if all sides or a pentagon were equal in length?

Question

at what angle would they intersect at??

Answer 1

There is no way to tell. If the pentagon is regular, then the angle is 108 degrees; otherwise, it can be any combination of numbers between 0 and 180 that adds up to 540.

Answer 2

A polygon with equal length sides and equal interior angles is called a regular polygon.
3*60=180,1=180*(3-2)
4*90=360=180*2=180*(4-2)
6*120=720=180*4=180*(6-2)
8*135=1080=180*6=180*(8-2)
n*A=180*(n-2)
The internal angle A of any n-sided regular polygon =
A = 180(n-2)/n
For a pentagon,
A = 180(5-2)/5

Note that the triangle is the only polygon with equal length sides that MUST have all interior angles equal.

Answer 3

The sum of interior angles of a polygon is equal to (n-2)*180, where n is the number of sides. Therefore, the sum of interior angles of a pentagon is 540 degrees.

For each vertex of a regular polygon, the interior angle is 540/5, or 108 degrees.

Answer 4

for an equilateral triangle this is (3-2)*180/3 = 60
for a square this is (4-2)*180/4 = 90
for pentagon this will be (5-2)*180/5 = 108

Answer 5

108

Answer 6

sum of interior angles (polygon) = (n-2)*180,
in it, n = number of sides
Hence, for pentogon (sum of angles) = 540 degrees.
In regular polygon, the interior angle = 108degree(as 540/5)

but in other cases the all polygon with 5 sides is considered as the pentagon.
obviously due to difference in length gives combinations of various angles

Answer 7

if allsides are equal then all angles would be 108 but otherwise they would always differ!!
cheers

Answer 8

that would be 108 degrees