Question #5c832

3 Answers
YoshiMan007
Oct 22, 2017

#s=36sqrt(2)-36#

Explanation:

(PS, it's best to draw this octagon. The explanation will make much more sense).

Imagine a regular octagon #ABCDEFGH#, where #A# is the point at the top left and you move clockwise to letter the points. We can use the following formula to determine each angle in an octagon:

#180*("num sides"-2)/("num sides")#

#=180*(8-2)/8#

#=180*6/8#

#=135^@#

Now, on your octagon, draw the line #AF#. Based on the problem, we know that #AF# is #36# inches. Let's also assume that every side length of the octagon has length #s#. Then, draw the line from point #G# to the line #AF#. Call this point of intersection #I#. Notice that #GI# and #AF# are perpendicular and #GI# and #GH# are perpendicular. This means that angle #HGI# is #90^@#, so angle #FGI# is #135-90=45^@#.

Since angle #GIF# is #90^@#, triangle #GIF# is a #45-45-90# triangle. Since #GF# has length #s#, we can use trigonometry to determine that #FI# has length #s/sqrt(2)#.

Note that if you draw the line from #H# to #AF# which intersect at point #J#, then you get rectangle #GHJI#. Since #GH# has side length #s#, so does #IJ#.

Finally, we can see that triangle #HJA# is congruent to #GIF#, so #JA# also has side length #s/sqrt(2)#.

We know that #AF# has length #36#. But #AF# can be broken down to three parts: #AJ#, #JI#, and #IF#. Thus:

#s/sqrt(2)+s+s/sqrt(2)=36#

#s+ssqrt(2)+s=36sqrt(2)#

#s(2+sqrt(2))=36sqrt(2)#

#s=(36sqrt(2))/(2+sqrt(2))#

#s=(36sqrt(2))/(2+sqrt(2))*(2-sqrt(2))/(2-sqrt(2))#

#s=(72sqrt(2)-72)/(2)#

#s=36sqrt(2)-36#

EZ as pi
Oct 22, 2017

The sides are #14.9# inches long.

Explanation:

The distance between two parallel sides of the octagon is given as #36# inches.

A regular octagon can be divided up into two isosceles trapeziums and a rectangle. Therefore the distance of #36# inches is made up of:

Height + height + side. (height being the distance between the parallel sides of the isosceles trapeziums.

As each angle of the octagon is #135°# , the base angles of the trapezium are #135°-90° = 45°#

We therefore need to find the heights of two right-angled isosceles triangles.

With #s# as the length of the sides of the octagon and #h# as the equal sides of the triangles, by Pythagoras' we have:

#h^2+h^2 = s^2#

#2h^2 = s^2#

#h^2 = s^2/2#

#h = sqrt(s^2/2)#

Now make an equation from: height + height + side = #36#

#sqrt(s^2/2)+sqrt(s^2/2)+s =36#

#(2sqrts^2)/sqrt2 = 36-s#

#(2sqrts^2)/sqrt2 xxsqrt2/sqrt2= 36-s#

#(cancel2sqrt(2s^2))/cancel2=36-s" "larr# square both sides

#2s^2 = 1296-72s+s^2#

#s^2 +72s -1296=0" "larr# solve by completing the square

#s^2 +72s +36^2 = 1296+36^2#

#(s+36)^2 = 2592#

#s+36 = sqrt2592" "larr# only positive root is valid

#s = 50.911688 -36#

#s=14.9# inches

Somebody N.
Oct 22, 2017

#14.91# (2.d.p.)

Explanation:

There is a really quick way of solving this.

#180/8=45# This is the angle at the apex of the triangles that are form when dividing the octagon into sectors. #36/2=18# is the length of the perpendicular bisector of the angle at the apex and meets the base line you want to find. So using this:

#tan(22.5)= 1/2#base line/18:

So:

18tan(22.5)= half the length of octagon side:

#36tan(22.5)= 14.91# (2.d.p.)

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