what is the degree of the exterior angle for a regular dodegon?
The degree of an exterior angle for a regular dodecagon is 30 degrees.
if there is 2 trangles and 1 triangle is labled ABE and the other one is labeld CBD and BE and BD are congruent. we can prove that ABE and CBD is congruent as well by what
AAS
AAA
ASA
SAS
none
The correct answer is SAS (Side-Angle-Side).
To prove that ABE and CBD are congruent, we need to show that:
- AB = CB (both triangles share side B)
- BE = BD (given)
- ∠ABE = ∠CBD (vertically opposite angles)
We have two pairs of congruent sides (BE = BD and AB = CB) and an angle that is shared by both triangles (∠ABE = ∠CBD). Therefore, we can use the SAS congruence criterion to prove that ABE and CBD are congruent.
asa
ASA (Angle-Side-Angle) is not a valid congruence criterion for these triangles.
While it is true that ∠ABE = ∠CBD (angle), and BE = BD (side), we cannot conclude that the third pair of corresponding angles are congruent. In fact, we don't know anything about the lengths of sides AE and CD and whether they are congruent or not.
Therefore, we cannot use ASA to prove that ABE and CBD are congruent. SAS is the correct congruence criterion for these triangles.
To find the degree of the exterior angle of a regular polygon, you can use the formula:
Degree of exterior angle = 360 degrees / Number of sides
In this case, for a regular dodecagon (a polygon with 12 sides), you can substitute the number of sides into the formula:
Degree of exterior angle = 360 degrees / 12 = 30 degrees
Therefore, the degree of the exterior angle for a regular dodecagon is 30 degrees.