Question

Can you help me solve this question?

Answer 1

Step1:

hence from the diagram above, the dimension of the rug is

`\begin{gathered} 34-2x\text{ and } \\ 24-2x \end{gathered}`

Since we have area of the rug,

then we have

`(34-2x)(24-2x)=704`

We now solve the equation above quadratically

`\begin{gathered} (34-2x)(24-2x)=704 \\ \text{expand the parenthesis } \\ 16-116x+4x^2=704 \\ \text{subtract 704 from both sides } \\ 816-116x+4x^2-704=704-704 \end{gathered}`

Then we have

`4x^2-116x+112=0`

Solve using factor method we obtain

`\begin{gathered} \text{Divide through by 4} \\ x^2-29x+28=0 \\ x^2-28x-x+28=0 \\ x(x-28)-1(x-28)=0 \\ (x-28)(x-1)=0 \end{gathered}`

Equating each factor to zero we have

`\begin{gathered} x-28=0,x-1=0 \\ x=28,1 \end{gathered}`

Since x can not be 28, the value of x is 1

The Dimension of the rug becomes

`\begin{gathered} 34-2x=34-2(1)=32 \\ \text{and } \\ 24-2x=24-2(1)=22 \end{gathered}`

Hence the dimension of the rug is 32 ft by 22ft (22,32)