Questions on Geometry: Length, distance, coordinates, metric length answered by real tutors!

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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?
: 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?

Answer by MRperkins(77) 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/n 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 360/8 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/a The tangent of an angle is 'tangent_angle'='opposite_leg'/'adjacent_leg'.
*NOTE: you use 22.5 because you bisected the central angle
a=(5/(tan22.5)) 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"
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contact justin.sheppard.tech@hotmail.com with any questions
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?
: 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?

Answer by Edwin McCravy(2190) 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(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