Geometric formulae, word problems, theorems and proofs, etc.
maroonblazer
Posts: 51
Joined: Thu Aug 12, 2010 11:16 am
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My text states the following:
$A_1 = \frac{1}{2}*h_1*b$

$A_2=\frac{1}{2}*h_2*e$

$\frac{A_1}{A_2}=\frac{\frac{1}{2}*h_1*b}{\frac{1}{2}*h_2*e}=\left(\frac{h_1}{h_2}\right) \left(\frac{b}{e}\right)=\left(\frac{b}{e}\right) \left(\frac{b}{e}\right)=\left(\frac{b^2}{e^2}\right)$

I'm not following how they're getting from:

$\left(\frac{h_1}{h_2}\right) \left(\frac{b}{e}\right)=\left(\frac{b}{e}\right) \left(\frac{b}{e}\right)$

I fear it's something painfully obvious.

mb

buddy
Posts: 197
Joined: Sun Feb 22, 2009 10:05 pm
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### Re: Question about similar triangles

$\frac{A_1}{A_2}=\frac{\frac{1}{2}*h_1*b}{\frac{1}{2}*h_2*e}=\left(\frac{h_1}{h_2}\right) \left(\frac{b}{e}\right)=\left(\frac{b}{e}\right) \left(\frac{b}{e}\right)=\left(\frac{b^2}{e^2}\right)$

I'm not following how they're getting from:

$\left(\frac{h_1}{h_2}\right) \left(\frac{b}{e}\right)=\left(\frac{b}{e}\right) \left(\frac{b}{e}\right)$
Me either. I mean, yeh, it's true because that's what areas do. If the shapes are similar then the sides are in proportion in terms of their lengths and the areas are in proportion in terms of the squares of the lengths. So really they should just do A1/A2=b^2/e^2. dunno why they did the rest. Yeh, it's confusing.

maroonblazer
Posts: 51
Joined: Thu Aug 12, 2010 11:16 am
Contact:

### Re: Question about similar triangles

I posted this over on StackExchange and got the following answer:
For two similar triangles, the ratio of the height of the first triangle to the height of the second triangle is equal to the ratio of the base of the first triangle to the base of the second triangle.
http://math.stackexchange.com/questions ... a-question

buddy
Posts: 197
Joined: Sun Feb 22, 2009 10:05 pm
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### Re: Question about similar triangles

Yeh, but that just moves the Q, it doesn't answer it. Why didn't they just start with that ratio in the 1st place instead of doing the magic change in t he middle??