What is the area of a rhombus whose vertices are (3,0), (4,5), (-1,4), and (-2,-1)?
Answer: B
The area of a rhombus can be calculated as half the product of its diagonals.
Let the vertices be A(3,0), B(4,5), C(-1,4), D(-2,-1).
Length of diagonal AC = \(\sqrt{(-1-3)^2 + (4-0)^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2}\).
Length of diagonal BD = \(\sqrt{(-2-4)^2 + (-1-5)^2} = \sqrt{(-6)^2 + (-6)^2} = \sqrt{36+36} = \sqrt{72} = 6\sqrt{2}\).
Area = \(\frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} = \frac{1}{2} \times 24 \times 2 = 24\) sq units.