the area of a rectangle is 40, the length of the rectangle is three times it's width. what is the length and width 

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

We know two things:
1. The area of the rectangle is 40, so the formula for the area is:
\[
l \times w = 40
\]
2. The length is three times the width:
\[
l = 3w
\]

Now, substitute \( l = 3w \) into the area formula:
\[
(3w) \times w = 40
\]
Simplifying:
\[
3w^2 = 40
\]
Now, solve for \( w \) by dividing both sides by 3:
\[
w^2 = \frac{40}{3} \approx 13.33
\]
Taking the square root of both sides:
\[
w \approx \sqrt{13.33} \approx 3.65
\]

Now, substitute \( w \approx 3.65 \) back into \( l = 3w \) to find the length:
\[
l = 3 \times 3.65 \approx 10.95
\]

Thus, the width is approximately \( 3.65 \) units, and the length is approximately \( 10.95 \) units.

Would you like more details or have any questions?

Here are 5 related questions to explore further:
1. What is the perimeter of this rectangle?
2. How would the dimensions change if the area were doubled?
3. What happens if the length were only twice the width?
4. Can you find the diagonal length of this rectangle?
5. How do you calculate the aspect ratio of this rectangle?

**Tip:** Always double-check your units when solving geometry problems to avoid confusion between measurements!

The area of a rectangle is 40, the length of the rectangle is three times its width. What is the length and width?

Learn how to find the length and width of a rectangle where the area is 40 and the length is three times the width. This problem involves using algebraic substitution and solving a quadratic equation to determine the dimensions. Suitable for students in Grades 6-8.

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