To find the perimeter of the right isosceles triangle, we first need to determine the coordinates of all three vertices.
We know two of the vertices:
- \( A (4, 2) \) (the right angle vertex)
- \( B (8, 2) \) (one vertex)
Since this is a right isosceles triangle, the lengths of the two legs will be equal. The length of one leg (along the x-axis) from point \( A \) to point \( B \) can be calculated as follows:
\[ AB = |x_2 - x_1| = |8 - 4| = 4 \text{ units} \]
Since this is an isosceles right triangle, the other leg will also be \( 4 \) units long, but it will go vertically from point \( A \) to the third vertex \( C \).
The coordinates of point \( C \) (the vertex perpendicular to the base \( AB \)) can either be:
- \( C (4, 2 + 4) = (4, 6) \) or
- \( C (4, 2 - 4) = (4, -2) \)
Now, we can calculate the lengths of all sides of the triangle.
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Side \( AB \) has been calculated: \[ AB = 4 \text{ units} \]
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Side \( AC \) can be computed: \[ AC = |y_2 - y_1| = |6 - 2| = 4 \text{ units} \quad \text{(for } C(4, 6) \text{)} \] or \[ AC = |y_2 - y_1| = |-2 - 2| = 4 \text{ units} \quad \text{(for } C(4, -2) \text{)} \]
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Side \( BC \): \[ BC = \sqrt{(8 - 4)^2 + (2 - y_C)^2} = \sqrt{4^2 + (2 - y_C)^2} \]
- For \( C(4, 6) \): \[ BC = \sqrt{(8 - 4)^2 + (2 - 6)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \approx 5.7 \text{ units} \]
- For \( C(4, -2) \): \[ BC = \sqrt{(8 - 4)^2 + (2 - (-2))^2} = \sqrt{4^2 + (2 + 2)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \approx 5.7 \text{ units} \]
Adding it all together for the perimeter: \[ P = AB + AC + BC = 4 + 4 + 4\sqrt{2} \approx 4 + 4 + 5.7 = 13.7 \text{ units} \]
So, the perimeter of the triangle rounded to the nearest tenths place is:
\[ \boxed{13.7 \text{ units}} \]
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