Question

un triángulo abc tiene sus vértices en a( -2 , -3) b( 2 , 1) y c( 5 , -1 )
cuál es el perímetro del triángulo abc ?
a. 22
b. $4\sqrt{2} + \sqrt{13} + \sqrt{53}$
c. 100
d. $4\sqrt{2} + \sqrt{6}$

Explanation:

To find the perimeter of triangle \( ABC \) with vertices \( A(-2, -3) \), \( B(2, 1) \), and \( C(5, -1) \), we need to calculate the lengths of the three sides \( AB \), \( BC \), and \( CA \) using the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) and then sum them up.

Step 1: Calculate the length of \( AB \)

For points \( A(-2, -3) \) and \( B(2, 1) \):
\[

$$\begin{align*} AB &= \sqrt{(2 - (-2))^2 + (1 - (-3))^2} \\ &= \sqrt{(2 + 2)^2 + (1 + 3)^2} \\ &= \sqrt{4^2 + 4^2} \\ &= \sqrt{16 + 16} \\ &= \sqrt{32} \\ &= 4\sqrt{2} \end{align*}$$

\]

Step 2: Calculate the length of \( BC \)

For points \( B(2, 1) \) and \( C(5, -1) \):
\[

$$\begin{align*} BC &= \sqrt{(5 - 2)^2 + (-1 - 1)^2} \\ &= \sqrt{3^2 + (-2)^2} \\ &= \sqrt{9 + 4} \\ &= \sqrt{13} \end{align*}$$

\]

Step 3: Calculate the length of \( CA \)

For points \( C(5, -1) \) and \( A(-2, -3) \):
\[

$$\begin{align*} CA &= \sqrt{(-2 - 5)^2 + (-3 - (-1))^2} \\ &= \sqrt{(-7)^2 + (-3 + 1)^2} \\ &= \sqrt{49 + (-2)^2} \\ &= \sqrt{49 + 4} \\ &= \sqrt{53} \end{align*}$$

\]

Step 4: Calculate the perimeter

The perimeter \( P \) of triangle \( ABC \) is the sum of the lengths of its three sides:
\[

$$\begin{align*} P &= AB + BC + CA \\ &= 4\sqrt{2} + \sqrt{13} + \sqrt{53} \end{align*}$$

\]

Answer:

\( 4\sqrt{2} + \sqrt{13} + \sqrt{53} \) (Option B)