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In the diagram below, $ADEF$ is a square with side length 5. If $AC = 12$ what is $BD?$

 Oct 14, 2019
 #1
avatar+23350 
+3

In the diagram below, ADEF is a square with side length 5. If AC = 12
what is BD?

 

\(\text{Let $AF=AD=EF=5$ } \\ \text{Let $AC=12$ } \\ \text{Let $FC=AC-AF= 12-5 = 7$ } \)

 

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{EF}{AC-AF}} &=& \mathbf{\dfrac{AD+BD}{AC}} \\\\ \dfrac{5}{12-5} &=& \dfrac{5+BD}{12} \\\\ \dfrac{5}{7} &=& \dfrac{5+BD}{12} \quad | \quad \cdot 12 \\\\ \dfrac{12\cdot 5}{7} &=& 5+BD \\\\ BD &=& \dfrac{12\cdot 5}{7} -5 \\\\ BD &=& \dfrac{12\cdot 5}{7} -5 \\\\ BD &=& \dfrac{60-5\cdot 7}{7} \\\\ BD &=& \dfrac{60-35}{7} \\\\ BD &=& \dfrac{25}{7} \\ \mathbf{BD} &=& \mathbf{3.5714285714} \\ \hline \end{array}\)

 

\(BD \approx 3.57 \)

 

laugh

 Oct 14, 2019
 #2
avatar+104962 
+1

Thanks, heureka!!!

 

An alternative approach :

 

Note that triangles EFC  and BDE  are similar......which means that

 

EF/FC  =  BD/DE

 

5 / 7  =  BD / 5      multiply both sides by  5

 

25/7  = BD

 

cool cool cool

 Oct 15, 2019
edited by CPhill  Oct 15, 2019
 #3
avatar+23350 
+2

Thank you, CPhill !

 

laugh

heureka  Oct 15, 2019

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