Question 165933
<font size = 7 color = "red"><b>Edwin's solution:
Warning: MRperkins's solution is correct up to the last step.
He apparently mis-pressed something on his calculator and got
the wrong answer. Here is my complete solution with drawings:</b></font>

I need help with a question could someone please help??
 
Find the length, to the nearest tenth, of the apothem of a regular octogon whose sides are each 10 inches long?
<pre><font size = 4 color = "indigo"><b>  
Draw the octagon, all sides of which are 10 inches.
I'll just indicate that the bottom side is 10:
 
{{{drawing(400,400, -2,2,-2,2,
locate(0,-1.1,10),
line(.414,1,1,.414),
line(1,.414,1,-.414),
line(1,-.414,.414,-1),
line(.414,-1,-.414,-1),
line(-.414,-1,-1,-.414),
line(-1,-.414,-1,.414),
line(-.414,1,.414,1),
line(.414,1,1,.414),
line(-1,.414,-.414,1) )}}}
Now temporarily, connect the vertices
to the center:
{{{drawing(400,400, -2,2,-2,2,
locate(0,-1.1,10),
line(.414,1,1,.414),
line(1,.414,1,-.414),
line(1,-.414,.414,-1),
line(.414,-1,-.414,-1),
line(-.414,-1,-1,-.414),
line(-1,-.414,-1,.414),
line(-.414,1,.414,1),
line(.414,1,1,.414),
line(-1,.414,-.414,1),
line(0,0,         .414,1),
line(0,0,1,.414), 
line(0,0,        .414,-1),
line(0,0,1,-.414),
line(0,0,         -.414,1),
line(0,0,-1,.414),
line(0,0,       -.414,-1),
line(0,0,-1,-.414)
 )}}}
I did that just to show that each
of those 8 angles at the center are
{{{360/8)}}}° = {{{45}}}°, so that if we
erase all but the bottom two, like this:

{{{drawing(400,400, -2,2,-2,2,
locate(0,-1.1,10),
line(.414,1,1,.414),
line(1,.414,1,-.414),
line(1,-.414,.414,-1),
line(.414,-1,-.414,-1),
line(-.414,-1,-1,-.414),
line(-1,-.414,-1,.414),
line(-.414,1,.414,1),
line(.414,1,1,.414),
line(-1,.414,-.414,1),
line(0,0,        .414,-1),
line(0,0,       -.414,-1)
)}}}

Now we know that the angle in the
above is {{{45}}}°
{{{drawing(400,400, -2,2,-2,2,
locate(0,-1.1,10),
line(.414,1,1,.414),
line(1,.414,1,-.414),
line(1,-.414,.414,-1),
line(.414,-1,-.414,-1),
line(-.414,-1,-1,-.414),
line(-1,-.414,-1,.414),
line(-.414,1,.414,1),
line(.414,1,1,.414),
line(-1,.414,-.414,1),
locate(-.1,-.33,"45°"),
line(0,0,        .414,-1),
line(0,0,       -.414,-1)

 )}}}

Now draw in an apothem, the line from
the center to the midpoint of the bottom
 side, and label it {{{a}}}.
{{{drawing(400,400, -2,2,-2,2,
locate(0,-1.1,10),
line(.414,1,1,.414),
line(1,.414,1,-.414),
line(1,-.414,.414,-1),
line(.414,-1,-.414,-1),
line(-.414,-1,-1,-.414),
line(-1,-.414,-1,.414),
line(-.414,1,.414,1),
line(.414,1,1,.414),
line(-1,.414,-.414,1),
locate(-.1,-.33,"45°"),
line(0,0,        .414,-1),
line(0,0,       -.414,-1),
line(0,0,0,-1), locate(0.05,-.4,a)

 )}}}

Since the sides
of the octagon are 10 each, the two parts of
the bottom side are 5 each. Also the 45° angle
is bisected into two angles which are 22.5° each


{{{drawing(400,400, -2,2,-2,2,
locate(-.2,-1,5),locate(.2,-1,5),
line(.414,1,1,.414),
line(1,.414,1,-.414),
line(1,-.414,.414,-1),
line(.414,-1,-.414,-1),
line(-.414,-1,-1,-.414),
line(-1,-.414,-1,.414),
line(-.414,1,.414,1),
line(.414,1,1,.414),
line(-1,.414,-.414,1),
locate(-.4,-.2,"22.5°"),
line(0,0,        .414,-1),
line(0,0,       -.414,-1),
line(0,0,0,-1),
locate(0.05,-.4,a)

 )}}}

So lets take away everything but
just this little right triangle:

{{{drawing(400,400, -2,2,-2,2,
locate(-.2,-1,5),
line(-.414,-1,0,-1),

line(-.414,-1,0,0),


locate(-.4,-.2,"22.5°"),

line(0,0,       -.414,-1),
line(0,0,0,-1),
locate(0.05,-.4,a)

 )}}}

Then we just do a little trig on that triangle:

The side opposite the 22.5° angle is 5 and the
side adjacent to it is a, so

{{{tan(22.5)=5/a}}}

Multiply both sides by a:

{{{a*tan(22.5)=5}}}

Divide both sides by tan(22.5°):

{{{a=5/tan(22.5)}}}

{{{a=12.07106781}}}

or, to the nearest tenth,

{{{a=12.1}}} inches.

Edwin</pre>