Question

find the length and width of a rectangle whose perimeter is 32 feet and whose area is 60 square feet.
the length (the longer side) is
and the width (the shorter side) is
(simplify your answers.)

Explanation:

Step1: Set up equations

Let the length be \( l \) and the width be \( w \).
Perimeter formula: \( 2(l + w)=32\), so \( l + w = 16\), then \( l=16 - w \).
Area formula: \( lw = 60\).

Step2: Substitute and solve

Substitute \( l = 16 - w \) into \( lw = 60\):
\((16 - w)w=60\)
\(16w-w^{2}=60\)
\(w^{2}-16w + 60=0\)
Factor the quadratic equation: \(w^{2}-16w + 60=(w - 6)(w - 10)=0\)
So \( w = 6\) or \( w = 10\).
If \( w = 6\), then \( l=16 - 6=10\).
If \( w = 10\), then \( l=16 - 10 = 6\). Since length is the longer side.

Answer:

The length (the longer side) is \(10\) feet and the width (the shorter side) is \(6\) feet.