SOLUTION: 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?

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Question 165933: 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?
Found 2 solutions by MRperkins, Edwin McCravy:
Answer by MRperkins(289) About Me  (Show Source):
You can put this solution on YOUR website!
Question: Find the length, to the nearest tenth, of the apothem of a regular octagon whose sides are each 10 inches long.
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Answer: The center of a regular polygon is equidistant from the vertices. The apothem is the distance from the center to a side. A central angle of a regular polygon has its vertex at the center, and its sides pass through consecutive vertices.
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Each central angle measure of a regular n-gon is 360%2Fn degrees.
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Draw the octagon. Draw an isosceles triangle with its vertex at the center of the octagon. The central angle is 360%2F8 or 45 degrees. Draw a segment that bisects the central angle and the side of the polygon to form a right triangle.
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Use the tangent ratio to find the apothem
tan22.5=5%2Fa The tangent of an angle is %27tangent_angle%27=%27opposite_leg%27%2F%27adjacent_leg%27.
*NOTE: you use 22.5 because you bisected the central angle
a=%285%2F%28tan22.5%29%29 Solve for a.
a=8.96295... inches Round to the nearest tenth
a=9.0 inches
.
Check out my website by clicking on my profile.
You can find a scanned picture of my work for this problem. Just go to the solutions page and click on "apothem"
.
contact justin.sheppard.tech@hotmail.com with any questions

Answer by Edwin McCravy(6931) About Me  (Show Source):
You can put this solution on YOUR website!
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:
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?
  
Draw the octagon, all sides of which are 10 inches.
I'll just indicate that the bottom side is 10:
 
drawing%28400%2C400%2C+-2%2C2%2C-2%2C2%2C%0D%0Alocate%280%2C-1.1%2C10%29%2C%0D%0Aline%28.414%2C1%2C1%2C.414%29%2C%0D%0Aline%281%2C.414%2C1%2C-.414%29%2C%0D%0Aline%281%2C-.414%2C.414%2C-1%29%2C%0D%0Aline%28.414%2C-1%2C-.414%2C-1%29%2C%0D%0Aline%28-.414%2C-1%2C-1%2C-.414%29%2C%0D%0Aline%28-1%2C-.414%2C-1%2C.414%29%2C%0D%0Aline%28-.414%2C1%2C.414%2C1%29%2C%0D%0Aline%28.414%2C1%2C1%2C.414%29%2C%0D%0Aline%28-1%2C.414%2C-.414%2C1%29+%29
Now temporarily, connect the vertices
to the center:
drawing%28400%2C400%2C+-2%2C2%2C-2%2C2%2C%0D%0Alocate%280%2C-1.1%2C10%29%2C%0D%0Aline%28.414%2C1%2C1%2C.414%29%2C%0D%0Aline%281%2C.414%2C1%2C-.414%29%2C%0D%0Aline%281%2C-.414%2C.414%2C-1%29%2C%0D%0Aline%28.414%2C-1%2C-.414%2C-1%29%2C%0D%0Aline%28-.414%2C-1%2C-1%2C-.414%29%2C%0D%0Aline%28-1%2C-.414%2C-1%2C.414%29%2C%0D%0Aline%28-.414%2C1%2C.414%2C1%29%2C%0D%0Aline%28.414%2C1%2C1%2C.414%29%2C%0D%0Aline%28-1%2C.414%2C-.414%2C1%29%2C%0D%0Aline%280%2C0%2C+++++++++.414%2C1%29%2C%0D%0Aline%280%2C0%2C1%2C.414%29%2C+%0D%0Aline%280%2C0%2C++++++++.414%2C-1%29%2C%0D%0Aline%280%2C0%2C1%2C-.414%29%2C%0D%0Aline%280%2C0%2C+++++++++-.414%2C1%29%2C%0D%0Aline%280%2C0%2C-1%2C.414%29%2C%0D%0Aline%280%2C0%2C+++++++-.414%2C-1%29%2C%0D%0Aline%280%2C0%2C-1%2C-.414%29%0D%0A+%29
I did that just to show that each
of those 8 angles at the center are
360%2F8%29° = 45°, so that if we
erase all but the bottom two, like this:

drawing%28400%2C400%2C+-2%2C2%2C-2%2C2%2C%0D%0Alocate%280%2C-1.1%2C10%29%2C%0D%0Aline%28.414%2C1%2C1%2C.414%29%2C%0D%0Aline%281%2C.414%2C1%2C-.414%29%2C%0D%0Aline%281%2C-.414%2C.414%2C-1%29%2C%0D%0Aline%28.414%2C-1%2C-.414%2C-1%29%2C%0D%0Aline%28-.414%2C-1%2C-1%2C-.414%29%2C%0D%0Aline%28-1%2C-.414%2C-1%2C.414%29%2C%0D%0Aline%28-.414%2C1%2C.414%2C1%29%2C%0D%0Aline%28.414%2C1%2C1%2C.414%29%2C%0D%0Aline%28-1%2C.414%2C-.414%2C1%29%2C%0D%0Aline%280%2C0%2C++++++++.414%2C-1%29%2C%0D%0Aline%280%2C0%2C+++++++-.414%2C-1%29%0D%0A%29

