Need Help Regular Pentagon

• Dec 31st 2009, 12:39 PM
Mike Clemmons
Need Help Regular Pentagon
Hi,
I'm looking for help with a geometry/trig problem that has my son and me stumped. The problem basically requires determining the area of a regular pentagon, and finding the length of a perpendicular inside the pentagon.
We are coming with a set of answers that are very different from the solution the book has. Our answer to the question of how long the internal perpendicular is, is square root of 200....14.14. The book gets 13.8.
I've attached a copy of the problem as a word file. (I tried to put the problem sketch in this window, but it doesn't seem to take).
Mike Clemmons

[IMG]file:///C:/Users/MICHAE%7E1/AppData/Local/Temp/moz-screenshot-2.jpg[/IMG] [IMG]file:///C:/Users/MICHAE%7E1/AppData/Local/Temp/moz-screenshot.jpg[/IMG][IMG]file:///C:/Users/MICHAE%7E1/AppData/Local/Temp/moz-screenshot-1.jpg[/IMG]
• Dec 31st 2009, 01:13 PM
sym0110
Quote:

Originally Posted by Mike Clemmons
Hi,
I'm looking for help with a geometry/trig problem that has my son and me stumped. The problem basically requires determining the area of a regular pentagon, and finding the length of a perpendicular inside the pentagon.
We are coming with a set of answers that are very different from the solution the book has. Our answer to the question of how long the internal perpendicular is, is square root of 200....14.14. The book gets 13.8.
I've attached a copy of the problem as a word file. (I tried to put the problem sketch in this window, but it doesn't seem to take).
Mike Clemmons

[IMG]file:///C:/Users/MICHAE%7E1/AppData/Local/Temp/moz-screenshot-2.jpg[/IMG] [IMG]file:///C:/Users/MICHAE%7E1/AppData/Local/Temp/moz-screenshot.jpg[/IMG][IMG]file:///C:/Users/MICHAE%7E1/AppData/Local/Temp/moz-screenshot-1.jpg[/IMG]

If you connect OA-OE respectively, each angle at the center is 360/5=72
hence the base of each triangle is (180-72)/2=54
take triangle ABO, AB=20, let M be the midpoint of AB, hence AM=MB=10
so OM is simply tan(54)*10=13.8
Now knowing the base and side of the triangle, whose area is 1/5 of the pentagon, the rest is straight-forward
• Dec 31st 2009, 01:20 PM
Plato
Here is a good link: Pentagon -- from Wolfram MathWorld
You want formula (10).
$R=2\sqrt{25+10\sqrt{5}~}$