The problem from the image reads:

"A piece of land is 800 square meters. If the length is twice its width, what is the dimension of the land? (Ans: width = 20 m; length = 40 m)"

Let's break down the solution step-by-step:

1. Let the width of the land be $w$.

2. The length is twice the width, so the length is $2w$.

3. The area of a rectangle is given by:
   $\text{Area} = \text{Length} \times \text{Width}$ Substituting the values, we get: $800 = 2w \times w$ $800 = 2w^2$

4. Solving for $w$:
   $w^2 = \frac{800}{2}$ $w^2 = 400$ $w = \sqrt{400}$ $w = 20 \text{ meters}$

5. Length:
   $\text{Length} = 2w = 2 \times 20 = 40 \text{ meters}$

So, the dimensions of the land are:

• Width = 20 meters
• Length = 40 meters

Would you like more details or have any questions?

Related Questions:

1. If the length were three times the width, what would the dimensions be for the same area?
2. How would you find the dimensions if only the perimeter and area were given?
3. What would be the dimensions if the area was doubled, keeping the same proportion between length and width?
4. How do you calculate the diagonal of the land using the length and width?
5. What is the perimeter of this rectangular piece of land?

Tip: Always double-check your units to ensure consistency in calculations.