1. What perimeter and area measure
The total length of a shape's boundary. If you walked along the edge of the shape once, the perimeter is the distance you walked.
The amount of two-dimensional surface a shape covers. If you tiled the interior with unit squares, the area is the number of squares it takes.
That definition gap is also a units gap. Perimeter is a length, so it lives in linear units: metres, feet, centimetres. Area is the number of unit squares, so it lives in square units: m², ft², cm². The little superscript $^2$ is not decoration — it's telling you that this number was built by multiplying two lengths together.
Perimeter is the fence; area is the lawn. You buy fence by the metre, and you buy sod by the square metre. Different stores, different products — even though both are answered from the same set of measurements.
One consequence: doubling a shape's linear size doubles its perimeter but quadruples its area. A 2 m × 2 m square has perimeter $8$ m and area $4$ m². A 4 m × 4 m square has perimeter $16$ m and area $16$ m². The fence cost grows linearly; the lawn cost grows like the square. This is the same scaling law that makes ants strong and elephants slow.
2. Rectangle and square
Start with the rectangle, because every other area formula on this page is going to reduce to it. Let $\ell$ be the length and $w$ the width. The perimeter is the sum of all four sides:
$$ P = 2\ell + 2w = 2(\ell + w) $$The area is the number of unit squares that tile its interior. With $\ell$ rows of $w$ squares each:
$$ A = \ell \cdot w $$A square is the special case where $\ell = w = s$. Both formulas collapse:
$$ P = 4s, \qquad A = s^2 $$3. Triangle
For a triangle with base $b$ and perpendicular height $h$:
$$ A = \tfrac{1}{2}\,b\,h $$The perimeter is just the sum of the three sides — no clever formula, you add them up. The interesting part is the area, and that $\tfrac{1}{2}$ deserves an explanation.
The cleanest way to see it: every triangle is exactly half of a parallelogram. Take any triangle, make a copy, rotate it 180°, and glue it to the original along one side. The two triangles fit together into a parallelogram with base $b$ and height $h$, whose area is $bh$ (we'll prove that in the next section). Each triangle is half of that:
$$ A_{\triangle} = \tfrac{1}{2}\,A_{\text{parallelogram}} = \tfrac{1}{2}\,b\,h $$In $A = \tfrac{1}{2}bh$, $h$ is the perpendicular distance from the base to the opposite vertex — not the length of one of the slanted sides. For an obtuse triangle, $h$ may even land outside the triangle. Drop the altitude every time; never substitute a side length.
4. Parallelogram and trapezoid
Parallelogram
A parallelogram has two pairs of parallel sides. Its area looks deceptively simple:
$$ A = b\,h $$where $b$ is one side (the base) and $h$ is the perpendicular distance to the opposite side. The proof is a single scissor cut: slice a right triangle off one end of the parallelogram and slide it to the other end. What you get is a rectangle with the same base and height — and we already know its area is $bh$.
Trapezoid
A trapezoid has exactly one pair of parallel sides — call them $b_1$ and $b_2$, separated by perpendicular height $h$. Its area is the average of the parallel sides times the height:
$$ A = \tfrac{1}{2}(b_1 + b_2)\,h $$To see why, copy the trapezoid, flip the copy upside down, and glue it to the original. The two together form a parallelogram with base $b_1 + b_2$ and height $h$, whose area is $(b_1 + b_2)h$. Each trapezoid is half:
5. Circles (recap)
The circle plays by slightly different rules — its boundary curves, so the linear-units measurement gets a special name: circumference. With radius $r$:
$$ C = 2\pi r, \qquad A = \pi r^2 $$Both formulas come down to the same constant, $\pi$. Notice that circumference scales with $r$ (linear units) and area scales with $r^2$ (square units) — the same square-vs-linear distinction we saw with rectangles. Double the radius and the circle's area quadruples; its circumference only doubles.
If you want the derivation — slicing the disk into thin pie wedges and rearranging them into a near-rectangle of dimensions $\pi r \times r$ — see the dedicated Circles topic. For this page we'll just use the formulas.
Some problems give you the diameter $d = 2r$ instead. Don't plug the diameter in where the formula asks for $r$ — halve it first. $A = \pi (d/2)^2 = \pi d^2 / 4$ is the diameter version if you'd rather memorise it.
6. Composite shapes
Most shapes in the wild aren't named polygons — they're rooms with bay windows, parking lots with rounded corners, gardens with a flower bed cut out of them. The trick is always the same:
- Decompose the shape into pieces you recognise — rectangles, triangles, half-circles.
- Compute the area of each piece using its formula.
- Combine: add the pieces, and subtract any holes.
The same approach works for perimeter, with one caution: only the outer boundary counts. Internal cut lines between sub-pieces are not part of the perimeter; they exist only in your decomposition, not in the actual shape.
After computing, glance at the answer and ask: is this a reasonable amount of area for this shape? If your L-shaped room came out to 12 m² but the bounding rectangle alone is only 8 m², you've added when you should have subtracted. The arithmetic catches itself if you're paying attention.