Find the area of a regular hexagon with a 48-inch perimeter.

2 answers:

7 0

$A= \cfrac{3 \sqrt{3} }{2}*a^2 \ \ \ \ [a=sidelength] \\ \\ a=48/6=8 \ in \\ \\ \\ A= \cfrac{3 \sqrt{3} }{2}*(8)^2=A= \cfrac{3 \sqrt{3} }{2}*64=3 \sqrt{3}*32= 96 \sqrt{3} \ \ in^2$

Oduvanchick [21]2 years ago

7 0

The answer is: 96√3 in2

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