How this instrument works
A rectangle's perimeter is nothing more than the sum of its four sides, two lengths and two widths: P = l + w + l + w = 2(l + w). That identity is definitional, not derived from any theorem — it's just what 'distance around' means for a four-sided figure with two pairs of equal sides. Halve both sides and the two adjacent edges together add up to P ⁄ 2, the semiperimeter; move the known length across the equals sign and what's left is w = P ⁄ 2 − l, the whole formula this instrument runs.
Because that rearrangement is linear, the trade-off between the two sides is perfectly one-for-one for a fixed perimeter: add a unit to length and width gives up exactly a unit in return, no more and no less. Plot width against length for a constant perimeter and the result is a straight line of slope −1, not a curve — there is no quadratic here, unlike the sibling problem of recovering both sides from area and perimeter together, which needs two equations and a square root because neither side starts out known.
The formula also polices its own domain. Width cannot be negative for a shape that actually exists, so length is capped at half the perimeter; push length past that ceiling and the subtraction returns a negative number, this calculator's way of saying no rectangle matches those two figures. At the ceiling itself, length equals P ⁄ 2 exactly and width falls to zero — the rectangle has flattened into a doubled line segment with no enclosed area left.
- Enter the total distance around the shape into the Perimeter field.
- Enter the one side length you already know into the Length field.
- Read Width for the other side — the value that makes those two figures consistent with each other.
- If Width comes back negative, Length is longer than half of Perimeter and no rectangle fits those two numbers; shorten Length or raise Perimeter.
Worked example — 20 metres of fencing, one 6-metre edge already set
A community garden plot gets fenced with exactly 20 metres of railing, and one edge is already fixed at 6 metres because it runs flush against an existing brick path. Half the perimeter is P ⁄ 2 = 20 ⁄ 2 = 10, and subtracting the known 6-metre length leaves width = 10 − 6 = 4 metres for the pair of remaining sides.
Check it by rebuilding the total: two 6-metre lengths plus two 4-metre widths gives 2×6 + 2×4 = 12 + 8 = 20 metres, matching the railing exactly. Had the path instead run 11 metres long — more than half of 20 — the same subtraction would hand back a negative width, the calculator's flag that no rectangle can carry those two figures together.
Questions
How is the formula width = P ⁄ 2 − length derived?
It falls straight out of what perimeter means: a rectangle has two lengths and two widths, so P = 2l + 2w = 2(l + w). Dividing by 2 gives l + w = P ⁄ 2, the semiperimeter; moving the known length to the other side leaves w = P ⁄ 2 − l, exactly the rearrangement this calculator performs.
Why doesn't this need the quadratic formula the way an area-and-perimeter solver does?
Because one side is already known here. A calculator starting from only area and perimeter has two unknowns and must solve x² − Sx + A = 0 for both roots at once; this one already has length, so w = P ⁄ 2 − l is a single subtraction, not a quadratic — the same relationship with one fewer missing piece.
What does a negative or impossible width mean?
It means the entered length is longer than half the perimeter, so two copies of that length alone would already exceed the whole fence. A real side can never be negative, so that pairing of Perimeter and Length describes no rectangle at all — a quick check against a mistyped measurement.
When does this calculator hand back a square instead of a rectangle?
Precisely when the entered Length sits at one quarter of the Perimeter, since w = P ⁄ 2 − l only matches l again at l = P ⁄ 4. A Perimeter of 20 paired with a Length of 5 returns a Width of 5 — the midpoint of the valid range, where the shape stops leaning either direction.
Does the unit I use for Perimeter and Length matter?
Yes, but only if Perimeter and Length share the same linear unit, because the formula is a bare rearrangement with no built-in conversion baked in. Feed it both in metres and Width comes back in metres; pair metres with feet and the arithmetic still runs, but the number it hands back no longer describes a real fence.
Could this same subtraction solve for length instead of width?
Yes, by symmetry — l = P ⁄ 2 − w solves for length given perimeter and width, the identical subtraction with the two known values swapped. This sheet is fixed to solve for width because Length is the field treated as the given input.