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?

centimeters by
centimeters

Answer 1

Let the length of the ink pad be \( l \) cm and the width be \( w \) cm. We have two equations based on the information given:

1. The area of the rectangle:
\[
l \times w = 36
\]

2. The perimeter of the rectangle:
\[
2l + 2w = 26
\]
Simplifying the perimeter equation gives:
\[
l + w = 13
\]

Now we can solve these two equations simultaneously. From the second equation, we can express \( w \) in terms of \( l \):
\[
w = 13 - l
\]

Now substitute this expression for \( w \) into the area equation:
\[
l \times (13 - l) = 36
\]

Expanding and rearranging:
\[
13l - l^2 = 36
\]
\[
l^2 - 13l + 36 = 0
\]

Now we can use the quadratic formula to solve for \( l \):
\[
l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = -13 \), and \( c = 36 \).

Calculating the discriminant:
\[
b^2 - 4ac = (-13)^2 - 4 \times 1 \times 36 = 169 - 144 = 25
\]

Now solving for \( l \):
\[
l = \frac{13 \pm \sqrt{25}}{2} = \frac{13 \pm 5}{2}
\]
This gives us two possible values for \( l \):
\[
l = \frac{18}{2} = 9 \quad \text{and} \quad l = \frac{8}{2} = 4
\]

So, the two possible dimensions are:
1. \( l = 9 \), \( w = 4 \)
2. \( l = 4 \), \( w = 9 \)

Thus, the dimensions of the ink pad are:
\[
\text{9 centimeters by 4 centimeters}
\]
or equivalently,
\[
\text{4 centimeters by 9 centimeters}
\]

Final answer:
**9 centimeters by 4 centimeters.**