To determine the perimeter of the right isosceles triangle with vertices at (4,2) and (8,2), let's identify the three vertices.
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Identifying the vertices:
- The right angle vertex is at (4, 2).
- One vertex (which we'll call \( B \)) is at (8, 2).
- Since this is a right isosceles triangle, the two legs must be equal, and they are both perpendicular to each other.
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Calculating the length of the base (horizontal leg):
- The distance from the point (4, 2) to (8, 2) is: \[ \text{Length of } AB = 8 - 4 = 4 \]
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Calculating the coordinates of the third vertex \( C \):
- Since the triangle is isosceles, the height from vertex \( A \) (4, 2) to point \( C \) must also be 4 units long (the same as the length of leg \( AB \)).
- The vertical leg would go either upwards or downwards from point \( A \). We can compute the coordinates for both scenarios, but it doesn't alter the length since both would give the same distance. Here, we take the vertex upward: \[ C = (4, 2 + 4) = (4, 6) \]
- So the three vertices of the triangle are \( A(4,2) \), \( B(8,2) \), and \( C(4,6) \).
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Calculating the lengths of the other sides:
- Length \( AC \) (from \( A(4, 2) \) to \( C(4, 6) \)): \[ AC = 6 - 2 = 4 \]
- Length \( BC \) (from \( B(8, 2) \) to \( C(4, 6) \)):
- To find this length, use the distance formula: \[ BC = \sqrt{(8 - 4)^2 + (2 - 6)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \]
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Perimeter of the triangle:
- The perimeter \( P \) is the sum of all sides: \[ P = AB + AC + BC = 4 + 4 + 4\sqrt{2} \]
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Calculating \( 4\sqrt{2} \):
- Approximating \( \sqrt{2} \) to about 1.414: \[ 4\sqrt{2} \approx 4 \times 1.414 \approx 5.656 \]
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Final perimeter computation:
- Adding it up: \[ P \approx 4 + 4 + 5.656 = 13.656 \]
- Rounding to the nearest tenths gives \( 13.7 \).
Thus, the perimeter of the triangle is 13.7 units.
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