Figure D is a regular polygon. If D’ is a 20 degree rotation of D and is concurrent to D, what is the smallest number of sides that D could have?

18

Step-by-step explanation:

I've never heard of concurrent polygons. I'm assuming that by rotating D by 20°, D keeps its original orientation. Therefore, an exterior angle of D is 20°.

number of sides = 360°/20° = 18

true

explanation:

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl