How do I solve for the area of a regular hexagon? One side is equal to 5 feet.

1 Answer
Jun 24, 2018

#color(blue)("Area"=(75sqrt(3))/2=64.95" ft"^2)# 2 d.p.

Explanation:

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Using diagram.

Take a hexagon with side length #bba#. We can form 6 congruent triangles within the hexagon. The angle formed at the apex of each triangle is:

#(360^@)/n#

Where #bbn# is the number of sides, in this case #n=6#

#:.#

#(360^@)/6=60^@#

The interior angles of a regular polygon are given by:

Where #bbn# is the number of sides.

#180^@n-360^@#

#180^@(6)-360^@=120^@#

Dividing this by 2:

#(120^@)/2=60^@#

Looking at the diagram we can see that all the triangles in the hexagon have equal angles i.e. #60^@#. This means that they are equilateral and therefore have equal sides, in this case #bba#.

Drop a perpendicular bisector #bbh#. We now have 2 right angled triangles with sides #1/2a, a and h#

The length of #bbh# can be found using Pythagoras' theorem.

#h^2=a^2-(1/2a)^2#

#h^2=a^2-(a^2)/4=(4a^2-a^2)/4=(3a^2)/4#

#h=(asqrt(3))/2#

We can now find the area of one equilateral triangle:

#"Area"=1/2"base"xx"height"#

#"Area"=1/2(a)(h)#

#"Area"=1/2(a)((asqrt(3))/2)=(a^2sqrt(3))/4#

This is the area of one triangle. Since we have six of these triangles in a regular hexagon, area of hexagon is:

#6((a^2sqrt(3))/4)=bb((3a^2sqrt(3))/2)#

This is the formula for the area of a regular hexagon with side length #bba#

For this problem, we have a side length of 5.

#a=5#

#"Area"=(3(5^2)sqrt(3))/2=(75sqrt(3))/2" ft"^2#

#(75sqrt(3))/2=64.95" ft"^2color(white)(88)# 2 d.p.

#color(blue)("Area"=(75sqrt(3))/2=64.95" ft"^2)#