How do you find the area of a regular hexagon with a radius of 5? Please show working.

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Answer 1 · 2018-05-09T05:04:16

$A = (75 sqrt(3))/2 ~~65 " units"^2$

Given: a regular hexagon with radius = 5

$A = 1/2 a P$ , where $a$ = apothem , $P$ = perimeter

The apothem is the perpendicular distance from the center to a side.

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$n =$ number of sides $= 6$

$s =$ side length $= 2x$

$P = ns = 6s = 6(2x) = 12x$

$A = 1/2a(12x) = 6ax$

$2 theta = (360^@)/n = 360/6 = 60^@$

$theta = (60^@)/2 = 30^@$

Use trigonometry to find $a$ and $x$ given $r$ and $theta$ :

$(sin 30^@)/1 = x/r = x/5; " " x = 5 sin 30^@ = 5 (1/2) = 5/2$

$(cos 30^@)/1 = a/r = a/5; " " a = 5 cos 30^@ = (5 sqrt(3))/2$

$A = 6ax = 6 ((5 sqrt(3))/2) (5/2) = (75 sqrt(3))/2 ~~64.95 " units"^2$