# Regular Pentagon

• Nov 2nd 2009, 06:52 AM
warriors837
Regular Pentagon
The perimeter of a regular pentagon is 100 mm. Find the perimeter of the regular pentagon formed by connecting it diagonals?
• Nov 2nd 2009, 10:17 AM
ukorov
tried to draw a regular pentagon but not easy(Clapping)
the pentagon produced by connecting the diagonals should be MNOPQ as in the attached pic.
the bigger pentagon ABCDE has each side 2cm and each interior angle 108 degrees.
in isoscles triangle ABE, both angles ABE and AEB are 36 degrees.
therefore, each interior angle of pentagon ABCDE can be proved to have been equally divided by 2 diagonals into 36 degrees x 3.
In triangle ABE, use Sine Law to find BE:
BE / sin108 = AB / sin36 = 2 / sin36
In triangles ABM and AEQ, find BM and QE by the same method.
you will see BM = QE
So you get the length of MQ = BE - 2BM and in the end, the perimeter of pentagon MNOPQ = 5MQ
• Nov 6th 2009, 09:18 AM
warriors837
The resulting perimeter is 5[(20sin108/sin36)-2(20sin36/sin108)] millimeters. So the perimeter is approximately 3.820 cm or 38.197 mm.
• Nov 6th 2009, 07:25 PM
ukorov
Quote:

Originally Posted by warriors837
The resulting perimeter is 5[(20sin108/sin36)-2(20sin36/sin108)] millimeters. So the perimeter is approximately 3.820 cm or 38.197 mm.

that's right, 3.81966cm.