Question

Perpendiculars are drawn from the vertex of the obtuse angles of a rhombus to its sides. The length of each perpendicular is equal to a units. The distance between their feet being equal to b units. The area of the rhombus is

(a) \(\frac{\sqrt{a^2+b^2}}{2\sqrt{b^2-a^2}}\)

(b) \(\frac{2ab}{2\sqrt{b^2-a^2}}\)

(c) \(\frac{ab^2}{2\sqrt{b^2-a^2}}\)

(d) \(\frac{2a^2b^2}{2\sqrt{b^2-a^2}}\)

Answer 1

 (c) \(\frac{ab^2}{2\sqrt{b^2-a^2}}\)

Let AE and CF be the perpendiculars drawn from vertex A and vertex C respectively on sides CD and AB. 

⇒ E and F are the feet of the perpendiculars on the sides.

Given, AE = CF = a and EF = b.

AF = EC = \(\sqrt{b^2-a^2}\)

∴ Area of AFCE = AE × AF = a x \(\sqrt{b^2-a^2}\)

Let FB = x. Then, CB = \(\sqrt{a^2+x^2}\)

As ABCD is a rhombus,

CB = AB

⇒ \(\sqrt{a^2+x^2}\) = \(\sqrt{b^2-a^2}\) + x

⇒ a² + x² = b² – a² + x² + 2x\(\sqrt{b^2-a^2}\)

⇒ 2a² – b² = 2x\(\sqrt{b^2-a^2}\)

⇒ x = \(\frac{2a^2-b^2}{2\sqrt{b^2-a^2}}\)

∴ Area of rhombus ABCD = base × height = AB × AE