+0

-3
144
3

1. G is the centroid of equilateral Triangle ABC. D,E, and F are midpointsof the sides as shown. P,Q, and R are the midpoints of line AG,line BG and line CG, respectively. If AB= sqrt 3, what is the perimeter of DREPFQ? Nov 17, 2019

#1
0

What have you tried so far?

Nov 17, 2019
#2
+2

Since ABC is isosceles....Angle FAP   = Angle EAP  = 30°

EA = FA

PA  = PA

So....by SAS....triangle FAP = triangle  EAP

And Angle EAP  = Angle ECR   =  30°

So angle CGA  = 120°

Draw DF   and angle  DQF  = 120°

And angle QDF  = angle QFD  = 30°

So triangles CGA and  DQF are similar

Draw GE

And EA  = sqrt (3)  / 2

And triangle GEA  is a 30 -60 -90  right triangle

So.....AG  = EA * 2 * sqrt (3)  =  sqrt (3) / 2 * 2 / sqrt (3)  = 1

And triangle BDF is similar to triangle BCA

And since BF = (1/2) BA, then DF = (1/2)CA

And since CGA and DQF  are similar....then  DF  = (1/2) CA  so   QF = (1/2) AG = 1/2

And AG  is the side of regular  hexagon DREPFQ

So.....the perimeter  is  6 (AG)  =  6 (1/2)   =  3   Nov 17, 2019
#3
+1

YES! That is what I was trying, proving that there are 30 60 90 triangles within the shape, then solving!

Nov 17, 2019