To find the dimensions of the rectangular waiting room given its area and perimeter, we need to solve the problem step-by-step with the following information:
- Area (A) = 120 square feet
- Perimeter (P) = 44 feet
Let's denote the length of the rectangle as [tex]\( L \)[/tex] and the width as [tex]\( W \)[/tex].
1. Formulate the equations:
From the definition of area:
[tex]\[
L \times W = 120
\][/tex]
The formula for the perimeter of a rectangle is:
[tex]\[
2L + 2W = 44
\][/tex]
Simplifying the perimeter equation by dividing everything by 2, we get:
[tex]\[
L + W = 22
\][/tex]
2. Solve the system of equations:
We now have the following system of equations:
[tex]\[
\begin{cases}
L \times W = 120 \\
L + W = 22
\end{cases}
\][/tex]
First, solve the second equation for one of the variables, say [tex]\( W \)[/tex]:
[tex]\[
W = 22 - L
\][/tex]
Substitute this expression for [tex]\( W \)[/tex] into the first equation:
[tex]\[
L \times (22 - L) = 120
\][/tex]
Simplify and solve the resulting quadratic equation:
[tex]\[
22L - L^2 = 120
\][/tex]
[tex]\[
L^2 - 22L + 120 = 0
\][/tex]
3. Solve the quadratic equation:
The quadratic equation [tex]\( L^2 - 22L + 120 = 0 \)[/tex] can be solved using the quadratic formula:
[tex]\[
L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = -22 \)[/tex], and [tex]\( c = 120 \)[/tex].
Calculate the discriminant:
[tex]\[
b^2 - 4ac = (-22)^2 - 4 \times 1 \times 120 = 484 - 480 = 4
\][/tex]
Thus,
[tex]\[
L = \frac{22 \pm \sqrt{4}}{2 \times 1} = \frac{22 \pm 2}{2}
\][/tex]
This gives us two solutions for [tex]\( L \)[/tex]:
[tex]\[
L = \frac{22 + 2}{2} = 12 \quad \text{and} \quad L = \frac{22 - 2}{2} = 10
\][/tex]
4. Find the corresponding widths:
For [tex]\( L = 12 \)[/tex]:
[tex]\[
W = 22 - 12 = 10
\][/tex]
For [tex]\( L = 10 \)[/tex]:
[tex]\[
W = 22 - 10 = 12
\][/tex]
Therefore, the dimensions of the rectangular room are either:
- Length = 12 feet and Width = 10 feet
- Length = 10 feet and Width = 12 feet
In summary, the dimensions of the waiting room are 12 feet by 10 feet. Since the length and width can be interchanged, both pairs of dimensions are correct.
