A rectangle is 4 m longer than it is wide. Its perimeter is 48 m. Find its area.

Step-by-step solution

1. 1

   Name the smaller side. Let the width be $x$ metres. Because the length exceeds the width by $4$ m, the length is $x+4$ metres. Choosing the smaller side as the variable keeps the algebra free of negative signs.

2. 2

   Translate the perimeter into an equation.

   $2\big((x+4)+x\big)=48$

3. 3

   Simplify and solve.

   $2(2x+4)=48\;\Longrightarrow\;4x+8=48\;\Longrightarrow\;4x=40\;\Longrightarrow\;x=10$

   Width $=10$ m, length $=14$ m.

4. 4

   Multiply to get the area.

   $A=14\times 10=140\ \text{m}^2$

   $\boxed{140\ \text{m}^2}$

5. 5

   Verify and compare. Perimeter: $2(14+10)=48$ m, and the length exceeds the width by $4$ m — both conditions hold. Interesting comparison: the sister problem with perimeter $52$ m and a $6$ m gap also has width $10$ m but a larger area, $160\ \text{m}^2$; a longer perimeter buys more area even though the shape is less square.