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Difficulty: Very hardPerimeter and Area of Plane Shapes

A regular hexagon has a side length of 6 cm6\text{ cm}. At each vertex of the hexagon, a circular sector of radius 3 cm3\text{ cm} is formed inside the figure. Taking π=227\pi = \frac{22}{7} and 3=1.732\sqrt{3} = 1.732, what is the area of the remaining region inside the hexagon not covered by the sectors, in cm2\text{cm}^2, correct to two decimal places?

Answer: 36.96 cm^2

Answer

The area of the remaining region inside the hexagon is 36.96 cm236.96\text{ cm}^2.
The total area of the regular hexagon is computed by multiplying the area of one equilateral triangle of side 6 cm6\text{ cm} by 6, yielding 543=54×1.732=93.528 cm254\sqrt{3} = 54 \times 1.732 = 93.528\text{ cm}^2. Each interior angle of a regular hexagon is 120∘120^\circ, so each vertex sector has a central angle of 120∘120^\circ and radius 3 cm3\text{ cm}. The area of one sector is 120∘360∘×227×32=667 cm2\frac{120^\circ}{360^\circ} \times \frac{22}{7} \times 3^2 = \frac{66}{7}\text{ cm}^2. The total area for all six sectors is 6×667=3967≈56.5714 cm26 \times \frac{66}{7} = \frac{396}{7} \approx 56.5714\text{ cm}^2. Subtracting this from the total area gives 93.528−56.5714=36.9566 cm293.528 - 56.5714 = 36.9566\text{ cm}^2, which rounds to 36.96 cm236.96\text{ cm}^2.

Step-by-Step Solution

1
Determine the interior angle of the regular hexagon.
Each interior angle is 120∘120^\circ.
The formula for the interior angle of a regular polygon with nn sides is (n−2)×180∘n\frac{(n-2) \times 180^\circ}{n}.
2
Calculate the total area of the 6 circular sectors at the vertices.
Total sector area is 3967≈56.5714 cm2\frac{396}{7} \approx 56.5714\text{ cm}^2.
Each sector has a central angle of 120∘120^\circ and radius 3 cm3\text{ cm}. With 6 sectors, the total area is 6×120∘360∘×227×32=18×227=3967 cm26 \times \frac{120^\circ}{360^\circ} \times \frac{22}{7} \times 3^2 = 18 \times \frac{22}{7} = \frac{396}{7}\text{ cm}^2.
3
Calculate the total area of the regular hexagon.
Hexagon area is 93.528 cm293.528\text{ cm}^2.
A regular hexagon consists of 6 equilateral triangles of side length 6 cm6\text{ cm}. Area = 6×(34×62)=543=54×1.732=93.528 cm26 \times \left(\frac{\sqrt{3}}{4} \times 6^2\right) = 54\sqrt{3} = 54 \times 1.732 = 93.528\text{ cm}^2.
4
Subtract the sector area from the total hexagon area.
93.528−56.5714=36.9566 cm2≈36.96 cm293.528 - 56.5714 = 36.9566\text{ cm}^2 \approx 36.96\text{ cm}^2.
The remaining area is the total area minus the area occupied by the six corner sectors.

Key Concept

Area of regular polygons and circular sectors
Estimated Time:2m 30s
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Topics

Perimeter and Area of Plane Shapes

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