Question:
If $\Delta$ denotes the area of triangle $ABC$ and $s$ is its semi-perimeter, then
If $\Delta$ denotes the area of triangle $ABC$ and $s$ is its semi-perimeter, then
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For a fixed perimeter, the equilateral triangle gives the maximum possible area.
Updated On: Jun 3, 2026
- \[ \Delta=\frac{s^2}{2} \]
- \[ \Delta\le\frac{s^2}{3\sqrt3} \]
- \[ \Delta>\frac{s^2}{\sqrt3} \]
- \[ \Delta=\frac{s^2}{2\sqrt3} \]
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The Correct Option is B
Solution and Explanation
Step 1: Concept
The maximum area of a triangle for a fixed perimeter occurs when the triangle is equilateral.
Step 2: Meaning
Let the perimeter be \[ P=2s. \] For an equilateral triangle, \[ a=\frac{2s}{3}. \]
Step 3: Analysis
Area of an equilateral triangle is \[ \Delta = \frac{\sqrt3}{4}a^2. \] Substituting \[ a=\frac{2s}{3}, \] we obtain \[ \Delta_{\max} = \frac{\sqrt3}{4} \left(\frac{2s}{3}\right)^2 = \frac{s^2}{3\sqrt3}. \] Therefore every triangle satisfies \[ \Delta\le\frac{s^2}{3\sqrt3}. \]
Step 4: Conclusion
Hence the correct relation is \[ \Delta\le\frac{s^2}{3\sqrt3}. \]
Final Answer: (B)
The maximum area of a triangle for a fixed perimeter occurs when the triangle is equilateral.
Step 2: Meaning
Let the perimeter be \[ P=2s. \] For an equilateral triangle, \[ a=\frac{2s}{3}. \]
Step 3: Analysis
Area of an equilateral triangle is \[ \Delta = \frac{\sqrt3}{4}a^2. \] Substituting \[ a=\frac{2s}{3}, \] we obtain \[ \Delta_{\max} = \frac{\sqrt3}{4} \left(\frac{2s}{3}\right)^2 = \frac{s^2}{3\sqrt3}. \] Therefore every triangle satisfies \[ \Delta\le\frac{s^2}{3\sqrt3}. \]
Step 4: Conclusion
Hence the correct relation is \[ \Delta\le\frac{s^2}{3\sqrt3}. \]
Final Answer: (B)
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