Question

The area of a rectangular ink pad is 36 square centimeters. The perimeter is 26 centimeters. What are the dimensions of the ink pad?

Answer 1

Let's denote the length of the rectangular ink pad as \( l \) and the width as \( w \).

We know two things:
1. The area of the rectangle:
\[
l \times w = 36
\]
2. The perimeter of the rectangle:
\[
2l + 2w = 26
\]
This can be simplified to:
\[
l + w = 13
\]

Now we have a system of equations:
1. \( l \times w = 36 \)
2. \( l + w = 13 \)

From the second equation, we can express \( w \) in terms of \( l \):
\[
w = 13 - l
\]

Substituting this expression for \( w \) into the first equation gives:
\[
l \times (13 - l) = 36
\]

Expanding and rearranging the equation, we get:
\[
13l - l^2 = 36
\]
\[
l^2 - 13l + 36 = 0
\]

Now we can solve this quadratic equation using the quadratic formula:
\[
l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = -13 \), and \( c = 36 \). Plugging in these values, we have:
\[
l = \frac{13 \pm \sqrt{(-13)^2 - 4 \cdot 1 \cdot 36}}{2 \cdot 1}
\]
\[
l = \frac{13 \pm \sqrt{169 - 144}}{2}
\]
\[
l = \frac{13 \pm \sqrt{25}}{2}
\]
\[
l = \frac{13 \pm 5}{2}
\]

This gives us two potential values for \( l \):
1.
\[
l = \frac{18}{2} = 9
\]
2.
\[
l = \frac{8}{2} = 4
\]

Now we will find the corresponding values of \( w \):
1. If \( l = 9 \):
\[
w = 13 - 9 = 4
\]
2. If \( l = 4 \):
\[
w = 13 - 4 = 9
\]

Thus, the dimensions of the ink pad are:
\[
\text{Length} = 9 \text{ cm}, \text{ Width} = 4 \text{ cm}
\]

The dimensions of the ink pad are \( 9 \, \text{cm} \times 4 \, \text{cm} \).