From the given information, we can draw the following picture:
where the widht is three-fourths the length, denoted by x. So, the perimeter (P) is given as
![P=x+x+(3)/(4)x+(3)/(4)x](https://img.qammunity.org/2023/formulas/mathematics/college/c6590yuhlq2ab1msu9yv92ike883yb5adw.png)
which gives
![\begin{gathered} P=2x+(6)/(4)x \\ P=(14)/(4)x \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/k7sfbvge4mesdye1oowmhqibdgfnleoymq.png)
Since the perimeter is 70 feet, we have
![(14)/(4)x=70](https://img.qammunity.org/2023/formulas/mathematics/college/17a0hma7qui3w3ea212d37xlngo1h31vny.png)
so x is given as
![x=(70*4)/(14)=20](https://img.qammunity.org/2023/formulas/mathematics/college/hnjeeit7k59hhnc7vcdt2rvnpsy664wh5a.png)
This means that the length measure 20 feet and the width measures:
![\text{ width=}(3)/(4)(20)=15\text{ ft}](https://img.qammunity.org/2023/formulas/mathematics/college/wdj6831y8jw95y5u5y5ajgu41a8dl0ss8s.png)
Therefore, the width measures 15 ft.