Question

1. a triangle abc has its vertices at a(-2, -3), b(2, 1), and c(5, -1). what is the perimeter of the triangle abc? a. 4√2+√6 b. 4√2+√13+√53 c. 22 d. 100

Explanation:

Step1: Use distance formula $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$ for side AB

Let $A(-2,-3)$ and $B(2,1)$. Then $AB=\sqrt{(2 + 2)^2+(1 + 3)^2}=\sqrt{16 + 16}=\sqrt{32}=4\sqrt{2}$.

Step2: Use distance formula for side BC

Let $B(2,1)$ and $C(5,-1)$. Then $BC=\sqrt{(5 - 2)^2+(-1 - 1)^2}=\sqrt{9+4}=\sqrt{13}$.

Step3: Use distance formula for side AC

Let $A(-2,-3)$ and $C(5,-1)$. Then $AC=\sqrt{(5 + 2)^2+(-1 + 3)^2}=\sqrt{49 + 4}=\sqrt{53}$.

Step4: Calculate perimeter

Perimeter $P=AB + BC+AC=4\sqrt{2}+\sqrt{13}+\sqrt{53}$.

Answer:

B. $4\sqrt{2}+\sqrt{13}+\sqrt{53}$