the length of a rectangle is one cm less than twice the width. The area is 45 square cm. Find the dimensions.

Let the width of the rectangle be \( w \) cm. Then, the length of the rectangle is given as "one cm less than twice the width," which translates to:

\[
\text{Length} = 2w - 1 \text{ cm}
\]

The area of the rectangle is given by the formula:

\[
\text{Area} = \text{Length} \times \text{Width}
\]

Substituting the expressions for length and width, we get:

\[
45 = (2w - 1) \times w
\]

Now, expand and solve for \( w \):

\[
45 = 2w^2 - w
\]

Rearranging the equation:

\[
2w^2 - w - 45 = 0
\]

We can solve this quadratic equation using the quadratic formula:

\[
w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation \( 2w^2 - w - 45 = 0 \), we have \( a = 2 \), \( b = -1 \), and \( c = -45 \). Substituting these into the quadratic formula:

\[
w = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-45)}}{2(2)}
\]
\[
w = \frac{1 \pm \sqrt{1 + 360}}{4}
\]
\[
w = \frac{1 \pm \sqrt{361}}{4}
\]
\[
w = \frac{1 \pm 19}{4}
\]

This gives two possible solutions:

\[
w = \frac{1 + 19}{4} = \frac{20}{4} = 5 \quad \text{or} \quad w = \frac{1 - 19}{4} = \frac{-18}{4} = -4.5
\]

Since the width must be positive, we have \( w = 5 \).

Now, find the length:

\[
\text{Length} = 2w - 1 = 2(5) - 1 = 10 - 1 = 9
\]

Thus, the dimensions of the rectangle are:

\[
\text{Width} = 5 \, \text{cm}, \quad \text{Length} = 9 \, \text{cm}
\]

### Final Answer:
- Width = 5 cm
- Length = 9 cm

Would you like more details or have any questions?

Here are 5 related questions to explore:
1. How would the formula change if the area was given as 60 square cm?
2. What if the length was three times the width minus 2 cm?
3. How would the problem change if it involved a square instead of a rectangle?
4. What is the perimeter of the rectangle given its dimensions?
5. Can you solve for the dimensions using factoring instead of the quadratic formula?

**Tip:** Always check for positive values when solving for real-world measurements like length and width.

The length of a rectangle is one cm less than twice the width. The area is 45 square cm. Find the dimensions.

Discover how to find the dimensions of a rectangle when the length is one cm less than twice the width, and the area is 45 square cm. This solution covers algebraic concepts, including the use of quadratic equations and the quadratic formula. Ideal for students in Grades 8-10.

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