what is the area of a rectangle with vertices at $(-6,3)$, $(-3,6)$, $(1,2)$, and $(-2,-1)$? enter your answer in the box. do not round any side lengths. units²

Question

Accepted Answer

# Explanation:
## Step1: Calculate the length of one side
Use the distance formula $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Let's take two adjacent vertices $(-6,3)$ and $(-3,6)$.
$$
\begin{align*}
d_1&=\sqrt{(-3+6)^2+(6 - 3)^2}\
&=\sqrt{3^2+3^2}\
&=\sqrt{9 + 9}\
&=\sqrt{18}=3\sqrt{2}
\end{align*}
$$
## Step2: Calculate the length of the adjacent side
Take vertices $(-3,6)$ and $(1,2)$
$$
\begin{align*}
d_2&=\sqrt{(1 + 3)^2+(2-6)^2}\
&=\sqrt{4^2+(-4)^2}\
&=\sqrt{16+16}\
&=\sqrt{32}=4\sqrt{2}
\end{align*}
$$
## Step3: Calculate the area of the rectangle
The area of a rectangle $A=\text{length}\times\text{width}$. Here $A = d_1\times d_2$
$$
\begin{align*}
A&=(3\sqrt{2})\times(4\sqrt{2})\
&=3\times4\times\sqrt{2}\times\sqrt{2}\
&=12\times2\
&=24
\end{align*}
$$
# Answer:
24

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate the length of one side

Step2: Calculate the length of the adjacent side

Step3: Calculate the area of the rectangle

Answer: