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 = $\frac{1}{2}$ (product of its diagonals)]

Let A(3, 0), B(4, 5), C(- 1, 4) and D(- 2, - 1) be the vertices of a rhombus ABCD.

Also, Area of a rhombus =1/2 × (product of its diagonals)

Hence we will calculate the values of the diagonals AC and BD.

We know that the distance between the two points is given by the distance formula,

Distance formula = √( x₂ ^_- x₁ )² + (y₂ - y₁)² 

Therefore, distance between A (3, 0) and C (- 1, 4) is given by

Length of diagonal AC =√ [3 - (-1)]² + [0 - 4]²

= √(16 + 16)

= 4√2 

The distance between B (4, 5) and D (- 2, - 1) is given by

Length of diagonal BD = √[4 - (-2)]² + [5 - (-1)]²

= √(36 + 36)

= 6√2

Area of the rhombus ABCD = 1/2 × (Product of lengths of diagonals) = 1/2 × AC × BD

Therefore, the area of the rhombus ABCD = 1/2 × 4√2 × 6√2 square units

= 24 square units