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# geometry problem

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ED = 2*CE and BA = CD = 50.  Find FG. Dec 7, 2020

#1
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Use the Pythagorean theorem to figure out the length of line AE and DB.

If you don't know what that is, it's \$A^2 + B^2 = C^2\$

Dec 7, 2020
edited by heartSTORM907  Dec 7, 2020
#2
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AB = BC = CD = 50

DE = 2/3 * 50           DE = 33.334

Using the law of sines we can calculate DF and EF.     (DF = 28.284, EF = 24.037)

When we have all 3 sides of ΔDFE, we can calculate the triangle's area.   [DFE] = 333.334 u2

Height  FH = 2[DFE] / DE    ==>    FH = 20

BG = FG = BC - FH                     FG = 30 units Dec 7, 2020
#3
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∠FEH = tan-1(DE / AD)     Incorrect! ∠FEH = tan-1(AD / DE)     Correct! jugoslav  Dec 7, 2020
edited by jugoslav  Dec 7, 2020
#4
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EC  = (1/3) 50 =   50/3

ED =  (2/3) 50   =  100/3

And triangle  DFE  ≈  triangle  BFA

So DF / BF  =   (100/3) / 50  =  2/3

And triangle DBC ≈  triangle  FBG

But  BF / BD  =  BF  / (BF + DF)  =  3 / (3 + 2)  =   3/5

Therefore

FG/ DC =  FB/ BD

FG  / 50  =  3/5

FG  =  50  (3/5)  =  30   Dec 7, 2020