Teacher2Teacher 
Q&A #193 
View entire discussion [<< prev] [ next >>]
From: Loyd <loydlin@aol> To: Teacher2Teacher Public Discussion Date: 2003030908:34:37 Subject: Re: Re: Trig Ratios; area of octagon using trig On 2003030813:02:49, sharon smith wrote: >if the radius of a circle is 4cm,and inside this circle there is an >octagon what is the area of the octagon > An Octagon has a central angle of 45 degrees. So, an octagon is composed of 8 triangles if you draw a line segment from the center to each vertex. If you bisect one of the eight triangles, you will have a 22.5, 67.5, 90 degree right triangle. Find the area of each triangle which there are now 16. Of course when you multiply the base times the height, you will have the area of two such triangles. So: Lets call the angle 22.5 degrees angle A. I can't draw a good graphic here but you can see the opposite side is the base=b of one of the 16 triangles and adjacent side is the altitude=a of one of the 16 triangles. I will abbreviate the hypotenuse as h. (You should draw a really good graphic is properly see this). b sin A =  h b= hsin A a cos A =  h a=h cos A Area of one of the 16 triangles is 1/2 ab which means that one of the 8 triangles is: Area = hsin A hcos A = h^2 sin A Cos A . If you plug in h=4 and A=22.5 degrees you will get an answer of about 45.25 sq cm. That sounds pretty close since the are of a circle with a radius of 45 is a about 50.27 sq cm. There is another formula that says the area of an Octagon is 2r^2(sqrt(2)). That works also. The latter is a geometric result that is based up on the following: The area of an inscribed regular octagon is equivalent to that of a rectangle whose dimensions are are the sides of the inscribed and circumscribed squares.
Post a reply to this message

[Privacy Policy] [Terms of Use]
Math Forum Home 
The Math Library 
Quick Reference 
Math Forum Search