Find the length of each side of the triangle ABC and determine whether the triangle is an isosceles triangle, a right triangle, both, or neither. A (-1, 0, 1), B (1, 1, -1), C (1, 1, 1)
A) 1, 1, $\sqrt { 2 }$ , both
B) 1, 2, $\sqrt { 5 }$ , neither
C) 3, 2, $\sqrt { 5 }$ , right
D) 1, 1, $$\sqrt { 2 }$$ , isosceles

Question

Quizplus · Accepted Answer

To find the length of each side of the triangle, we need to use the distance formula:

d = $\sqrt { (x2-x1)^{2} + (y2-y1)^{2} + (z2-z1)^{2} }$

AB = $\sqrt { (1-(-1))^{2} + (1-0)^{2} + (-1-1)^{2} }$ = $\sqrt { 4 + 1 + 4 }$ = $\sqrt { 9 }$ = 3

AC = $\sqrt { (1-(-1))^{2} + (1-0)^{2} + (1-1)^{2} }$ = $\sqrt { 4 + 1 }$ = $\sqrt { 5 }$

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

Therefore, the lengths of the sides are AC = $\sqrt { 5 }$, AB = 3, and BC = 2.

To determine whether the triangle is isosceles, we need to check if any two sides have the same length. Since none of the sides have the same length, the triangle is not isosceles.

To determine whether the triangle is a right triangle, we need to check if the sum of the squares of two sides is equal to the square of the third side. Checking each combination of sides, we find that:

AC^2 + BC^2 = ($\sqrt { 5 }$)^2 + 2^2 = 5 + 4 = 9
AB^2 = 3^2 = 9

Since AC^2 + BC^2 = AB^2, the triangle is a right triangle.

Therefore, the answer is C (3, 2, $\sqrt { 5 }$, right).

Find the Length of Each Side of the Triangle ABC 2\sqrt { 2 }2​

Find the Length of Each Side of the Triangle ABC $\sqrt { 2 }$