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)#
