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

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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

$A = \frac{75 \sqrt{3}}{2} \approx 65 {units}^{2}$ $A = (75 sqrt(3))/2 ~~65 " units"^2$

Explanation:

Given: a regular hexagon with radius = 5

$A = \frac{1}{2} a P$ $A = 1/2 a P$ , where $a$ $a$ = apothem , $P$ $P$ = perimeter

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

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$n =$ $n =$ number of sides $= 6$ $= 6$

$s =$ $s =$ side length $= 2 x$ $= 2x$

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

$A = \frac{1}{2} a (12 x) = 6 a x$ $A = 1/2a(12x) = 6ax$

$2 θ = \frac{360^{\circ}}{n} = \frac{360}{6} = 60^{\circ}$ $2 theta = (360^@)/n = 360/6 = 60^@$

$θ = \frac{60^{\circ}}{2} = 30^{\circ}$ $theta = (60^@)/2 = 30^@$

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

$\frac{{sin 30}^{\circ}}{1} = \frac{x}{r} = \frac{x}{5}; x = 5 {sin 30}^{\circ} = 5 (\frac{1}{2}) = \frac{5}{2}$ $(sin 30^@)/1 = x/r = x/5; " " x = 5 sin 30^@ = 5 (1/2) = 5/2$

$\frac{{cos 30}^{\circ}}{1} = \frac{a}{r} = \frac{a}{5}; a = 5 {cos 30}^{\circ} = \frac{5 \sqrt{3}}{2}$ $(cos 30^@)/1 = a/r = a/5; " " a = 5 cos 30^@ = (5 sqrt(3))/2$

$A = 6 a x = 6 (\frac{5 \sqrt{3}}{2}) (\frac{5}{2}) = \frac{75 \sqrt{3}}{2} \approx 64.95 {units}^{2}$ $A = 6ax = 6 ((5 sqrt(3))/2) (5/2) = (75 sqrt(3))/2 ~~64.95 " units"^2$

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

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