Answer:
Area of the rhombus ABCD = 16 square units
Step-by-step explanation:
Area of a rhombus = [tex]\frac{1}{2}(\text{Diagonal 1})(\text{Diagonal 2})[/tex]
From the graph attached,
Diagonal 1 = Distance between the points A and C
Diagonal 2 = Distance between the points B and D
Length of a segment between (x₁, y₁) and (x₂, y₂) = [tex]\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^2 }[/tex]
Diagonal 1 (AC) = [tex]\sqrt{(4-0)^2+(-1+1)^2}[/tex] = 4 units
Diagonal 2(BD) = [tex]\sqrt{(2-2)^2+(3+5)^2}[/tex] = 8 units
Now area of the rhombus ABCD = [tex]\frac{1}{2}(\text{AC})(\text{BD})[/tex]
= [tex]\frac{1}{2}\times 4\times 8[/tex]
= 16 units²
Therefore, area of the given rhombus is 16 units².
