P ≡ (–3, 2), Q ≡ (–5, –5), R ≡ (2, –3), S ≡ (4, 4)∴ PQ = \(\sqrt{(-5+3)^2+(-5-2)^2}\) = \(\sqrt{4+49}\) = \(\sqrt{53}\)QR = \(\sqrt{(2+5)^2+(-3+5)^2}\) = \(\sqrt{49+4}\) = \(\sqrt{53}\)RS = \(\sqrt{(4-2)^2+(4+3)^2}\) = \(\sqrt{4+49}\) = \(\sqrt{53}\)PS = \(\sqrt{(4+3)^2+(4-2)^2}\) = \(\sqrt{49+4}\) = \(\sqrt{53}\)∵ PQ = QR = RS = PS, therefore PQRS is a rhombus. For PQRS to be a square, diagonals PR and QS should be equal.PR = \(\sqrt{(2+3)^2+(-3-2)^2}\) = \(\sqrt{25+25}\) = \(\sqrt{50}\) = \(5\sqrt2\)QS = \(\sqrt{(4+5)^2+(4+5)^2}\) = \(\sqrt{81+81}\) = \(\sqrt{162}\) = \(9\sqrt2\)As PR ≠ QS, so PQRS is not a square.Area of rhombus = \(rac{1}{2}\) x (Product of length of diagonals) = \(rac{1}{2}\) x \(5\sqrt2\) x \(9\sqrt2\) = 45 sq. units.