Question

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)]

Answer 1

Let (3, 0), (4, 5), (−1, 4) and (−2, −1) are the vertices A, B, C, D of a rhombus ABCD.  
Length of diagonal AC=\(\sqrt{[3-(-1)]^2+(0-4)}\)
                                    = \(\sqrt{16+16}=4\sqrt2\)
Length of diagonal BD=\(\sqrt{[4-(-2)]^2+[5-(-1)]^2}\)
                                    =\(\sqrt{36+36}=6\sqrt2\)
Therefore the area of rhombus ABCD = \(\frac{1}{2}\times4\sqrt2\times6\sqrt2\)
                                                             = 24 square units