A regular hexagonal nut has outer side 6 cm and inner hollow side 3 cm. What is the material approximate area in the cross-section?

Correct — A. The material is the outer hexagon minus the hollow one. A regular hexagon of side s has area (3√3⁄2)s².

Outer: (3√3⁄2) × 6² = 54√3
Inner: (3√3⁄2) × 3² = 13.5√3
Material: 54√3 − 13.5√3 = 40.5√3
With √3 ≈ 1.732: 40.5 × 1.732 ≈ 70.1 cm² → option (a)

Why the others are wrong

• (b)
  60 cm² — 60 cm² would be about 64% of the outer face (93.5 cm²). A hollow of half the side removes one quarter of that area, so the material is 75% of it, about 70 cm².
• (c)
  56 cm² — 56 cm² is about 60% of the outer 93.5 cm², well under the 75% that remains when a hexagon of half the side is cut out.
• (d)
  66 cm² — 66 cm² is still about 4 cm² below 40.5√3. Even the rough √3 ≈ 1.7 gives 68.9 cm², which is closer to 70 than to 66.

A regular hexagon splits into 6 equilateral triangles of side s, each of area (√3⁄4)s². So its area is 6 × (√3⁄4)s² = (3√3⁄2)s² ≈ 2.598s².

For a hollow shape, the material is outer area − inner area. Area scales with the square of the side, so a hollow of half the side takes away exactly 1⁄4, leaving 3⁄4 × 54√3 = 40.5√3.

Key facts

• Area of a regular hexagon of side s = (3√3⁄2)s² ≈ 2.598s².
• Side 6 cm gives 54√3 ≈ 93.53 cm², and side 3 cm gives 13.5√3 ≈ 23.38 cm².
• Halving the side of any shape divides its area by 4.

Common traps

• Subtracting the sides first. A hexagon of side 6 − 3 = 3 cm has area 13.5√3 ≈ 23.4 cm², which is the hollow, not the material.
• Using (√3⁄4)s², the area of one equilateral triangle, and forgetting to multiply by 6.

No directly related past PYQ was found.