The problem asks us to find the length of a rectangle where the length is 6 cm more than its width, and the area is 216 cm².

Let's denote:

• $w$ as the width of the rectangle.
• The length, $l$, can be expressed as $l = w + 6$.

The area of a rectangle is given by: $\text{Area} = \text{length} \times \text{width}$ So, we have the equation: $w \times (w + 6) = 216$

Let's solve this quadratic equation for $w$, and then find the length $l$.The width of the rectangle is $12$ cm, and the length is $18$ cm.

Would you like more details or have any questions?

Here are 5 related questions to explore:

1. How would the solution change if the length were 8 cm more than the width instead of 6 cm?
2. What would the perimeter of this rectangle be?
3. How do you solve quadratic equations in general?
4. If the area was 300 cm² instead of 216 cm², what would the new dimensions be?
5. Can you find the diagonal of this rectangle using the Pythagorean theorem?

Tip: Quadratic equations often have two solutions, but in geometry, negative values are discarded as dimensions must be positive.