Question

A rectangular roof shingle has an area of 60 square inches. Its perimeter is 32 inches. What are the dimensions of the shingle?

inches by
inches

Answer 1

Let the length of the shingle be \( l \) inches and the width be \( w \) inches. We are given the following two equations based on the area and perimeter:

1. **Area equation**:
\[
l \times w = 60
\]

2. **Perimeter equation**:
\[
2l + 2w = 32
\]
Simplifying the perimeter equation, we can divide everything by 2:
\[
l + w = 16
\]

Now we have a system of equations:
1. \( lw = 60 \)
2. \( l + w = 16 \)

From the second equation, we can express \( w \) in terms of \( l \):
\[
w = 16 - l
\]

Now we can substitute this expression for \( w \) into the area equation:
\[
l(16 - l) = 60
\]
Expanding this, we get:
\[
16l - l^2 = 60
\]
Rearranging to form a standard quadratic equation:
\[
l^2 - 16l + 60 = 0
\]

Next, we can use the quadratic formula to find the values of \( l \):
\[
l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1 \), \( b = -16 \), and \( c = 60 \):
\[
l = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot 60}}{2 \cdot 1}
\]
\[
l = \frac{16 \pm \sqrt{256 - 240}}{2}
\]
\[
l = \frac{16 \pm \sqrt{16}}{2}
\]
\[
l = \frac{16 \pm 4}{2}
\]

This gives us two potential solutions for \( l \):
\[
l = \frac{20}{2} = 10 \quad \text{or} \quad l = \frac{12}{2} = 6
\]

Now, we can find \( w \) for both cases:

1. If \( l = 10 \):
\[
w = 16 - 10 = 6
\]
2. If \( l = 6 \):
\[
w = 16 - 6 = 10
\]

Thus, the dimensions of the shingle are \( 10 \) inches by \( 6 \) inches.

The final answer is:
**10 inches by 6 inches**.