What is the area of a regular hexagon with a 48-inch perimeter?

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What is the area of a regular hexagon with a 48-inch perimeter?

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Answer 1 · 2015-12-01T00:16:21

$16 \sqrt{3} \approx 27.71$ $16 sqrt(3) approx 27.71$ square inches.

Explanation:

First of all, if the perimeter of a regular hexagon measures $48$ $48$ inches, then each of the $6$ $6$ sides has to be $\frac{48}{6} = 8$ $48/6=8$ inches long.

To compute the area, you can divide the figure in equilateral triangles as follows.

Given the side $s$ $s$ , the area of an equilateral triangle is given by $A = \frac{\sqrt{3}}{4} s^{2}$ $A=sqrt(3)/4 s^2$ (you can prove this using the Pythagorean Theorem or trigonometry).

In our case $s = 8$ $s=8$ inches, so the area is $A = \frac{\sqrt{3}}{4} 8^{2} = 16 \sqrt{3} \approx 27.71$ $A=sqrt(3)/4 8^2=16 sqrt(3) approx 27.71$ square inches.

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What is the area of a regular hexagon with a 48-inch perimeter?

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