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RRB NTPC 2014 Paper

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Question : 55 of 100

Marks: +1, -0

A triangle is inscribed in a semicircle of radius R. What will be the perimeter of such a triangle which possesses the largest possible area?

3R
$\frac{1+\sqrt{2}}{2R}$
$2R(1+\sqrt{2})$
2 $R(1+\sqrt{2})$

Solution:

(4) For largest area, diameter should be the base of the triangle and radius should be its height.

Triangle ABC is the required triangle.

A B=A C=\sqrt{2} R

So, the perimeter

=\sqrt{2} R+\sqrt{2} R+

2 R=2 R+2 \sqrt{2}=2 R(1+\sqrt{2})

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