SOLUTION: Find the area of a regular nonagon of side 6cm

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Question 1142081: Find the area of a regular nonagon of side 6cm
Found 4 solutions by josgarithmetic, ikleyn, MathTherapy, Alan3354:
Answer by josgarithmetic(39625) About Me  (Show Source):
You can put this solution on YOUR website!
The nonagon has nine of these isosceles triangles:
Peak angle 40 degrees
Each base angle 70 degrees
Let r be distance peak to base angle
Let a be altitude of one isosceles triangle

r%2Fsin%2870%29=6%2Fsin%2840%29
r=6%28sin%2870%29%2Fsin%2840%29%29
-
a%2Fsin%2870%29=r%2Fsin%2890%29
a=r%2Asin%2870%29%2F1
a=6%28sin%2870%29%2Fsin%2840%29%29sin%2870%29
highlight_green%28a=6%2Asin%5E2%2870%29%2Fsin%2840%29%29
-
Total area of nonagon
highlight_green%289%281%2F2%29%2A6%2Aa%29
9%2A18%2Asin%5E2%2870%29%2Fsin%2840%29
%280.88302%2F0.64278%29%2A9%2A8
highlight%28highlight_green%28222.55%29%29



----------------------------BELOW STILL WRONG-------------------------------
It has nine of these isosceles triangles:
Peak angle, 40 degrees
Each base angle, 25 degrees--------MISTAKE(Really should be %28180-40%29%2F2=70)
Base length, 6 cm
Each congruent side, r, distance from a peak angle to base angle
-
sin%2840%29%2F6=sin%2870%29%2Fr
r%2Fsin%2870%29=6%2Fsin%2840%29
r=6%28sin%2870%29%2Fsin%2840%29%29
The original figure is made of 18 of these right triangles:
leg r
leg 6%2F2=3
-
AREA of original figure, 18%2A3%2Ar
OR
18%2A3%2A6%28sine%2870%29%2Fsin%2840%29%29

Answer by ikleyn(52855) About Me  (Show Source):
You can put this solution on YOUR website!
.

            The  "solution"  by @josgarithmetic is   W R O N G   from its third line to the end.

            So,  for your safety,  simply ignore it . . .

            I came to bring the correct solution. 


There are 9 congruent isosceles triangles covering the area of the nonagon.


The central angle of each triangle is  360%5Eo%2F9 = 40°.


Each of the base angle of such a triangle is  %28180%5E0-40%5E0%29%2F2 = 70°.


Half of the base length is 3,  and the height of each isosceles triangle is  h = 3*tan(70°).


Thus the area of each isosceles triangle is  A = %281%2F2%29%2A6%2A3%2Atan%2870%5Eo%29.


The area of the nonagon is  9A = %289%2F2%29%2A6%2A3%2Atan%2870%5Eo%29 = %289%2F2%29%2A18%2A2.742 = 222.1 cm^2  (approximately).    ANSWER

Completed and solved.


/\/\/\/\/\/\/\/\/


This quasi- and pseudo- tutor @josgarithmetic is a misfortune of this forum.

Having no systematic Math education, he makes errors and provides incorrect solution to each second, third, fourth problem . . .

So, be aware when you get "solutions" from him.

In many (toooooooo many cases) they are wrong.

Do not trust him and always wait for the reaction from other tutors . . .



Answer by MathTherapy(10556) About Me  (Show Source):
You can put this solution on YOUR website!
Find the area of a regular nonagon of side 6cm 
Correct answer: 

One answer is NOT NEARLY CLOSE to being correct - the person is LOST - while the other is close.

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Find the area of a regular nonagon of side 6cm
--------
n = 9
s = 6
===========
Area+=+ns%5E2%2Acot%28180%2Fn%29%2F4
---
You can use that, or you can do the 9 triangles thing.