Question

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

Answer 1

First you calculate the distance between B and D, A and C. 
We have the distance between B and D will be:  sqrt[(x2-x1)^2+(y2-y1)^2] = 8.5
The distance between A and C will be: 14.14 
We have the area of rhombus: 1/2 * 8.5 * 14.14 = 60 square units. 

Answer 2

The area of the rhombus is 60

Rhombus

A quadrilateral has four equal sides but angles are not 90 degrees.

Given

The corners of the rhombus are

A(-4,-2),

B(-2,6),

C(6,8), and

D(4,0)

To find the area of a rhombus

$\rm Area of rhombus = \dfrac{Product\ of\ diagonals}{2}$

The diagonals of a rhombus are AC and BD.

$\rm Distance\ between\ two\ poiunts=\sqrt{(x_{2} -x_{1} )^{2} +(y_{2} -y_{1} )^{2} }$

Diagonal AC will be

$AC=\sqrt{(x_{2} -x_{1} )^{2} +(y_{2} -y_{1} )^{2} }$

$AC=\sqrt{[6-(-4 )]^{2} +[8-(-2)]^{2} }$

$AC=\sqrt{[10]^{2} +[10]^{2} }\\ AC= \sqrt{100+100}\\ AC = \sqrt{200}\\ AC=10\sqrt{2}$

SImilarly for diagonal BD will be

$BD=\sqrt{(x_{2} -x_{1} )^{2} +(y_{2} -y_{1} )^{2} }$

$BD=\sqrt{[4-(-2 )]^{2} +[0-6]^{2} }$

$BD=\sqrt{[6]^{2} +[-6]^{2} }\\\\ BD= \sqrt{36+36}\\ \\ BD = \sqrt{72}\\\\ BD=6\sqrt{2}$

$\rm Area of rhombus = \dfrac{Product\ of\ diagonals}{2} \\ \rm Area of rhombus = \dfrac{AC*BD}{2}$

$\rm Area of rhombus = \dfrac{10\sqrt{2} * 6\sqrt{2} }{2} \\ Area of rhombus = 10*6\\ Area of rhombus=60$

Thus, the area of the rhombus is 60.

To more about the rhombus link is given below.

https://brainly.com/question/14462098