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    Topics || Problems

    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