the coordinates of rhombus abcd are a(-4,-2), b(-2,6), c(6,8), and d(4,0). what is the area of the rhombus? round to the nearest whole number, if necessary. 30 square units 60 square units 102 square units 120 square units

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Accepted Answer

# Explanation:
## Step1: Find the length of the diagonals
The formula for the distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
For diagonal $AC$ with $A(-4,-2)$ and $C(6,8)$:
$$
\begin{align*}
d_{AC}&=\sqrt{(6-(-4))^2+(8 - (-2))^2}\
&=\sqrt{(6 + 4)^2+(8+ 2)^2}\
&=\sqrt{10^2+10^2}\
&=\sqrt{100 + 100}\
&=\sqrt{200}=10\sqrt{2}
\end{align*}
$$
For diagonal $BD$ with $B(-2,6)$ and $D(4,0)$:
$$
\begin{align*}
d_{BD}&=\sqrt{(4-(-2))^2+(0 - 6)^2}\
&=\sqrt{(4 + 2)^2+(-6)^2}\
&=\sqrt{6^2+(-6)^2}\
&=\sqrt{36+36}\
&=\sqrt{72}=6\sqrt{2}
\end{align*}
$$
## Step2: Calculate the area of the rhombus
The area formula of a rhombus is $A=\frac{1}{2}d_1d_2$, where $d_1$ and $d_2$ are the lengths of the diagonals.
Substitute $d_1 = 10\sqrt{2}$ and $d_2=6\sqrt{2}$ into the formula:
$$
\begin{align*}
A&=\frac{1}{2}\times(10\sqrt{2})\times(6\sqrt{2})\
&=\frac{1}{2}\times10\times6\times(\sqrt{2}\times\sqrt{2})\
&=30\times2\
& = 60
\end{align*}
$$

# Answer:
B. 60 square units

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QUESTION IMAGE

Question

Explanation:

Step1: Find the length of the diagonals

Step2: Calculate the area of the rhombus

Answer: