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#