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zepdrix (zepdrix):
|dw:1407010829469:dw|We have similar triangles, yes?
If you call this lower piece something like ... \(\Large\rm a\)
Then you can setup your relation for similar triangles:\[\Large\rm \frac{2x-6}{a}=\frac{16}{a+a}\]
zepdrix (zepdrix):
|dw:1407011019543:dw|Understand how I setup that relationship? :o
There's a nice shortcut for this one actually, but I was hoping this might help you understand the concept.
OpenStudy (jessiegonzales):
okay yes im understanding!
zepdrix (zepdrix):
\[\Large\rm \frac{2x-6}{a}=\frac{16}{a+a}\]
\[\Large\rm \frac{2x-6}{a}=\frac{16}{2a}\]Multiply both sides by a,\[\Large\rm 2x-6=\frac{16}{2}\]And here we can see what the relationship is telling us.
The 2x-6 side is `half the length` of the 16 side.
Do you understand how to solve for x form that point?
OpenStudy (jessiegonzales):
so will it be 2x-6=8
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