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.You need to login to perform this action.
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