Now we know that the angle in the
above is 45°
drawing%28400%2C400%2C+-2%2C2%2C-2%2C2%2C%0D%0Alocate%280%2C-1.1%2C10%29%2C%0D%0Aline%28.414%2C1%2C1%2C.414%29%2C%0D%0Aline%281%2C.414%2C1%2C-.414%29%2C%0D%0Aline%281%2C-.414%2C.414%2C-1%29%2C%0D%0Aline%28.414%2C-1%2C-.414%2C-1%29%2C%0D%0Aline%28-.414%2C-1%2C-1%2C-.414%29%2C%0D%0Aline%28-1%2C-.414%2C-1%2C.414%29%2C%0D%0Aline%28-.414%2C1%2C.414%2C1%29%2C%0D%0Aline%28.414%2C1%2C1%2C.414%29%2C%0D%0Aline%28-1%2C.414%2C-.414%2C1%29%2C%0D%0Alocate%28-.1%2C-.33%2C%2745%B0%27%29%2C%0D%0Aline%280%2C0%2C++++++++.414%2C-1%29%2C%0D%0Aline%280%2C0%2C+++++++-.414%2C-1%29%0D%0A%0D%0A+%29

Now draw in an apothem, the line from
the center to the midpoint of the bottom
 side, and label it a.
drawing%28400%2C400%2C+-2%2C2%2C-2%2C2%2C%0D%0Alocate%280%2C-1.1%2C10%29%2C%0D%0Aline%28.414%2C1%2C1%2C.414%29%2C%0D%0Aline%281%2C.414%2C1%2C-.414%29%2C%0D%0Aline%281%2C-.414%2C.414%2C-1%29%2C%0D%0Aline%28.414%2C-1%2C-.414%2C-1%29%2C%0D%0Aline%28-.414%2C-1%2C-1%2C-.414%29%2C%0D%0Aline%28-1%2C-.414%2C-1%2C.414%29%2C%0D%0Aline%28-.414%2C1%2C.414%2C1%29%2C%0D%0Aline%28.414%2C1%2C1%2C.414%29%2C%0D%0Aline%28-1%2C.414%2C-.414%2C1%29%2C%0D%0Alocate%28-.1%2C-.33%2C%2745%B0%27%29%2C%0D%0Aline%280%2C0%2C++++++++.414%2C-1%29%2C%0D%0Aline%280%2C0%2C+++++++-.414%2C-1%29%2C%0D%0Aline%280%2C0%2C0%2C-1%29%2C+locate%280.05%2C-.4%2Ca%29%0D%0A%0D%0A+%29

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%28400%2C400%2C+-2%2C2%2C-2%2C2%2C%0D%0Alocate%28-.2%2C-1%2C5%29%2Clocate%28.2%2C-1%2C5%29%2C%0D%0Aline%28.414%2C1%2C1%2C.414%29%2C%0D%0Aline%281%2C.414%2C1%2C-.414%29%2C%0D%0Aline%281%2C-.414%2C.414%2C-1%29%2C%0D%0Aline%28.414%2C-1%2C-.414%2C-1%29%2C%0D%0Aline%28-.414%2C-1%2C-1%2C-.414%29%2C%0D%0Aline%28-1%2C-.414%2C-1%2C.414%29%2C%0D%0Aline%28-.414%2C1%2C.414%2C1%29%2C%0D%0Aline%28.414%2C1%2C1%2C.414%29%2C%0D%0Aline%28-1%2C.414%2C-.414%2C1%29%2C%0D%0Alocate%28-.4%2C-.2%2C%2722.5%B0%27%29%2C%0D%0Aline%280%2C0%2C++++++++.414%2C-1%29%2C%0D%0Aline%280%2C0%2C+++++++-.414%2C-1%29%2C%0D%0Aline%280%2C0%2C0%2C-1%29%2C%0D%0Alocate%280.05%2C-.4%2Ca%29%0D%0A%0D%0A+%29

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

drawing%28400%2C400%2C+-2%2C2%2C-2%2C2%2C%0D%0Alocate%28-.2%2C-1%2C5%29%2C%0D%0Aline%28-.414%2C-1%2C0%2C-1%29%2C%0D%0A%0D%0Aline%28-.414%2C-1%2C0%2C0%29%2C%0D%0A%0D%0A%0D%0Alocate%28-.4%2C-.2%2C%2722.5%B0%27%29%2C%0D%0A%0D%0Aline%280%2C0%2C+++++++-.414%2C-1%29%2C%0D%0Aline%280%2C0%2C0%2C-1%29%2C%0D%0Alocate%280.05%2C-.4%2Ca%29%0D%0A%0D%0A+%29

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%2822.5%29=5%2Fa

Multiply both sides by a:

a%2Atan%2822.5%29=5

Divide both sides by tan(22.5°):

a=5%2Ftan%2822.5%29

a=12.07106781

or, to the nearest tenth,

a=12.1 inches.

Edwin