question_answer

Of the given perimeter, the triangle having maximum area is

A)            Isosceles triangle

B)            Right angled triangle

C)            Equilateral

D)            None of these

Correct Answer: C

Solution :

           First we observe that triangle ABC having base AB, the area of \[\Delta ABC\]is greatest for which the altitude of C w.r.t. AB is greatest.            Let \[\theta \]be the semi-vertical angle of such a triangle ABC.            \[S=\]area of the \[t=2\]              \[=2\frac{{{a}^{2}}}{2}\sin (\pi -2\theta )+\frac{1}{2}{{a}^{2}}\sin 4\theta \]            \[={{a}^{2}}\sin 2\theta +\frac{{{a}^{2}}}{2}\sin 4\theta \]            Now S is maximum when \[\theta =\frac{\pi }{6}\]or \[2\theta =\frac{\pi }{3}\]            i.e., triangle is equilateral.