Proportional Parts- find the missing length. (see attachment)
help c:
Okie,just wait for a few minutes. I PMd someone who might be of assistance.
|dw:1383348113094:dw| \(\bf \cfrac{2}{4}=\cfrac{10}{x}\qquad then\\ \quad \\ \cfrac{2}{4}=\cfrac{y}{x} \)
i was typing a long explanation but jdoe got to iht .-.
|dw:1383348455873:dw|
You split up the triangles .. OHHHH c:
solving for x & y then finding the missing length?
there is no y though ._.
hehe,....right.. "y" is just the side you're after, since I used up "x' =) once you find "x", using the firsts ratio set, you'd find the corresponding side for the missing side on the 2nd triangle, I just happen to name it "y" =)
\[\frac{ 2 }{ 4 } = \frac{ 10 }{ x }\] cross multiplication I think, my memory of geometry is rusty..
\(\bf \cfrac{2}{4}=\cfrac{10}{\square }\qquad then\\ \quad \\ \cfrac{2}{4}=\cfrac{x}{\square }\) =)
2x = 40 find x then put put the result of that into \[\frac{ 2 }{ 4 } = \frac{ y }{ x }\]
and cross multiply and solve for y.
|dw:1383348172933:dw| That's one of my example problems.
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