To solve for the perimeter, area, and the diagonal of a rectangle with sides of length [tex]\(\sqrt{3}+1\)[/tex] and [tex]\(\sqrt{3}-1\)[/tex], we can follow these steps:
### Perimeter
The perimeter [tex]\(P\)[/tex] of a rectangle is calculated using the formula:
[tex]\[ P = 2 \times ( \text{length} + \text{width} ) \][/tex]
Given the side lengths [tex]\(\sqrt{3}+1\)[/tex] and [tex]\(\sqrt{3}-1\)[/tex]:
[tex]\[ P = 2 \times \left( (\sqrt{3}+1) + (\sqrt{3}-1) \right) \][/tex]
Simplify inside the parentheses:
[tex]\[ (\sqrt{3}+1) + (\sqrt{3}-1) = \sqrt{3} + 1 + \sqrt{3} - 1 = 2\sqrt{3} \][/tex]
So the perimeter simplifies to:
[tex]\[ P = 2 \times 2\sqrt{3} = 4\sqrt{3} \][/tex]
### Area
The area [tex]\(A\)[/tex] of a rectangle is calculated using the formula:
[tex]\[ A = \text{length} \times \text{width} \][/tex]
Given the side lengths [tex]\(\sqrt{3}+1\)[/tex] and [tex]\(\sqrt{3}-1\)[/tex]:
[tex]\[ A = (\sqrt{3}+1) \times (\sqrt{3}-1) \][/tex]
This is a difference of squares:
[tex]\[ (\sqrt{3}+1)(\sqrt{3}-1) = (\sqrt{3})^2 - (1)^2 = 3 - 1 = 2 \][/tex]
### Diagonal
The length of the diagonal [tex]\(d\)[/tex] can be calculated using the Pythagorean theorem:
[tex]\[ d = \sqrt{(\text{length})^2 + (\text{width})^2} \][/tex]
Given the side lengths [tex]\(\sqrt{3}+1\)[/tex] and [tex]\(\sqrt{3}-1\)[/tex]:
[tex]\[ d = \sqrt{(\sqrt{3}+1)^2 + (\sqrt{3}-1)^2} \][/tex]
First, square each term:
[tex]\[ (\sqrt{3}+1)^2 = (\sqrt{3})^2 + 2 \times \sqrt{3} \times 1 + 1^2 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3} \][/tex]
[tex]\[ (\sqrt{3}-1)^2 = (\sqrt{3})^2 - 2 \times \sqrt{3} \times 1 + 1^2 = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3} \][/tex]
Now add the results:
[tex]\[ (\sqrt{3}+1)^2 + (\sqrt{3}-1)^2 = (4 + 2\sqrt{3}) + (4 - 2\sqrt{3}) = 4 + 4 = 8 \][/tex]
So the length of the diagonal is:
[tex]\[ d = \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} \][/tex]
### Summary
- Perimeter: [tex]\(4\sqrt{3}\)[/tex]
- Area: [tex]\(2\)[/tex]
- Diagonal: [tex]\(2\sqrt{2}\)[/tex]
These are the simplest forms of the answers expressed as surds.
