Question

A rectangle is 6 times as long as it is wide. The perimeter is 60 cm. Find the dimensions of the rectangle. Round to the nearest tenth if necessary.

Answer 1

We can take

That is, x be the length and and the width of the rectangle. So we have the following system of linear equations:

`\begin{gathered} \mleft\{\begin{aligned}x=6y\text{ (1)} \\ x+y+x+y=60\text{ (2)}\end{aligned}\mright. \\ \mleft\{\begin{aligned}x=6y\text{ (1)} \\ 2x+2y=60\text{ (2)}\end{aligned}\mright. \end{gathered}`

Since to obtain the perimeter all the sides of the rectangle must be added.

`\begin{gathered} 2(6y)+2y=60\text{ (2)} \\ 12y+2y=60 \\ 14y=60 \\ y=(60)/(14) \\ y=4.29 \end{gathered}`

Now that we have the value of y we can plug it into equation (1)

`\begin{gathered} x=6\cdot(60)/(14)\text{ (1)} \\ x=25.71 \end{gathered}`

Finally, the dimensions of the rectangle are 25.7 cm long and 4.3 cm wide.