers exactly that.
Case $s = 0$ (degenerate — no slant side). $A = b \cdot (0 \cdot sin θ) = 0 m^{2}$ Why this step? With zero side length the parallelogram collapses to a line segment, which covers no area.

Verify: $70 > 60.62 > 0$ : the $θ = 60°$ answer from Ex 4 sits between the flat ( $0$ ) and upright ( $70$ ) limits, exactly where it should. ✓ Units all m². ✓

Ex 6 — Trapezium, real-world word problem (cell Tr1)

Worked example A garden plot from a story

A garden is bounded by a straight north fence $12 m$ long and a straight south fence $8 m$ long, and these two fences are parallel. The straight-across (perpendicular) distance between them is $5 m$ . The two side fences are slanted. Find the garden's area.

Forecast: Will the area equal a $10 \times 5$ rectangle? (Notice $10$ is the average of $8$ and $12$ …)

Extract the parallel sides and height from the words. Parallel sides $a = 12$ , $b = 8$ ; perpendicular height $h = 5$ . Why this step? The formula only wants the two parallel edges (these are the $a$ and $b$ defined at the top) plus the perpendicular gap. The slanted side fences are decoys — we never use their lengths.
Apply the trapezium formula. $A = \frac{1}{2} (a + b) h = \frac{1}{2} (12 + 8) \times 5 = \frac{1}{2} \times 20 \times 5 = 50 m^{2}$ Why this step? $\frac{a + b}{2} = 10$ is the average width; times height $5$ gives area.

Verify: A rectangle of width $10$ (the average) and height $5$ is $50 m^{2}$ — matches. ✓ Units m². ✓

Figure s05 — which fences to use. The two parallel fences are the blue horizontal edges labelled $a = 12$ (top) and $b = 8$ (bottom); the pink dashed line between them is $h = 5$ . Glean: only these three chalk measurements enter the formula — the two slanted side fences are drawn but carry no number.

Ex 7 — Trapezium limit: parallel sides equal (cell Tr2)

Ex 8 — Composite by adding (cell C1)

Worked example House front = wall + roof

A house front is a rectangle $8 m$ wide $\times 3 m$ tall, topped by an isosceles triangle roof of base $8 m$ and height $2 m$ . Find the total area.

Forecast: The roof is thinner than the wall block — will it add closer to $8$ or to $24$ m²?

Split into known shapes. Wall = rectangle; roof = triangle. They meet along the $8 m$ top edge with no overlap. Why this step? Divide-and-conquer: sum areas only when pieces share an edge but don't overlap.
Rectangle area. $A_{rect} = 8 \times 3 = 24 m^{2}$ Why this step? A rectangle's area is width × height, and for the wall both are given directly with no slant to correct for — so we multiply the two perpendicular sides straight away.
Triangle area. $A_{△} = \frac{1}{2} \times 8 \times 2 = 8 m^{2}$ Why this step? Roof height $2 m$ is the perpendicular height for the triangle formula.
Add. $A = 24 + 8 = 32 m^{2}$ Why this step? The two pieces sit side by side with no shared interior, so their areas simply sum.

Verify: Roof $8 m^{2}$ is a third of the wall's $24$ — matches the forecast (much closer to $8$ ). Total $32 m^{2}$ , units consistent. ✓

Figure s06 — stack, don't overlap. The blue-shaded rectangle is the wall; the pink-shaded triangle is the roof sitting exactly on the shared $8 m$ top edge (pink dashed line = roof height $2$ ). Glean: the two shaded regions touch only along that edge, so no area is counted twice — you simply add $24 + 8$ .

Ex 9 — Composite by subtracting (cutout) (cell C2)

Worked example A picture frame

A rectangular frame has outer size $20 cm \times 15 cm$ ; a rectangular hole $16 cm \times 11 cm$ is cut from its centre. Find the area of the frame (the material only).

Forecast: Roughly what fraction of the $300 cm^{2}$ board is left as frame?

Outer area. $A_{outer} = 20 \times 15 = 300 cm^{2}$ Why this step? We first measure the whole board as a plain rectangle (width × height, both perpendicular) before removing anything — you cannot subtract a hole until you know the full sheet.
Inner (hole) area. $A_{inner} = 16 \times 11 = 176 cm^{2}$ Why this step? The hole is empty space — its area must be removed, not added.
Subtract. $A_{frame} = 300 - 176 = 124 cm^{2}$ Why this step? Frame material = whole board minus the punched-out window.

Verify: Frame is $124/300 \approx 41%$ of the board — plausible for a fairly wide border. Units cm². ✓

Ex 10 — Exam twist: run the formula backwards (cell X)

Recall Quick self-test (reveal after guessing)

A parallelogram has base 10 and slant 7 at angle 30°. Its area? ::: $10 \times 7 sin 30° = 10 \times 7 \times 0.5 = 35 cm^{2}$ A trapezium's parallel sides are 4 and 4, height 9. Which simpler formula applies? ::: The parallelogram/rectangle form: $A = 4 \times 9 = 36$ . Frame: outer $10 \times 10$ , hole $6 \times 6$ . Area of material? ::: $100 - 36 = 64$ . What is another name for a "trapezium"? ::: A "trapezoid" (same shape; the word varies by country).

See also: 1.2.11-Perimeter · 1.2.13-Circle-area-and-circumference · 3.4.2-Integrationas-area · 1.2.1-Basic-shapes