am I on the right track here?

[allegansveritatem]

Here is the problem (I am focusing on part a for now):


Here is what I did:


I think now that I didn't need to square the a and the b so that where I have a squared variable the exponent can be ignored. No? But is this what is being asked for here?

 

[Dr.Peterson]

I think now that I didn't need to square the a and the b so that where I have a squared variable the exponent can be ignored. No? But is this what is being asked for here?

You're right that it's wrong to square -- in similar triangles, we're talking about the ratio of lengths, not the ratio of a length to an area (which is meaningless).

So, what answer do you really get? y = hb/(a-b)? Then you're right.

 

[MarkFL]

Yes, I would use:

[MATH]\frac{y}{b}=\frac{y+h}{a}[/MATH]
And thus:

[MATH]y=\frac{b}{a-b}h[/MATH]

 

[allegansveritatem]

You're right that it's wrong to square -- in similar triangles, we're talking about the ratio of lengths, not the ratio of a length to an area (which is meaningless).

So, what answer do you really get? y = hb/(a-b)? Then you're right.

I think the reason I squared those items with some reference to the pythagorean theorem. But later it occurred to me that theorem doesn't apply here. As long as the two triangles are similar, all is well without Pythagorus.

 

[allegansveritatem]

Yes, I would use:

[MATH]\frac{y}{b}=\frac{y+h}{a}[/MATH]
And thus:

[MATH]y=\frac{b}{a-b}h[/MATH]

Thanks. I will work this transformation out for myself when my brain is resurrected.