What is the area of rhombus ABCD? Enter your answer in the box. Do not round at any steps.

Mathematics

1 answer:

Mrac [35]3 years ago

5 0

Answer:

Area of the rhombus ABCD = 16 square units

Step-by-step explanation:

Area of a rhombus = $\frac{1}{2}(\text{Diagonal 1})(\text{Diagonal 2})$

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) = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^2 }$

Diagonal 1 (AC) = $\sqrt{(4-0)^2+(-1+1)^2}$ = 4 units

Diagonal 2(BD) = $\sqrt{(2-2)^2+(3+5)^2}$ = 8 units

Now area of the rhombus ABCD = $\frac{1}{2}(\text{AC})(\text{BD})$

= $\frac{1}{2}\times 4\times 8$

= 16 units²

Therefore, area of the given rhombus is 16 units².

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