Question

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

Answer 1

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

Explanation:

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)`