The problem shown in the image asks for the following:

1. Calculating the area (luas sawah) of a rectangular piece of land with dimensions $(6 + \sqrt{5}) \, \text{m}$ and $(6 - \sqrt{5}) \, \text{m}$.
2. Finding the length of the diagonal (panjang diagonal) of the same rectangle.

Let's solve both parts:

a) Calculating the area (Luas sawah):

The area $A$ of a rectangle is given by the formula: $A = \text{length} \times \text{width}$ Substituting the given dimensions: $A = (6 + \sqrt{5}) \times (6 - \sqrt{5})$ This is a difference of squares, so: $A = 6^2 - (\sqrt{5})^2 = 36 - 5 = 31 \, \text{square meters}$

b) Finding the diagonal length (Panjang diagonal):

The diagonal $d$ of a rectangle can be found using the Pythagorean theorem: $d = \sqrt{(\text{length})^2 + (\text{width})^2}$ Here, the length and width are $(6 + \sqrt{5})$ and $(6 - \sqrt{5})$. First, calculate the squares of each:

$(6 + \sqrt{5})^2 = 36 + 2 \times 6 \times \sqrt{5} + (\sqrt{5})^2 = 36 + 12\sqrt{5} + 5 = 41 + 12\sqrt{5}$

$(6 - \sqrt{5})^2 = 36 - 2 \times 6 \times \sqrt{5} + (\sqrt{5})^2 = 36 - 12\sqrt{5} + 5 = 41 - 12\sqrt{5}$

Now, adding the two expressions: $(6 + \sqrt{5})^2 + (6 - \sqrt{5})^2 = (41 + 12\sqrt{5}) + (41 - 12\sqrt{5}) = 82$ Therefore, the diagonal length is: $d = \sqrt{82} \approx 9.06 \, \text{meters}$

Final Answers:

• a) Area = 31 square meters
• b) Diagonal length ≈ 9.06 meters

Would you like more details or have any questions?

Related Questions:

1. How is the formula for the area of a rectangle derived?
2. What is the difference of squares, and how is it applied in this problem?
3. How is the Pythagorean theorem used to find the diagonal of a rectangle?
4. Can the area of a rectangle be negative?
5. How does the presence of square roots affect algebraic calculations?

Tip:

When using the Pythagorean theorem, always square the sides first before summing them up to avoid errors.