Question

Can I get some help, it is due in 15 minutes and I need help. Thanks, any help appreciated

Answer 1

Well, I'm way past the 15 min mark, but here's how to do the question.

With this, you will need to use the distance formula, $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$, on XY, YZ, and ZX.

XY: $\sqrt{(3-1)^2+(1-6)^2}$

Firstly, solve inside the parentheses: $\sqrt{(2)^2+(-5)^2}$

Next, solve the exponents: $\sqrt{4+25}$

Next, solve the addition, and XY's distance will be √29

(The process is the same with the other 2 sides, so I'll go through them real quickly)

YZ:

$\sqrt{(6-3)^2+(3-1)^2}\\ \sqrt{(3)^2+(2)^2}\\ \sqrt{9+4}\\ \sqrt{13}$

ZX:

$\sqrt{(1-6)^2+(6-3)^2}\\ \sqrt{(-5)^2+(3)^2}\\ \sqrt{25+9}\\ \sqrt{34}$

Now that we got the 3 sides, we can add them up: $\sqrt{29}+\sqrt{13} +\sqrt{34} =14.8$

In short, your answer is 14.8, or the second option.