Question 1142081
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The &nbsp;"solution" &nbsp;by @josgarithmetic is &nbsp;&nbsp;<U>W R O N G</U> &nbsp;&nbsp;from its third line to the end.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;So, &nbsp;for your safety, &nbsp;simply ignore it . . . 


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;I came to bring the correct solution.



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


The central angle of each triangle is  {{{360^o/9}}} = 40°.


Each of the base angle of such a triangle is  {{{(180^0-40^0)/2}}} = 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 = {{{(1/2)*6*3*tan(70^o)}}}.


The area of the nonagon is  9A = {{{(9/2)*6*3*tan(70^o)}}} = {{{(9/2)*18*2.742}}} = 222.1 cm^2  (approximately).    <U>ANSWER</U>
</pre>

Completed and solved.



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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 . . .