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# HELP HELP HELP with geometry

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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
+24388
+3

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

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

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

$$BD \approx 3.57$$

Oct 14, 2019
#2
+109450
+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

Oct 15, 2019
edited by CPhill  Oct 15, 2019
#3
+24388
+2

Thank you, CPhill !

heureka  Oct 15, 2019