2006-12-04 08:17:38 · 13 answers · asked by Robbie P 1 in Science & Mathematics ➔ Mathematics
∑ incl. ∆’s in a polygon = (A-(B/n)*n
This is applicable to all polygonal shapes (shapes having equilateral sides and angles. Including triangles and squares.
Where:
A = ∑ incl. ∆’s of a triangle
B = Number of ° in a circle
n = Number of sides to the polygon.
The ∑ incl. ∆’s for a pentagon is 540°.
2006-12-04 08:36:16 · answer #1 · answered by God all Mighty 3 · 0⤊ 0⤋
Let the two ends of one of the sides be called point A and point B and well take an arbitrary point also on the line called point C. Note that since A,B and C are collinear, angle ABC =0 degrees
Therefore each side of a pentagon measures upto 0 much degrees.
Moral of the story: Ask a stupid question get a stupid answer or GIGO to an Informatist.
2006-12-05 11:58:41 · answer #2 · answered by fretty 3 · 0⤊ 0⤋
There are a lot of previous answers that are just wrong! Angles of a polygon do not have to add up to 360 e.g a triangle is 180!
All internal angles of a polygon add up to
(2n - 4) x 90 where n is the number of sides
e.g. for pentagon, it's
(2x5-4)x90
=6x90 = 540 so each angle is 540/5 = 108 degrees
e.g. for a triangle, it's
(2x3 - 4) x 90
= 2 x 90 = 180 so each angle of an equilateral triangle is 180/3 = 60.
2006-12-04 16:31:26 · answer #3 · answered by goulash 2 · 0⤊ 0⤋
A pentagon has 5 sides.
The sum of the exterior angles of all polygons measure 360 degrees. Hence each of the exterior angles are 360/5 = 72 degrees.
So each of the interior angles is 180 - 72 = 108 degrees.
2006-12-04 18:57:45 · answer #4 · answered by Kemmy 6 · 0⤊ 0⤋
All the answers I have read to this question have been wrong so far. Think of a regular pentagon. The centre of the regular pentagon can be joined to the angles on the outside by straight lines that form isosceles triangles (as the distance from centre to one angle is the same as to any other) The angle of at the centre of each of these isosceles triangles is 72 (360/5 [angles at a point]) the other two angles must be 54 ((180-72)/2 [angles in an isosceles triangle]) Each angle of the pentagon is made up of two of these angles so it is 108 degrees for each angle. (Editting: Can I apologise to "goulash" who must have posted her answer at the same time as me. Your answer is correct (obviously as you have a degree in mathematics) but the I am intrigued by your formula. Can you prove it as I haven't seen it before? Can you understand my working?)
2006-12-04 16:32:04 · answer #5 · answered by f+v=e+2 Euler's a genius!!! 1 · 0⤊ 0⤋
There are five sides to a pentagon and five angles forming it, hence its name. The total of these angles is 360 degrees, therefore each must be 360 divided by five. You work it out.
2006-12-04 16:33:51 · answer #6 · answered by Bunnylove 2 · 0⤊ 0⤋
108.
Forumula: Number of sides of sahpe - 2. Times that by 180. Then divide by the number of sides to get the degrees.
5-2=3*180=540
540/5=108
2006-12-04 16:22:04 · answer #7 · answered by B-B@!! P!@Y@ 4 · 2⤊ 1⤋
A regular pentagon has internal angles at 360/5 = 72 degrees
(Total internal angles are always 360 degrees)
2006-12-04 16:19:57 · answer #8 · answered by Anonymous · 0⤊ 4⤋
sides aren't measured in degrees. corners are.if (counter clockwise) La = 90 and Lb = 90 then Lc and Le have to be greater than 90deg. Total sum is greater than or equal to 360deg.
2006-12-04 16:33:54 · answer #9 · answered by Master M 3 · 0⤊ 0⤋
108 degrees
2006-12-04 17:57:00 · answer #10 · answered by lenpol7 7 · 0⤊ 0⤋