2. Find the dimensions of a rectangle where the altitude is four sevenths of the base and the perimeter is 330feet
Solutions:
let a be the altitude, b be the base and P be the perimeter.
P = 330
a = \(\frac{4}{7}\) b
P = 2a+ 2b ; Formula for the perimeter of a rectangle
330 = (2)\(\frac{4}{7}\)b +2b
330 = \(\frac{8}{7}\)b + 2b
\(b = \frac{{330\left( 7 \right)}}{{8 + 14}}\)
b = 105 feet
a = \(\frac{4}{7}\)(105)
a = 60 feet