If a triangle has sides a, b, c, and fixed sum a+b+c=18, then the maximum possible area (in cm²) of the triangle is

Question

If a triangle has sides a, b, c, and fixed sum a+b+c=18, then the maximum possible area (in cm²) of the triangle is

1. A. 6√3
2. B. 3√3
3. C. 9√3 (Correct answer)
4. D. 12√3

Correct Answer

Option C — 9√3

Detailed Solution & Explanation

The correct answer is $9\sqrt{3}$.

Shortcut Trick

• Among all triangles with a fixed perimeter, the one with the largest area is equilateral. This is a standard result and turns the question into a single formula.
• Perimeter 18 gives a side of $18 \div 3 = 6$ cm.
• Area of an equilateral triangle $= \dfrac{\sqrt{3}}{4}a^2 = \dfrac{\sqrt{3}}{4}(36) = \textbf{9}\sqrt{\textbf{3}}$ cm².

Alternate Method

• Using Heron's formula with $s = \dfrac{18}{2} = 9$:
  • $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$
  • The product $(s-a)(s-b)(s-c)$ is largest when the three factors are equal, i.e. $a = b = c = 6$.
  • $\text{Area} = \sqrt{9 \times 3 \times 3 \times 3} = \sqrt{243} = 9\sqrt{3}$

Additional Information

• The general principle is isoperimetric: for a fixed perimeter the most symmetric figure encloses the greatest area — equilateral among triangles, square among rectangles, and the circle among all plane figures.
• $\sqrt{3} \approx 1.732$, so $9\sqrt{3} \approx 15.59$ cm².
• Useful companions for an equilateral triangle of side $a$: height $= \dfrac{\sqrt{3}}{2}a$, inradius $= \dfrac{a}{2\sqrt{3}}$, circumradius $= \dfrac{a}{\sqrt{3}}$.