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# 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) are the vertices of a rhombus ABCD.

Length of diagonal AC
$$=\sqrt{[3-(-1)]^2 + (0-4)^2}$$
$$= \sqrt{16+16}$$
$$= \sqrt{32}$$
$$= 4\sqrt{2}$$
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 of rhombus = $$\frac{1}{2}$$ × product of diagonals
= $$\frac{1}{2} × 4\sqrt{2}×6\sqrt{2} = 24$$sq unit