Question

2-1 Discussion: Models and Applications. 15 is 1cm A Rectangle has perimeter 140 cm and to langth than twice its width more

Answer 1

The length of the rectangle is 47 cm while its width is 23 cm

Here, we want to get the width and the length of the rectangle from the given information

Since we do not know the measure of these, we start by labeling the width and the length using variables

Let the width be w and the length be l

Firstly, we write the expresssion that calculates the perimeter

The expression is calculated as;

`P\text{ = 2(l + w)}`

where P in this case is 140 cm

Thus, we have;

`\begin{gathered} 140\text{ = 2(l + w)} \\ l\text{ + w = 70 }\ldots\ldots.(i) \end{gathered}`

We use the extra information to get another equation

The extra information is that the length is 1 cm more than twice its width

Mathematically, we have the equation as;

`l\text{ = 1 + 2w }\ldots\ldots\ldots\ldots(ii)`

So, now we have two equations to solve so as to get the width and the length

We can directly substitute the expression for l into equation 1

We have this as;

`\begin{gathered} 1\text{ + 2w + w = 70 } \\ 1\text{ + 3w = 70} \\ \\ 3w\text{ = 70-1} \\ \\ 3w\text{ = 69} \\ \\ w\text{ = }(69)/(3) \\ \\ w\text{ = 23 cm} \end{gathered}`

To get l, we simply substitute the value of w into the second equation

That would be;

`\begin{gathered} l\text{ = 1 + 2(23)} \\ l\text{ = 1 + 46} \\ l\text{ = 47 cm } \end{gathered}`