Question

The length of a rectangle is 2 more than thrice its width. The perimeter of the rectangle is 52 cm. Find the area of the rectangle.

Answer 1

SOLUTION

Let the length of the rectangle be L

and the width of the rectangle be w

We are told that the length of a rectangle is 2 more than thrice its width.

That is

$\begin{gathered} L=2+(3\times w) \\ L=2+3w \end{gathered}$

So, the perimeter of the rectangle is 52 cm, and perimeter P of a rectangle is calculated as

$\begin{gathered} P=2(L+w) \\ 52=2(L+w) \end{gathered}$

now we will substitute the 2 + 3w for L into the equation, we have

$\begin{gathered} 52=2(L+w) \\ 52=2(2+3w+w) \\ 52=2(2+4w) \\ 52=4+8w \\ 8w=52-4 \\ 8w=48 \\ w=\frac{48}{8} \\ w=6\text{ cm } \end{gathered}$

So, the width is 6 cm, the length becomes

$\begin{gathered} L=2+3w \\ L=2+(3\times6) \\ L=2+18 \\ L=20cm\text{ } \end{gathered}$

The area A becomes

$\begin{gathered} A=L\times w \\ A=20\times6 \\ A=120cm^2 \end{gathered}$

Hence the answer is 120 square centimeters