Rectangle Dimensions: Find Length & Width

Length & Width from Area & Perimeter

Enter the rectangle's area (A) and perimeter (P). The calculator will find the length (L), width (W), and diagonal (d).

Rectangle Parameters

Perimeter (P) (e.g., 14)

Area (A) (e.g., 12)

Display Options

Decimal Places (0-10)

Width (W) --

Length (L) --

Diagonal (d) --

✓ 2(L+W) = -- ✓ L×W = --

Formulas and Solution Method

Given the perimeter \( P = 2L + 2W \) and the area \( A = L \times W \) of a rectangle, we can find its length \(L\) and width \(W\).

Step-by-Step Derivation

From the perimeter: \( \displaystyle L + W = \frac{P}{2} \)
Let \( S = \frac{P}{2} \). Then \( W = S - L \).
Substitute into the area equation: \( \displaystyle L \times (S - L) = A \)
This simplifies to the quadratic equation: \( \displaystyle L^2 - S\,L + A = 0 \)
Solving for \(L\) using the quadratic formula: \[ L = \frac{S \pm \sqrt{S^2 - 4A}}{2} \]
The width is then \( W = S - L \). The diagonal is found using the Pythagorean theorem: \( d = \sqrt{L^2 + W^2} \).

Existence Condition

A rectangle with given area \(A\) and perimeter \(P\) exists only if the discriminant \( \; S^2 - 4A \; \) is non-negative: \[ \left(\frac{P}{2}\right)^2 - 4A \ge 0 \quad \text{or equivalently} \quad P^2 \ge 16A \]

If this condition is not met, no real rectangle has those dimensions.

Example

For \(P = 14\) and \(A = 12\) (the default values):

\(S = \frac{P}{2} = 7\)
Discriminant: \(7^2 - 4 \times 12 = 49 - 48 = 1\)
\(\sqrt{\Delta} = 1\)
Length = \(\frac{7 + 1}{2} = 4\), Width = \(\frac{7 - 1}{2} = 3\)
Check: Perimeter = \(2(4+3) = 14\) ✓, Area = \(4 \times 3 = 12\) ✓
Diagonal = \(\sqrt{4^2 + 3^2} = 5\)

More Geometry Resources

Table of Geometry Formulas
More Online Geometry Calculators and Solvers