Length and Width of Rectangle Given Area Calculator
Find the length and width of a rectangle when area and perimeter are known
Calculate Rectangle Dimensions
Total area of the rectangle
Total distance around the rectangle
Rectangle Dimensions
Example Calculation
Garden Plot Example
Problem: A rectangular garden has an area of 24 m² and a perimeter of 20 m. What are its dimensions?
Given: Area = 24 m², Perimeter = 20 m
Solution Steps
1. From P = 2L + 2W: W = P/2 - L = 10 - L
2. Substitute in A = L × W: 24 = L × (10 - L)
3. Expand: 24 = 10L - L²
4. Rearrange: L² - 10L + 24 = 0
5. Solve: L = (10 ± √(100 - 96))/2 = (10 ± 2)/2
6. Solutions: L = 6 m, W = 4 m
Key Formulas
Rectangle Area
A = L × W
Rectangle Perimeter
P = 2L + 2W
Quadratic Solution
L = (P/2 ± √((P/2)² - 4A))/2
Diagonal Length
d = √(L² + W²)
Quick Tips
For a solution to exist: P ≥ 4√A
When P = 4√A, the rectangle is a square
The quadratic always gives two solutions that add up to P/2
Length is conventionally the longer dimension
Understanding Rectangle Dimension Calculations
Mathematical Approach
To find the length and width of a rectangle given its area and perimeter, we need to solve a system of equations. This leads to a quadratic equation that can be solved using the quadratic formula.
The Problem Setup
Given:
- • Area (A) = L × W
- • Perimeter (P) = 2L + 2W
Find: Length (L) and Width (W)
Solution Method
Step 1: W = P/2 - L
Express width in terms of length
Step 2: A = L × (P/2 - L)
Substitute into area equation
Step 3: L² - L×P/2 + A = 0
Rearrange to quadratic form
Step 4: Use quadratic formula
Solve for L, then find W
When Does a Solution Exist?
Condition for Real Solutions
For the quadratic equation to have real solutions, the discriminant must be non-negative:
(P/2)² - 4A ≥ 0
This simplifies to: P ≥ 4√A
Special Cases
- • When P = 4√A: Rectangle is a square
- • When P > 4√A: Two different solutions (length ≠ width)
- • When P < 4√A: No real solution exists