1.

The points (a, a), (–a, –a) and (\(-\sqrt{3a},+\sqrt{3a}\))  are the vertices of.......triangle whose area is.......(a) Isosceles, \(2\sqrt2a^2\) sq. units (b) Equilateral, \(2\sqrt3a^2\) sq. units (c) Scalene, \(4\sqrt3a^2\) sq. units (d) None of these

Answer»

(b) Equilateral, \(2\sqrt3a^2\) sq. units

Let A(a, a), B(–a, –a) and C \((-\sqrt3a,\sqrt3a)\) be the vertices of ΔABC. Then,

AB = \(\sqrt{(a+a)^2+(a+a)^2}\) = \(\sqrt{4a^2+4a^2}\) = \(\sqrt{8a^2}\) = \(2\sqrt2a\)

BC = \(\sqrt{(-a+\sqrt3a)^2+(-a-\sqrt3a)^2}\)

= \(\sqrt{a^2-2\sqrt3a+3a^2+a^2+2\sqrt3a+3a^2}\) = \(\sqrt{8a^2}\) = \(2\sqrt2a\)

AC = \(\sqrt{(-a+\sqrt3a)^2+(-a-\sqrt3a)^2}\)

= \(\sqrt{a^2+2\sqrt3a+3a^2+a^2-2\sqrt3a+3a^2}\) = \(\sqrt{8a^2}\) = \(2\sqrt2a\)

∵ AB = BC = AC, ΔABC is equilateral.

Area = \(\frac{\sqrt3}{4}\) (side)2 = \(\frac{\sqrt3}{4}\) x (\(2\sqrt2a\))2

= \(\frac{\sqrt3}{4}\) x 8a2 = \(2\sqrt3a^2\).


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