A = lw is the shortest formula in geometry, so the interest here is not the multiplication. It is the thing almost everyone gets wrong about it: area and perimeter are independent. A 25 × 25 square and a 1 × 49 strip both have a perimeter of 100, and their areas are 625 and 49. Knowing the distance ...
Area and perimeter are independent, and the animation below shows it. A 25 × 25 square and a 1 × 49 strip both have a perimeter of 100, and their areas are 625 and 49. Five ways in are supported, including perimeter with area — which is a quadratic, and which refuses when the area asked for is more than the perimeter can enclose. That refusal is not a limitation of the method: no such rectangle exists.
Any two facts about the rectangle. Whole numbers, decimals and fractions all work.
LENGTH
WIDTH
AREA
40
A square with the same perimeter would enclose 42.25 — this shape reaches 94.7% of that.
| Length | 8 | 8 |
| Width | 5 | 5 |
| Perimeter | 26 | 26 |
| Diagonal | √89 | 9.433981 |
| Area | 40 | 40 |
THE ANSWER PUT BACK
l × w = 8 × 5 = 40 ✓
l² + w² = 89 → d = √89 ✓
PERIMETER
26
twice the length plus the width
DIAGONAL
√89
√(l² + w²) — the longest line inside
SHAPE
1.6 : 1
long side against short side
SAME AREA AS A SQUARE
of side 2√10
which would need less edging round it
WHAT IS HAPPENING
A = 8 × 5 = 40
The area counts the unit squares that tile the rectangle: w rows of l squares each. That is all the multiplication is doing, and it is why the units end up squared.
P = 2 × (8 + 5) = 26
Area and perimeter are independent. Two rectangles with the same perimeter can enclose very different amounts — a fact worth holding on to, since it is the commonest misconception about this shape.
Why the same fence encloses different fields. The perimeter fixes only the sum of the two sides, and a sum can be split many ways. Their product — the area — is largest when the two are equal and shrinks as they separate, towards nothing at the extremes. A 2:1 rectangle still keeps 88.9% of the square's area, so the loss is gentle at first; a 25:1 strip keeps 14.8%. This is the whole reason a field, a room or a pen is worth squaring up when the fencing is fixed, and the reason perimeter tells you almost nothing about area on its own.
Exact values throughout · surds kept as surds, never rounded away
Pick which two facts you have — two sides, the area with a side, the perimeter with a side, a diagonal with a side, or a perimeter with an area — and fill in the boxes.
Read the decimal first and the exact value beneath it. Where a length is a surd, the surd is the answer and the decimal is a rounding of it.
Press the slide control and watch the two readouts. The perimeter figure in the caption never changes while the area figure below the shape does — that is the whole point.
Find the navy tick on the scale beneath: it marks where your own rectangle sits between a long strip and the square.
Take a rectangle 8 by 5. The area is 8 × 5 = 40. Think of it as unit squares: 5 rows of 8, which is why the units end up squared — centimetres by centimetres gives square centimetres. The perimeter is 2 × (8 + 5) = 26. Each side is counted twice, so half the perimeter, 13, is one length plus one width. Forgetting that halving is the usual slip and it doubles the answer. The diagonal is √(8² + 5²) = √89, about 9.434. It is exact as √89 and not as any decimal. Now the point of this page. A rectangle 12.5 by 0.5 also has a perimeter of 26 — and an area of 6.25, a sixth of the first one. The same fence, a sixth of the field. Perimeter constrains only the sum of the sides, and a fixed sum can be split many ways; the product is largest when the two are equal and shrinks as they separate. How fast does it shrink? With a perimeter of 100 the square is 25 × 25 with area 625. A 2:1 rectangle is 33.33 × 16.67, area 555.56 — still 88.9% of the best. A 3:1 is 37.5 × 12.5, area 468.75, or 75%. A 10:1 is 45.45 × 4.55, area 206.61, only 33.1%. And a 25:1 is 48.08 × 1.92, area 92.46, just 14.8%. The loss is gentle near the square and then falls off a cliff. Now working from a perimeter and an area together. Given P = 26 and A = 40: the sides add to 13 and multiply to 40, so they are the roots of x² − 13x + 40 = 0. That factors as (x − 8)(x − 5), giving 8 and 5 — the rectangle we started with. Try P = 20 with A = 23 instead. The sides add to 10 and multiply to 23, so x² − 10x + 23 = 0 and x = 5 ± √2. Both sides are irrational, and yet (5 + √2)(5 − √2) = 25 − 2 = 23 exactly. The surds cancel in the product. Now try P = 20 with A = 30. The discriminant is 100 − 120, which is negative, so there are no real roots. That is not a failure of the method: a perimeter of 20 encloses at most 5 × 5 = 25, so an area of 30 is simply impossible. Strip that check out and the calculator will happily report sides of 5 + √−5, which are not lengths at all. One last trap, and it costs real money. Converting an area between units squares the conversion factor. One square metre is 100 × 100 = 10,000 square centimetres, not 100. Using the length factor leaves an answer a hundred times too small, and the same applies to square feet from square yards, or acres from square metres.
| Rule | What it says | Why |
|---|---|---|
| Area | A = l × w | The unit squares that tile it: w rows of l each. That is why units square. |
| Perimeter | P = 2(l + w) | Each side counted twice. Half the perimeter is one length plus one width. |
| They are independent | same P, very different A | A 25×25 square and a 1×49 strip both have perimeter 100. |
| The square is the best | A ≤ P² ÷ 16 | No rectangle of perimeter P encloses more than the square of side P/4. |
| The fall-off | 2:1 keeps 88.9% | Gentle near the square, severe at the extremes: 25:1 keeps only 14.8%. |
| From the area | w = A ÷ l | An area divided by a length gives a length — a quick check on the direction. |
| From the perimeter | w = P ÷ 2 − l | Halve first. Forgetting to halve doubles the answer. |
| The diagonal | d = √(l² + w²) | It cuts the rectangle into two right triangles, so Pythagoras applies. |
| From both P and A | x² − (P/2)x + A = 0 | A sum and a product means a quadratic. The sides are its two roots. |
| When that has no answer | A > P² ÷ 16 | Not a limitation of the method — no such rectangle exists. |
| Units square as well | 1 m² = 10,000 cm² | Not 100. Using the length factor leaves an area 100 times too small. |
| Doubling both sides | quadruples the area | Two lengths are involved, so the scale factor is squared. |
| Rounding compounds | 3.2% average error | Measuring both sides to the nearest unit; up to 75% on small rectangles. |
| The same area, less edge | a square of side √A | Any rectangle needs at least as much edging as the square holding the same area. |
Multiply the two sides: A = l × w. The result counts the unit squares that tile the rectangle — w rows of l squares each — which is why the units come out squared. A rectangle 8 by 5 has an area of 40, and if those measurements were centimetres the area is 40 square centimetres.
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Last updated: September 2, 2026 · Exact values throughout, surds kept as surds · The same perimeter can enclose very different areas.