Q. 326 - Intermediate Algebra

Question

A rectangular retaining wall has area $15$ square feet. The height of the wall is two feet less than its length. Find the height and the length of the wall.

Step-by-Step Solution

Answer

The height and the length of the wall is $3 feet$ and $5 feet$.

Step 1. Assumption of integers.

Assume the length to be $x feet$ and height of the wall to be $x-2 feet$.

Area of rectangular retaining wall$=length\times height$

Substitute the values in the equation of area,

$$\begin{gathered} 15=\left(x-2\right)x \\ {x}^2-2x=15 \\ {x}^2-2x-15=0 ...... \left(i\right) \end{gathered}$$

Step 2. Factor the equation.

On factoring,

$$\begin{gathered} x^2-5x+3x-15=0 \\ x\left(x-5\right)+3\left(x-5\right)=0 \\ \left(x-5\right)\left(x+3\right)=0 \end{gathered}$$

Use the zero product property,

$$\begin{gathered} x-5=0 \\ x=5 \end{gathered}$$

Then,

$$\begin{gathered} x+3=0 \\ x=-3 \end{gathered}$$

As we know, the length of the wall cannot be negative, so $x=5$.

Then, the height of the wall$$\begin{gathered} =x-2 \\ =5-2 \\ =3 \end{gathered}$$

Step 3. Check the answers.

Substitute the height and the length in equation (i),

$$\begin{gathered} 15=3\times 5 \\ 15=15 \end{gathered}$$

This is true.

Hence, the height and the length of the wall is  $3 feet$ and $5 feet$.