The length of the rug is 4 ft.
The width of the rug is 2.5 ft.
Explanation:
The area of the rug is 10 ft.
The length of the rug be l.
Let us convert the inches to feet.
Thus, 
Thus, the length of the rug is 
Let the width of the rug be w.
Substituting these values in the formula of area of the rectangle, we get,


Solving the expression using the quadratic formula,

Substituting the values, we have,



Thus,
and 
Since, the value of w cannot be negative, the value of w is 2.5ft
Thus, the width of the rug is 2.5ft
Substituting
in
, we get,

Thus, the length of the rug is 4 ft.