Question

The triangle of maximum area that can be inscribed in a given circle of radius 'r' is :

• A. An equilateral triangle having each of its side of length \(\sqrt{3}\,r\).
• B. An equilateral triangle of height \(\frac{2r}{3}\).
• C. A right angle triangle having two of its sides of length \(2r\) and \(r\).
• D. An isosceles triangle with base equal to \(2r\).

Hint:

For a given base the area is largest for the isosceles triangle; maximise \(r^2\sin\theta(1+\cos\theta)\) to get \(\theta = 60^\circ\), the equilateral triangle.

Answer 1

The Correct Option is A

For a fixed base chord, the area is maximum when the apex is the farthest point on the circle, so the triangle is isosceles. Let the half-angle at the centre subtended by the base be \(\theta\): base \(= 2r\sin\theta\), height \(= r + r\cos\theta\).
Area \(A = r^2\sin\theta\,(1+\cos\theta)\). \(\frac{dA}{d\theta} = r^2(\cos\theta + \cos 2\theta) = 0\) gives \(2\cos^2\theta + \cos\theta - 1 = 0\), so \(\cos\theta = \frac12\), \(\theta = 60^\circ\).
Base \(= \sqrt3\,r\), height \(= \frac{3r}{2}\); all sides equal \(\sqrt3\,r\), an equilateral triangle (area \(\frac{3\sqrt3}{4}r^2\)).
Answer: (1) An equilateral triangle having each of its side of length \(\sqrt{3}\,r\).