Final answer:
To solve for the dimensions of a rectangle with a perimeter of 68 feet where the width is 8/9 of the length, we set up an equation and solve for the length, finding it to be 18 feet. The width is then calculated to be 16 feet.
Step-by-step explanation:
To find the dimensions of a rectangle when the perimeter is given as 68 ft and the width is 8/9 times its length, we set up two equations from the details provided.
Let the length be L and the width be W, so according to the width relation W = (8/9)L. The perimeter of the rectangle is given by P = 2(L + W) = 68 ft.
Now, substituting W in the perimeter equation:
2(L + (8/9)L) = 68
Multiplying all terms by 9 to clear out fractions:
9[2(L + (8/9)L)] = 9(68)
18L + 16L = 612
34L = 612
L = 612 / 34
L = 18 ft
Now using the relation for W:
W = (8/9)L = (8/9)(18 ft) = 16 ft
So, the dimensions of the rectangle are 18 feet in length and 16 feet in width.