# Geometry:Similar Triangles

• Feb 3rd 2011, 12:12 PM
vaironxxrd
Geometry:Similar Triangles
Just want to know if i got this right guys. Im trying very hard now and i got serious with school im getting good grades :)! thanks guys.
But here i got this similar triangles questions. They are done... is just to know if im right

http://img809.imageshack.us/img809/2...artriangle.jpg
• Feb 3rd 2011, 12:20 PM
topsquark
These triangles are not similar. 4/12 is not equal to 20/15.

-Dan

PS Okay, I appear to be wrong. I'll check out why later.

Ah. I see now. That is a "y" not a 4. Please excuse.
• Feb 3rd 2011, 12:23 PM
vaironxxrd
Quote:

Originally Posted by topsquark
These triangles are not similar. 4/12 is not equal to 20/15.

-Dan

Ratios
• Feb 3rd 2011, 12:25 PM
Yes, that's fine.

You can think of the sides of LMN are magnified by the exact same amount
to be the sides of PQR.

Let k be the magnification factor

$k[MN]=[QR]\Rightarrow\ k=\frac{[QR]}{[MN]}$

$k[LM]=[PQ]\Rightarrow\ k=\frac{[PQ]}{[LM]}$

$k[LN]=[PR]\Rightarrow\ k=\frac{[PR]}{[LN]}$

Therefore

$\frac{y}{12}=\frac{x}{18}=\frac{20}{15}$

or

$\frac{y}{20}=\frac{12}{15}$

$\frac{x}{20}=\frac{18}{15}$
• Feb 3rd 2011, 12:27 PM
vaironxxrd
Quote:

Yes, that's fine.

You can think of the sides of LMN are magnified by the exact same amount
to be the sides of PQR.

Let k be the magnification factor

$k[MN]=[QR]\Rightarrow\ k=\frac{[QR]}{[MN]}$

$k[LM]=[PQ]\Rightarrow\ k=\frac{[PQ]}{[LM]}$

$k[LN]=[PQ]\Rightarrow\ k=\frac{[PQ]}{[LN]}$

Therefore

$\frac{y}{12}=\frac{x}{18}=\frac{20}{15}$

or

$\frac{y}{20}=\frac{12}{15}$

$\frac{x}{20}=\frac{18}{15}$

Thanks i knew i was getting good at this. Sorry for not pointing out it was a fraction
• Feb 3rd 2011, 12:29 PM
Note there was a typo in my post.
• Feb 3rd 2011, 12:35 PM
vaironxxrd
Quote: