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# In the diagram below, we have DE = 2EC and AB = DC = 20. Find the length of FG.

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In the diagram below, we have DE = 2EC and AB = DC = 20. Find the length of FG. Thanks!

EDIT: its 12 :)

Feb 15, 2018
edited by AnonymousConfusedGuy  Feb 15, 2018

### 5+0 Answers

#1
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FG=10

Feb 15, 2018
#2
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Sorry its not 10

AnonymousConfusedGuy  Feb 15, 2018
#3
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Angle AFB  =  Angle DFE

Angle DBA  = Angle FDE

Therefore, by AA congruency  Triangle DEF  is similar to Triangle BAF

But  DE  is  2/3  of DC.... and since AB  = DC, then  DE  is also 2/3 of AB

Then all  parts of triangle DEF  are 2/3  of  their corrresponding parts in triangle BAF

Thus....the altitude of triangle  DEF  is 2/3  of the altitude of triangle BAF

But the altitude of triangle DEF  = CG since they are parallel

And for the same reason, the altitude of triangle BAF   = BG

Thus  CG   =  2/3  of BG.....so ...  there are 5 parts of BC......and BG  is 3 of these

So  BG  =  3/5   of BC

And triangle BGF  is similar to  triangle BCD

But  BG  =  3/5  of BC....so  GF  =  3/5 of CD

So    GF  = FG  =   (3/5)  20    =   12   Feb 15, 2018
edited by CPhill  Feb 15, 2018
#4
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In the diagram below, we have DE = 2EC and AB = DC = 20. Find the length of FG. $$\text{Let DC = 20 } \\ \text{Let DE = \dfrac23\cdot 20 } \\ \text{Let EC = \dfrac13\cdot 20 } \\ \text{Let FG = x + EC  or x=FG-EC  }\\ \text{Let BG = p  } \\ \text{Let GC = q  }$$

1. intercept theorem

$$\begin{array}{|rcll|} \hline \dfrac{q}{x} &=& \dfrac{p}{DE-x} \quad & \quad DE= \dfrac23\cdot 20 \\\\ \dfrac{q}{x} &=& \dfrac{p}{\dfrac23\cdot 20-x} \\\\ \mathbf{\dfrac{p}{q}} & \mathbf{=} & \mathbf{\dfrac{40-3x}{3x} \qquad (1)} \\ \hline \end{array}$$

2. intercept theorem

$$\begin{array}{|rcll|} \hline \dfrac{q}{DE-x} &=& \dfrac{p}{x+EC} \quad & \quad DE= \dfrac23\cdot 20 \qquad EC=\dfrac13\cdot 20 \\\\ \dfrac{q}{ \dfrac23\cdot 20-x} &=& \dfrac{p}{x+\dfrac13\cdot 20} \\\\ \mathbf{\dfrac{p}{q}} & \mathbf{=} & \mathbf{\dfrac{3x+20}{40-3x} \qquad (2)} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline (1)=(2): & \mathbf{\dfrac{p}{q}} = \mathbf{\dfrac{40-3x}{3x} }&\mathbf{=}& \mathbf{\dfrac{3x+20}{40-3x}} \\\\ & \dfrac{40-3x}{3x} & = & \dfrac{3x+20}{40-3x} \\\\ & (40-3x)^2 & = & 3x(3x+20) \\\\ & 1600-240x+\not{9x^2} & = & \not{9x^2} +60x \\\\ & 1600-240x & = & 60x \\\\ & 300x &=& 1600 \quad & | \quad : 300 \\\\ & x &=& \dfrac{1600}{300} \\\\ & x &=& \dfrac{16}{3} \\\\ & FG &=& x + EC \qquad EC=\dfrac{20}{3} \\\\ & &=& \dfrac{16}{3} + \dfrac{20}{3} \\\\ & &=& \dfrac{36}{3} \\\\ & \mathbf{FG} &\mathbf{=} & \mathbf{12} \\ \hline \end{array}$$ Feb 16, 2018
#5
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Thanks so much you guys! I love this site and its members, you are all so helpful!

Feb 22, 2018