Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order.

Let co-ordinates of rhombus be A(3, 0), B(4, 5), C(-1, 4) and D(-2, -1).

By distance formula,

Distance between two points = $\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}

Substituting values we get :

$\Rightarrow AC = \sqrt{(-1 - 3)^2 + (4 - 0)^2} \\[1em] = \sqrt{(-4)^2 + 4^2} \\[1em] = \sqrt{16 + 16} \\[1em] = \sqrt{32} \\[1em] = 4\sqrt{2} \text{ units}. \\[1em] \Rightarrow BD = \sqrt{(-2 - 4)^2 + (-1 - 5)^2} \\[1em] = \sqrt{(-6)^2 + (-6)^2} \\[1em] = \sqrt{36 + 36} \\[1em] = \sqrt{72} \\[1em] = 6\sqrt{2} \text{ units}.$

By formula,

Area of rhombus (A) = $\dfrac{1}{2} \times$ Product of diagonals

Substituting values we get :

$A = \dfrac{1}{2} \times AC \times BD \\[1em] = \dfrac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} \\[1em] = \dfrac{1}{2} \times 24 \times 2 \\[1em] = \dfrac{1}{2} \times 48 \\[1em] = 24 \text{ sq. units}.$

Hence, area of rhombus = 24 sq. units.