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Chapter 13: Problem 27

Find the lengths of the sides of the triangle with the given vertices, and determine whether the triangle is a right triangle, an isosceles triangle, or neither of these. $$ (-2,2,4),(-2,2,6),(-2,4,8) $$

Short Answer

Expert verified
The lengths of the sides of the triangle are 2, \(2\sqrt{2}\) and \(2\sqrt{2}\). The triangle is a right triangle but not an isosceles triangle.

Step by step solution

01

Calculate the distances between vertices

First, calculate the distances (lengths of sides) between each pair of points using the distance formula. Let's denote the three vertices as A(-2,2,4), B(-2,2,6), C(-2,4,8). The lengths will be: \[ AB= \sqrt{(-2-(-2))^2+(2-2)^2+(6-4)^2}=2 \] \[ BC= \sqrt{(-2-(-2))^2+(4-2)^2+(8-6)^2}=2\sqrt{2} \] \[ AC= \sqrt{(-2-(-2))^2+(4-2)^2+(8-4)^2}=2\sqrt{2} \]
02

Determine the type of triangle

Now that each of the lengths is determined, use these lengths to figure out the triangle's type. Here, AB^2 + AC^2 = BC^2, so it鈥檚 a right triangle. Also, as AB 鈮� AC 鈮� BC, it鈥檚 not an isosceles triangle.

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