Question

The area of a rectangular street sign is 84 square inches. The perimeter is 40 inches. What are the dimensions of the sign?

inches by
inches

Answer 1

To find the dimensions of the rectangular street sign, we can use the formulas for area and perimeter of a rectangle.

Let the length be \( l \) and the width be \( w \).

1. The area \( A \) of the rectangle is given by:
\[
A = l \times w = 84 \quad \text{(1)}
\]

2. The perimeter \( P \) of the rectangle is given by:
\[
P = 2l + 2w = 40 \quad \text{(2)}
\]

From equation (2), we can simplify it:
\[
l + w = 20 \quad \text{(3)}
\]

Now, we have two equations:
1. \( l \times w = 84 \)
2. \( l + w = 20 \)

We can express \( w \) in terms of \( l \) from equation (3):
\[
w = 20 - l
\]

Substituting \( w \) into equation (1):
\[
l(20 - l) = 84
\]
Expanding this gives:
\[
20l - l^2 = 84
\]
Rearranging the equation gives us:
\[
l^2 - 20l + 84 = 0
\]

Now we can use the quadratic formula \( l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where:
- \( a = 1 \)
- \( b = -20 \)
- \( c = 84 \)

Calculating the discriminant:
\[
b^2 - 4ac = (-20)^2 - 4 \cdot 1 \cdot 84 = 400 - 336 = 64
\]

Now substituting back into the quadratic formula:
\[
l = \frac{20 \pm \sqrt{64}}{2} = \frac{20 \pm 8}{2}
\]

This gives us two possible values for \( l \):
1. \( l = \frac{28}{2} = 14 \)
2. \( l = \frac{12}{2} = 6 \)

Using \( l \) to find \( w \):
1. If \( l = 14 \):
\[
w = 20 - 14 = 6
\]

2. If \( l = 6 \):
\[
w = 20 - 6 = 14
\]

Thus, the dimensions of the street sign are:
\[
\text{14 inches by 6 inches.}
\]

Therefore, the final answer is:
**14 inches by 6 inches.**