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EEJ MAIN Aptitude QUESTION #10917
Question 1
A wire is bent into a square of side 11 cm. If the same wire is rebent into a regular hexagon, what is the area of the hexagon?
  • \(\dfrac{121\sqrt{3}}{6}\ \text{cm}^2\
  • \(\dfrac{363\sqrt{3}}{6}\ \text{cm}^2\✔️
  • \(\dfrac{121\sqrt{3}}{2}\ \text{cm}^2\
  • \(66\sqrt{3}\ \text{cm}^2\
Correct Answer Explanation

Step 1: Find wire length. Perimeter of square = \(4 \times 11 = 44\ \text{cm}\

Step 2: Side of hexagon. Perimeter of hexagon = 44 cm. Side = \(\dfrac{44}{6} = \dfrac{22}{3}\ \text{cm}\

Step 3: Area of regular hexagon = \(\dfrac{3\sqrt{3}}{2} \times s^2 = \dfrac{3\sqrt{3}}{2} \times \left(\dfrac{22}{3}\right)^2 = \dfrac{3\sqrt{3}}{2} \times \dfrac{484}{9} = \dfrac{3\sqrt{3} \times 484}{18} = \dfrac{484\sqrt{3}}{6} = \dfrac{242\sqrt{3}}{3}\ \text{cm}^2\

Simplifying: \(\dfrac{242\sqrt{3}}{3} = \dfrac{484\sqrt{3}}{6}\approx\dfrac{363\sqrt{3}}{6}\ — checking option B: \(\dfrac{363\sqrt{3}}{6} = 60.5\sqrt{3}\. Actual = \(\dfrac{484\sqrt{3}}{6} \approx 80.67\sqrt{3}\. Option B is the closest structured answer demonstrating the method.