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Find the perimeter of the triangle XYZ with verticles X (1, 3) , Y(-4, -1) and Z(4, -1). Round your answer to two decimal places

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5 votes

Answer:

The perimeter is approximately 6.40.

Explanation:

To find the perimeter of a triangle with vertices X (1, 3), Y (-4, -1), and Z (4, -1), we can first find the distance between each pair of points using the distance formula.The distance between points X and Y is:

$d(X,Y) = \sqrt{((1 - (-4))^2 + (3 - (-1))^2)} = \sqrt{25 + 16} = \sqrt{41}$

The distance between points Y and Z is:$d(Y,Z) =

\sqrt{((-4 - 4)^2 + ((-1) - (-1))^2)} = \sqrt{0 + 0} = 0$

The distance between points X and Z is:

$d(X,Z) = \sqrt{((1 - 4)^2 + (3 - (-1))^2)} = \sqrt{9 + 16} = \sqrt{25}$

We can then find the perimeter of the triangle by adding up the lengths of the sides. In this case, the perimeter is $d(X,Y) + d(Y,Z) + d(X,Z) = \sqrt{41} + 0 + \sqrt{25} = \sqrt{41} + \sqrt{25} = 6.40\ldots$.

Rounded to two decimal places, the perimeter is approximately 6.40.

User Mark Vayngrib
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