A rectangle is 6 m longer than it is wide. Its perimeter is 52 m. Find its area.

Step-by-step solution

1. 1

   Use one variable, not two. Two unknowns would need two equations, but the phrase "$6$ m longer than it is wide" already links them. Let the width be $x$ metres; then the length is $x+6$ metres, and only one equation is needed.

2. 2

   Write the perimeter equation. A rectangle's perimeter is twice the sum of one length and one width:

   $2\big((x+6)+x\big)=52$

3. 3

   Solve for the width.

   $2(2x+6)=52\;\Longrightarrow\;4x+12=52\;\Longrightarrow\;4x=40\;\Longrightarrow\;x=10$

   So the width is $10$ m and the length is $10+6=16$ m.

4. 4

   Answer the question that was actually asked. The problem wants the area, not the side lengths, so multiply:

   $A=\text{length}\times\text{width}=16\times 10=160\ \text{m}^2$

   $\boxed{160\ \text{m}^2}$

5. 5

   Check both given conditions. The difference is $16-10=6$ m as required, and the perimeter is $2(16+10)=2(26)=52$ m as required. Note also the unit change: lengths are in metres, area in square metres.