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The area of an isosceles triangle is a, when the angle included between the two equal sides is 60º. What will be the area if the angle included between the two equal sides becomes 120º? (Keeping the length of equal sides same as before) 

(a) \(\frac{a}{2}\)

(b) a 

(c) \(\frac{3a}{2}\)

(d) 2a

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Best answer

(b) a

In isosceles ΔABC, \(\frac{BC}{x}\) = sin 30º

⇒ BC = \(x\) sin 30º 

⇒ BD = 2x sin 30°

and \(\frac{AC}{AB}\) = cos 30º

⇒ AC = \(x\) cos 30°

∴ Area of ΔABC = a = \(\frac{1}{2}\) x BD x AC

\(\frac{1}{2}\) x 2\(x\) sin 30° x \(x\) cos 30°

\(x\)\(\frac{1}{2}\)\(\frac{\sqrt3}{2}\) = \(\frac{\sqrt3}{4}\)\(x\)= a (given)

⇒ \(x\)\(\frac{4}{\sqrt3}a\)

Now angle between equal sides = 120°

∴ Area of triangle = \(\frac{1}{2}\) \(x\)2  sin 120°

[Using Area of Δ = \(\frac{1}{2}\) AB AD sin 120°]

\(\frac{1}{2}\) x \(\bigg(\frac{4a}{\sqrt3}\bigg)^2\) x \(\frac{\sqrt3}{2}\) = a.

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