Let P(–3, 2), Q(–5, –5), R(2, –3) and S(4, 4) be four points i

1.	Let P(–3, 2), Q(–5, –5), R(2, –3) and S(4, 4) be four points in a plane. Then show that PQRS is a rhombus. Is it a square ? Also find the area of the rhombus.
Answer» 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 = \(\frac{1}{2}\) x (Product of length of diagonals) = \(\frac{1}{2}\) x \(5\sqrt2\) x \(9\sqrt2\) = 45 sq. units.

Let P(–3, 2), Q(–5, –5), R(2, –3) and S(4, 4) be four points in a plane. Then show that PQRS is a rhombus. Is it a square ? Also find the area of the rhombus.

Answer»

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 = \(\frac{1}{2}\) x (Product of length of diagonals)

= \(\frac{1}{2}\) x \(5\sqrt2\) x \(9\sqrt2\) = 45 sq. units.