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Mathematics 14 Online
OpenStudy (anonymous):

http://puu.sh/34HQy.png

OpenStudy (whpalmer4):

Looks like a job for similar triangles...

OpenStudy (anonymous):

Nope, that is the entire question right there.

OpenStudy (whpalmer4):

What is the distance between yachts 2 and 3?

OpenStudy (whpalmer4):

When you know that, you can set up similar triangles between harbor / yacht 2 / yacht 3 and yacht 1 / yacht 2 / harbor

OpenStudy (anonymous):

49.

OpenStudy (whpalmer4):

Okay, so how would you use that as I proposed?

OpenStudy (anonymous):

Well actually there is a property of rigth triangles wicch is prooven with similarity. Let x be the distance between Yacht 1 and harbor and in this case happens that x^2 = 130*81 Now let y be the distance between yacht 2 and harbor x^2=81^2 +y^2 (pythagorean theorem) x^2=81^2 +y^2 130*81-81^2 = y^2 81(130-81)=y^2 y^2=81*49 y =9*7 y=63. And that's the answer

OpenStudy (anonymous):

The opposite is 49 from what I would see. That is all the information I have so far.

OpenStudy (whpalmer4):

no need for Pythagoras: if we call distance between yacht 2 and harbor \(x\), then because we have similar triangles, \[\frac{81}{x} = \frac{x}{49}\]Cross multiply: \[x^2=81*49 = 9^2*7^2\]\[x=7*9=63\]

OpenStudy (anonymous):

Give me a second for my brain to process these. xD

OpenStudy (anonymous):

I don't understand where you got 9^2 and 7^2 from...

OpenStudy (anonymous):

Oh wait, I see now... Nevermind. So what is the rule for you to be cross multiplying?

OpenStudy (whpalmer4):

cross-multiplying is just a shortcut to solving: \[\frac{a}{b}=\frac{c}{d}\]Multiply both sides by \(b\): \[b\frac{a}{b} = b\frac{c}{d}\]Multiply both sides by \(d\): \[bd\frac{a}{b} = bd\frac{c}{d}\]\[ad=bc\] In other words, if the fractions are equal, then ad=bc.

OpenStudy (anonymous):

Damn, I know how mathematical laws are just that. Laws. So I can just follow them without thinking about the logic behind them but I can't seem to solder it into my brain. :/

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