What is the area of rhombus ABCD ?Enter your answer in the box. Do not round at any steps.
Accepted Solution
A:
Answer:Area of the rhombus ABCD = 16 square unitsStep-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 CDiagonal 2 = Distance between the points B and DLength 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 unitsDiagonal 2(BD) = [tex]\sqrt{(2-2)^2+(3+5)^2}[/tex] = 8 unitsNow area of the rhombus ABCD = [tex]\frac{1}{2}(\text{AC})(\text{BD})[/tex] = [tex]\frac{1}{2}\times 4\times 8[/tex] = 16 units²Therefore, area of the given rhombus is 16 units².