Question

Find the measures of the sides of triangle ABC with vertices A(1, 5), B(3, -2), and C(-3, 0). Then classify the triangle by its sides.

Answer 1

Answer:

The measure of the sides are;

$\begin{gathered} AB=7.28 \\ BC=6.32 \\ AC=6.40 \end{gathered}$

Classifying the triangle by its sides. The triangle is a Scalene triangle

Explanation:

We want to find the length of the triangle ABC.

Recall that the distance between two points can be calculated using the formula;

$d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}$

Given;

$\begin{gathered} A(1,5) \\ B(3,-2) \\ C(-3,0) \end{gathered}$

For the given points A, B and C, the distance between the points are;

For side AB;

$\begin{gathered} AB=\sqrt[]{(3-1)^2+(-2-5)^2} \\ AB=\sqrt[]{2^2+(-7)^2} \\ AB=\sqrt[]{4+49} \\ AB=\sqrt[]{53} \\ AB=7.28 \end{gathered}$

For side BC;

$\begin{gathered} BC=\sqrt[]{(-3-3)^2+(0-(-2))^2} \\ BC=\sqrt[]{(-6)^2+(2)^2} \\ BC=\sqrt[]{36+4} \\ BC=\sqrt[]{40} \\ BC=6.32 \end{gathered}$

For side AC;

$\begin{gathered} AC=\sqrt[]{(-3-1)^2+(0-5)^2} \\ AC=\sqrt[]{(-4)^2+(-5)^2} \\ AC=\sqrt[]{16+25} \\ AC=\sqrt{41} \\ AC=6.40 \end{gathered}$

So, the measure of the sides are;

$\begin{gathered} AB=7.28 \\ BC=6.32 \\ AC=6.40 \end{gathered}$

From the derived length of the sides of the triangle, classifying the triangle by its sides. The triangle is a Scalene triangle.

Because it has no equal sides.