Find the area of a rhombus if its vertices are (3, 0), (4, 5), (−1, 4) and (−2, −1) taken in order. (Hint: Area of a rhombus \(= \dfrac{1}{2}\) × product of its diagonals.)
24 sq. units
Let the vertices of the rhombus taken in order be \(A(3, 0)\), \(B(4, 5)\), \(C(-1, 4)\) and \(D(-2, -1)\).
In a rhombus, the area is given by \(\tfrac{1}{2} \times (\text{product of its diagonals})\), so we first find the lengths of the diagonals \(AC\) and \(BD\).
Diagonal \(AC\) joins \(A(3, 0)\) and \(C(-1, 4)\). Using the distance formula, \( AC = \sqrt{(-1 - 3)^2 + (4 - 0)^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \).
Diagonal \(BD\) joins \(B(4, 5)\) and \(D(-2, -1)\). Then \( BD = \sqrt{(-2 - 4)^2 + (-1 - 5)^2} = \sqrt{(-6)^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \).
Now, area of the rhombus \(= \tfrac{1}{2} \times AC \times BD = \tfrac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} \).
Compute the product: \(4\sqrt{2} \times 6\sqrt{2} = 24 \times 2 = 48\). So the area is \( \tfrac{1}{2} \times 48 = 24 \).
Therefore, the area of the rhombus is \(24\) square units.