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The problem in the image asks for the dimensions of a piece of land that has an area of 800 square meters, with the condition that the length is twice the width. Here's how we can solve it:

Let:
- \( w \) be the width of the land (in meters),
- \( l = 2w \) be the length of the land (since the length is twice the width).

The area \( A \) of the land is given by:
\[
A = \text{length} \times \text{width}
\]
Substituting the given values:
\[
800 = 2w \times w
\]
This simplifies to:
\[
800 = 2w^2
\]
Now solve for \( w \):
\[
w^2 = \frac{800}{2} = 400
\]
\[
w = \sqrt{400} = 20 \text{ meters}
\]

Since the length is twice the width:
\[
l = 2 \times 20 = 40 \text{ meters}
\]

Thus, the dimensions of the land are:
- Width = 20 meters,
- Length = 40 meters.

Would you like further clarification or more details on any part of the solution?

Here are 5 related questions to explore:

1. How would the dimensions change if the area were 1600 square meters instead?
2. What would happen to the dimensions if the length were three times the width instead of two?
3. Can you generalize a formula for the width if you are given the area and a different length-to-width ratio?
4. How would the solution change if the total perimeter of the land were given instead of the area?
5. What are the implications of the dimensions if the land is subdivided into two equal parts?

**Tip:** When solving problems involving areas and dimensions, make sure to express everything in terms of a single variable when possible, as this simplifies solving the equation.

A piece of land is 800 square meters. If the length is twice its width, what is the dimension of the land?

Solve a problem where the area of a rectangular piece of land is 800 square meters, and the length is twice its width. Learn how to use basic algebra and geometry to find the dimensions of the land, suitable for students in Grades 7-8.

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