We are given that the width $w$ of a rectangle is one-fourth of its length $l$, and its area is 9 square units.

The area $A$ of a rectangle is given by the formula:

$A = l \times w$

Since the width $w$ is one-fourth of the length $l$, we can express this as:

$w = \frac{l}{4}$

Substitute this into the area formula:

$A = l \times \frac{l}{4} = \frac{l^2}{4}$

We are told that the area is 9 square units, so:

$\frac{l^2}{4} = 9$

Multiply both sides by 4 to eliminate the denominator:

$l^2 = 36$

Now, solve for $l$ by taking the square root of both sides:

$l = \sqrt{36} = 6$

Thus, the length of the rectangle is 6 units.

Would you like further clarification or details?

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Here are 5 related questions to consider:

1. How would you find the width of the rectangle based on its length?
2. What is the perimeter of the rectangle if its length is 6 units?
3. If the length were doubled, what would be the new area?
4. Can you solve the equation for length if the area is changed to 16 square units?
5. What is the relationship between the length and width in this type of rectangle?

Tip: Always check if you can express one variable in terms of another to simplify solving geometry problems.