Rectangle Diagonal via A & P

Statement

The diagonal of a rectangle can be calculated if its area and perimeter are known, without needing its length and width directly:

\[ d = \sqrt{\left(\frac{P}{2}\right)^2 - 2A}. \]

Why it’s true

• Let the rectangle have length \(l\) and width \(w\).
• Perimeter: \(P = 2(l+w)\), so \(l+w = \tfrac{P}{2}\).
• Area: \(A = lw\).
• The diagonal: \(d = \sqrt{l^2 + w^2}\) (by Pythagoras).
• But \(l^2 + w^2 = (l+w)^2 - 2lw = (\tfrac{P}{2})^2 - 2A\).
• So \(d = \sqrt{(\tfrac{P}{2})^2 - 2A}\).

Recipe (how to use it)

1. Find perimeter \(P\) and area \(A\) of the rectangle.
2. Compute \((P/2)^2\).
3. Subtract \(2A\).
4. Take the square root to find \(d\).

Spotting it

Use this formula when a problem gives you only the area and perimeter of a rectangle, but asks for the diagonal.

Common pairings

• Pythagoras’ theorem in rectangles.
• Mixed geometry questions involving area, perimeter, and diagonals.
• Applications in tiling, fencing, and optimization problems.

Mini examples

1. Rectangle with \(A=24\), \(P=20\): \(d=\sqrt{(10)^2 - 48}=\sqrt{52}≈7.21\).
2. Rectangle with \(A=30\), \(P=22\): \(d=\sqrt{(11)^2 - 60}=\sqrt{61}≈7.81\).

Pitfalls

• Forgetting to halve the perimeter before squaring.
• Mixing up formula with \(d = \sqrt{l^2+w^2}\) (this one is a shortcut version).
• Negative inside the square root means given area and perimeter are inconsistent (watch out for exam trick questions).

Exam strategy

• Always write the formula clearly before substituting values.
• Check if exact square root is needed or decimal approximation.
• If area and perimeter don’t fit, state that no such rectangle exists.

Summary

This formula is a neat identity linking area, perimeter, and diagonal of a rectangle. It avoids calculating length and width separately, making problem-solving faster.