+0  
 
0
93
3
avatar

In triangle $DEF,$ $\angle D = 30^\circ,$ $\angle F = 60^\circ,$ and $EF = 6.$ Find $DE + DF.$ [asy] unitsize(2 cm); pair D, E, F; D = (0,sqrt(3)); E = (0,0); F = (1,0); draw(D--E--F--cycle); label("$D$", D, N); label("$E$", E, SW); label("$F$", F, SE); label("$6$", (E + F)/2, S); [/asy]

 Dec 7, 2020
 #1
avatar+323 
+1

A drawing to accompany the asymptote or do the answerers have to do it themselves?

 Dec 8, 2020
 #2
avatar
0

Construct the perpendicular from E to hypotneuse DF, and apply Pythagoras.  Then DE + DF = 6 + 2*sqrt(3).

 Dec 8, 2020
 #3
avatar+1019 
0

DE + DF = 6√3 + 12  smiley

 Dec 8, 2020

48 Online Users

avatar
avatar
avatar