The perimeters of a circle, a square and an equilateral triangle are equal. Which one of the following statements is true?
Options:
(A) The circle has the largest area.
(B) The square has the largest area.
(C) The equilateral triangle has the largest area.
(D) All the three shapes have the same area.
Official Answer Key:
A
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2018 Q.4)**
**1. Intuitive Physical Setup & Given Parameters**:
- Let the common perimeter of all three geometric figures be $P$.
- Shapes: Circle ($P = 2\pi r$), Square ($P = 4a$), Equilateral Triangle ($P = 3s$).
**2. Step-by-Step Mathematical Derivation**:
1. **Area of the Circle ($A_c$)**:
$r = \frac{P}{2\pi} \implies A_c = \pi r^2 = \pi \left(\frac{P}{2\pi}\right)^2 = \frac{P^2}{4\pi} \approx \frac{P^2}{12.566} \approx 0.0796 P^2$
2. **Area of the Square ($A_s$)**:
$a = \frac{P}{4} \implies A_s = a^2 = \left(\frac{P}{4}\right)^2 = \frac{P^2}{16} = 0.0625 P^2$
3. **Area of the Equilateral Triangle ($A_t$)**:
$s = \frac{P}{3} \implies A_t = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \left(\frac{P}{3}\right)^2 = \frac{\sqrt{3} P^2}{36} \approx \frac{1.732}{36} P^2 \approx 0.0481 P^2$
4. **Comparison**:
$A_c (0.0796 P^2) > A_s (0.0625 P^2) > A_t (0.0481 P^2)$
By the classical Isoperimetric Theorem, for any given perimeter, the circle encloses the maximum possible area.
**3. Comprehensive Options & Distractor Analysis**:
- **(A)** *The circle has the largest area*: ✅ **CORRECT**. Directly validated by the isoperimetric inequality.
- **(B)** *The square has the largest area*: ❌ Incorrect. Square area is smaller than the circle.
- **(C)** *The equilateral triangle has the largest area*: ❌ Incorrect. The triangle encloses the smallest area among regular polygons.
- **(D)** *All have the same area*: ❌ Incorrect. Area depends strongly on shape geometry.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **A**
- **Isoperimetric Principle**: Among all plane closed figures with the same perimeter, the circle has the maximum area ($A_{circle} > A_{regular-n-gon} > \dots > A_{triangle}$).
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