What is the perimeter of a regular hexagon that has an area of #54sqrt3# units squared?

3 Answers
Dec 21, 2017

The perimeter of the regular hexagon is #36# unit.

Explanation:

The formula for the area of a regular hexagon is

#A = (3sqrt3 s^2)/2# where #s# is the length of a side of the

regular hexagon. #:. (3cancel(sqrt3) s^2)/2= 54 cancel(sqrt3)# or

#3 s^2=108 or s^2=108/3 or s^2=36 or s=6#

The perimeter of the regular hexagon is #P=6*s=6*6=36#

unit. [Ans]

Dec 21, 2017

Perimeter: #6# units

Explanation:

A hexagon can be decomposed into 6 equilateral triangles:
enter image source here
If we let #x# represent the length of each side of such equilateral triangle.

The area of a triangle with sides of length #x# is
#color(white)("XXX")A_triangle=sqrt(3)/4x^2#
#color(white)("XXXXXXXXXXXXXXXXXXXX")#(See below for derivation)

The area of the hexagon is #6A_triangle# which we are told is #54sqrt(3)# square units.

#6 * sqrt(3)/4x^2=54sqrt(3)#

#rarr sqrt(3)/4x^2=9sqrt(3)#

#rarr 1/4x^2=9#

#rarr x^2=4 * 9=2^2 * 3^2=6^2#

#rarr x =6color(white)("XXX")#Note since #x# is a geometric length #x>=0#

The perimeter of the hexagon is #6x#
#rarr# Perimeter of hexagon #= 36#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Finding the perimeter of an equilateral triangle with sides of length #x#:

Heron'[s formula for the area of a triangle tells us that if the semi-perimeter of a triangle is #s# and the triangle has sides of lengths, #x#, #x#, and #x#, then
#"Area"_triangle =sqrt(s(s-x)(s-x)(s-x))#

The semi-perimeter is #s=(x+x+x)/2=(3x)/2#
So #(x-s)=x/2#
and
#"Area"_triangle=sqrt((3x)/2 * (x/2) * (x/2) * (x/2))=sqrt(3)/4x^2#
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Dec 21, 2017

#36#

Explanation:

Let's start from an equilateral triangle with side #2#...

enter image source here

Bisecting the triangle results in two right angled triangles, with sides #1#, #sqrt(3)# and #2# as we can deduce from Pythagoras:

#1^2 + (sqrt(3))^2 = 2^2#

The area of the equilateral triangle is the same as a rectangle with sides #1# and #sqrt(3)# (just rearrange the two right angled triangles for one way to see that), so #1 * sqrt(3) = sqrt(3)#.

Six such triangles can be assembled to form a regular hexagon with side #2# and area #6 sqrt(3)#.

In our example, the hexagon has area:

#54 sqrt(3) = color(blue)(3)^2 * (6 sqrt(3))#

So the length of each side is:

#color(blue)(3) * 2 = 6#

and the perimeter is:

#6 * 6 = 36#