Sure, let's solve the problem step-by-step. We will find the perimeter, the area, and the diagonal length of the given rectangle.
### Given Data:
The sides of the rectangle are:
- Side 1: [tex]\(\sqrt{3} + 1\)[/tex]
- Side 2: [tex]\(\sqrt{3} - 1\)[/tex]
### Step-by-Step Solution:
1. Perimeter of the Rectangle:
The perimeter [tex]\(P\)[/tex] of a rectangle is calculated by the formula:
[tex]\[
P = 2 \times (\text{Length} + \text{Breadth})
\][/tex]
Here, we have:
[tex]\[
P = 2 \times ((\sqrt{3} + 1) + (\sqrt{3} - 1))
\][/tex]
Simplifying inside the parentheses:
[tex]\[
(\sqrt{3} + 1) + (\sqrt{3} - 1) = \sqrt{3} + \sqrt{3} + 1 - 1 = 2\sqrt{3}
\][/tex]
Therefore, the perimeter is:
[tex]\[
P = 2 \times 2\sqrt{3} = 4\sqrt{3}
\][/tex]
2. Area of the Rectangle:
The area [tex]\(A\)[/tex] of a rectangle is calculated by the formula:
[tex]\[
A = \text{Length} \times \text{Breadth}
\][/tex]
Here, the area [tex]\(A\)[/tex] will be:
[tex]\[
A = (\sqrt{3} + 1) \times (\sqrt{3} - 1)
\][/tex]
We can use the difference of squares formula: [tex]\((a + b)(a - b) = a^2 - b^2\)[/tex], where [tex]\(a = \sqrt{3}\)[/tex] and [tex]\(b = 1\)[/tex]:
[tex]\[
(\sqrt{3})^2 - (1)^2 = 3 - 1 = 2
\][/tex]
Therefore, the area is:
[tex]\[
A = 2
\][/tex]
3. Length of the Diagonal:
The length of the diagonal [tex]\(d\)[/tex] of a rectangle can be found using the Pythagorean theorem:
[tex]\[
d = \sqrt{(\text{Length})^2 + (\text{Breadth})^2}
\][/tex]
Substituting the side lengths:
[tex]\[
d = \sqrt{(\sqrt{3} + 1)^2 + (\sqrt{3} - 1)^2}
\][/tex]
Expanding both squares:
[tex]\[
(\sqrt{3} + 1)^2 = (\sqrt{3})^2 + 2(\sqrt{3})(1) + 1^2 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3}
\][/tex]
[tex]\[
(\sqrt{3} - 1)^2 = (\sqrt{3})^2 - 2(\sqrt{3})(1) + 1^2 = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3}
\][/tex]
Adding these results:
[tex]\[
4 + 2\sqrt{3} + 4 - 2\sqrt{3} = 8
\][/tex]
Thus, the length of the diagonal is:
[tex]\[
d = \sqrt{8} = 2\sqrt{2}
\][/tex]
### Summary of Results:
- The perimeter of the rectangle is: [tex]\(4\sqrt{3}\)[/tex]
- The area of the rectangle is: 2
- The length of the diagonal is: [tex]\(2\sqrt{2}\)[/tex]
