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Mathematics
General Math
-gre.ge.2 distance by similiar triangles
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[QUOTE="karush, post: 6782629, member: 703413"] [ATTACH type="full"]311760._xfImport[/ATTACH] $\textit{not to scale}$ A summer camp counselor wants to find a length, x, The lengths represented by AB, EB BD,CD on the sketch were determined to be 1800ft, 1400ft, 7000ft, 800 ft respectfully Segments $AC$ and $DE$ intersect at $B$, and $\angle AEB$ and $\angle CDE$ have the same measure What is the value of $x$? looks easy but still tricky $\dfrac{x}{EB}=\dfrac{CD}{BD} =\dfrac{x}{1400}=\dfrac{800}{700}$ multiple thru by 1400 then simplify $x=\dfrac{800(1400)}{700}=(800)(2)=1600$ hopefully I thot I posted this problem some time ago here but didn't see the solution in my overall document🕶 [/QUOTE]
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Mathematics
General Math
-gre.ge.2 distance by similiar triangles
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