Question

A rectangular ink pad has an area of 28 square centimeters. Its perimeter is 22 centimeters. What are the dimensions of the ink pad?

centimeters by
centimeters

Answer 1

Let the length and width of the rectangular ink pad be \( l \) and \( w \), respectively. We are given two equations based on the area and perimeter of the rectangle:

1. Area:
\[
l \times w = 28
\]

2. Perimeter:
\[
2l + 2w = 22
\]

We can simplify the perimeter equation:
\[
l + w = 11
\]

Now we have the system of equations:
1. \( l \times w = 28 \)
2. \( l + w = 11 \)

From equation 2, we can express \( w \) in terms of \( l \):
\[
w = 11 - l
\]

Now we can substitute this expression for \( w \) into equation 1:
\[
l \times (11 - l) = 28
\]

Expanding this gives:
\[
11l - l^2 = 28
\]

Rearranging gives us a standard quadratic equation:
\[
l^2 - 11l + 28 = 0
\]

We can use the quadratic formula to solve for \( l \):
\[
l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = -11 \), and \( c = 28 \).

Calculating the discriminant:
\[
b^2 - 4ac = (-11)^2 - 4 \times 1 \times 28 = 121 - 112 = 9
\]

Now substituting into the quadratic formula:
\[
l = \frac{11 \pm \sqrt{9}}{2} = \frac{11 \pm 3}{2}
\]

This gives us two possible solutions for \( l \):
\[
l = \frac{14}{2} = 7 \quad \text{and} \quad l = \frac{8}{2} = 4
\]

So, the dimensions \( l \) and \( w \) are:
- If \( l = 7 \), then \( w = 11 - 7 = 4 \).
- If \( l = 4 \), then \( w = 11 - 4 = 7 \).

Thus, the dimensions of the ink pad are:
\[
7 \text{ cm by } 4 \text{ cm}
\]

So the final answer is:
\[
\boxed{7 \text{ centimeters by } 4 \text{ centimeters}}
\]