To solve for the length of the third side and the area of a right-angled triangle with a hypotenuse of [tex]\(2 + \sqrt{2}\)[/tex] and one side [tex]\(1 + \sqrt{2}\)[/tex], we will use the Pythagorean theorem and the formula for the area of a triangle.
### Step 1: Use the Pythagorean theorem to find the third side
The Pythagorean theorem states:
[tex]\[
(hypotenuse)^2 = (side1)^2 + (side2)^2
\][/tex]
Given:
[tex]\[ hypotenuse = 2 + \sqrt{2} \][/tex]
[tex]\[ side1 = 1 + \sqrt{2} \][/tex]
First, square the hypotenuse:
[tex]\[
(2 + \sqrt{2})^2 = 4 + 4\sqrt{2} + 2 = 6 + 4\sqrt{2}
\][/tex]
Next, square the known side:
[tex]\[
(1 + \sqrt{2})^2 = 1 + 2\sqrt{2} + 2 = 3 + 2\sqrt{2}
\][/tex]
Using the Pythagorean theorem, solve for the unknown side [tex]\(side2\)[/tex]:
[tex]\[
side2^2 = (hypotenuse)^2 - (side1)^2 = (6 + 4\sqrt{2}) - (3 + 2\sqrt{2})
\][/tex]
[tex]\[
side2^2 = 6 + 4\sqrt{2} - 3 - 2\sqrt{2} = 3 + 2\sqrt{2}
\][/tex]
Hence, the square of the third side is given by:
[tex]\[
side2^2 = 3 + 2\sqrt{2}
\][/tex]
Therefore,
[tex]\[
side2 = \sqrt{3 + 2\sqrt{2}}
\][/tex]
### Step 2: Find the area of the triangle
The area [tex]\(A\)[/tex] of a right-angled triangle is given by:
[tex]\[
A = \frac{1}{2} \times base \times height
\][/tex]
In this case, we use [tex]\(side1\)[/tex] and [tex]\(side2\)[/tex] as the base and height:
[tex]\[
base = 1 + \sqrt{2}
\][/tex]
[tex]\[
height = \sqrt{3 + 2\sqrt{2}}
\][/tex]
Therefore, the area [tex]\(A\)[/tex] of the triangle is:
[tex]\[
A = \frac{1}{2} \times (1 + \sqrt{2}) \times \sqrt{3 + 2\sqrt{2}}
\][/tex]
As a result, the length of the third side and the area of the triangle in their simplest forms are:
[tex]\[
\text{Third side: } \sqrt{3 + 2\sqrt{2}}
\][/tex]
[tex]\[
\text{Area: } \frac{(1 + \sqrt{2}) \sqrt{3 + 2\sqrt{2}}}{2}
\][/tex]
Thus, the solutions are:
1. The length of the third side is [tex]\( \sqrt{3 + 2\sqrt{2}} \)[/tex].
2. The area of the triangle is [tex]\( \frac{(1 + \sqrt{2}) \sqrt{3 + 2\sqrt{2}}}{2} \)[/tex].
