Vertices of a quadrilateral ABCD are A(0, 0), B(4, 5), C(9, 9) and D(5, 4). What is the shape of the quadrilateral?

Explanation: Calculate the length of the sides: \(A(0, 0), B(4, 5), C(9, 9), D(5, 4)\). Distance formula: \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\). \(AB = \sqrt{(4-0)^2 + (5-0)^2} = \sqrt{16 + 25} = \sqrt{41}\). \(BC = \sqrt{(9-4)^2 + (9-5)^2} = \sqrt{25 + 16} = \sqrt{41}\). \(CD = \sqrt{(5-9)^2 + (4-9)^2} = \sqrt{(-4)^2 + (-5)^2} = \sqrt{16 + 25} = \sqrt{41}\). \(DA = \sqrt{(0-5)^2 + (0-4)^2} = \sqrt{(-5)^2 + (-4)^2} = \sqrt{25 + 16} = \sqrt{41}\). All four sides are equal: it's a Rhombus or a Square. Calculate the length of the diagonals: \(AC = \sqrt{(9-0)^2 + (9-0)^2} = \sqrt{81 + 81} = \sqrt{162}\). \(BD = \sqrt{(5-4)^2 + (4-5)^2} = \sqrt{(1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}\). Since the diagonals are NOT equal (\(\sqrt{162} \neq \sqrt{2}\)), the quadrilateral is a **Rhombus** but not a Square.