Q. 442 - Intermediate Algebra

Question

The area of a rectangular shaped patio $432$

square feet. The length of the patio is $6$ feet more than its width. Find the length and width.

Step-by-Step Solution

Answer

The length and width are $24$ feet and $18$ feet respectively.

Step 1. Given

The area of a rectangular shaped patio $432$

square feet.

The length of the patio is $6$ feet more than its width.

Step 2. Form equations for both the conditions, and solve them.

Let length=$l$

and width=$w$

$A$(Area)=$432$ square feet

$$\begin{gathered} A=432 \\ l\times w=432 \\ l=\frac{432}{w} \end{gathered}$$

$$\begin{gathered} l=w+6 \\ l-w=6 \\ \frac{432}{w}-w=6 \\ 432-w^2=6w \\ {w}^2+6w-432=0 \\ {w}^2+24w-18w-432=0 \\ w(w+24)-18(w+24)=0 \\ (w+24)(w-18)=0 \end{gathered}$$

So,

$w=-24,18$

But width can't be negative.

So, width is $18$

Length is

$$\begin{gathered} l+6 \\ =18+6 \\ =24 \end{gathered}$$

Step 3. Required answer

The length and width are $24$ feet and $18$ feet respectively.