Question

AB is a diameter of a circle and C is any point on the circumference of the circle. Then

• A. Area of ∆ABC is maximum when it is isosceles.
• B. Area of ∆ABC is minimum when it is isosceles.
• C. Perimeter of ∆ABC is minimum when it is isosceles.
• D. None

Answer 1

Answer: (A)

Area of ∆ABC is maximum when it is isosceles.

Answer 2

Let AB be the diameter of the circle with radius R. Thus, AB = 2R. Since C is any point on the circumference, the triangle ABC is inscribed in a semicircle. The angle in a semicircle is a right angle, so ∠ACB = 90°. Therefore, triangle ABC is a right-angled triangle.

The area of ∆ABC is given by $\frac{1}{2} \times \text{base} \times \text{height}$. Taking AB as the base, the height is the perpendicular distance from C to AB, let's call it h. Area(∆ABC) = $\frac{1}{2} \times AB \times h = \frac{1}{2} \times (2R) \times h = Rh$. The maximum value for the height h occurs when C is furthest from AB, which is the radius R. This happens when C is at the midpoint of the arc AB, directly above the center O. When h = R, the area is $R \times R = R^2$. In this case, the triangle ABC is isosceles, with AC = BC. Thus, the area of ∆ABC is maximum when it is isosceles.

The perimeter of ∆ABC is $P = AB + AC + BC = 2R + AC + BC$. Let ∠CAB = $\alpha$. Then AC = $2R \cos \alpha$ and BC = $2R \sin \alpha$. $P = 2R (1 + \cos \alpha + \sin \alpha)$. To find the extrema of the perimeter, we analyze $f(\alpha) = \cos \alpha + \sin \alpha = \sqrt{2} \sin(\alpha + \frac{\pi}{4})$. For $0 < \alpha < \frac{\pi}{2}$, the maximum value of $f(\alpha)$ is $\sqrt{2}$ when $\alpha = \frac{\pi}{4}$, which corresponds to an isosceles triangle. The maximum perimeter is $2R(1 + \sqrt{2})$. The minimum value of $f(\alpha)$ approaches 1 as $\alpha \to 0$ or $\alpha \to \frac{\pi}{2}$, corresponding to degenerate triangles with perimeter approaching $4R$. Therefore, the perimeter is maximum when the triangle is isosceles, not minimum.