# Similar triangles - calculating the length

• Sep 23rd 2010, 01:47 AM
PythagorasNeophyte
Similar triangles - calculating the length
Attachment 19012
AS=2cm, SB=4cm, BC=15cm

*DQ is a typo. It should be SQ. Many apologies.

Greetings everyone. PythagorasNeophyte here.

I am doubtful of part 1 - I don't know what formula(s) must be used in finding the answer in part 1. But I am very categorical that there must be similar sides in this diagram. For example, SQ is similar to BC. That is true. Nevertheless, SQ cannot be found.

Is there any way to find SQ?

As for part 2, the answer would be triangle SQA and triangle RQA.
• Sep 23rd 2010, 02:29 AM
sa-ri-ga-ma
Where is D in the diagram?

If SQ is parallel to BC, then the triangles ASQ and ABC are similar. Then

SQ/BC = AS/AB.

Find SQ
• Sep 23rd 2010, 04:13 AM
Wilmer
Quote:

Originally Posted by sa-ri-ga-ma
Where is D in the diagram?

Just a typo, I'm sure: DQ is really SQ.

You can go this way also:
SQ / AS = BC / AB : SQ = AS * BC / AB ; P.N. did you look up "similar triangles"?
• Sep 23rd 2010, 04:51 AM
PythagorasNeophyte
Oh! I am very sorry that I had made a typo error in naming SQ as DQ. Please omit DQ. I'm very sorry.

So:

$\dfrac{SQ}{BC} = \dfrac{AS}{AB}$

$
\dfrac{SQ}{15} = \dfrac{2}{4}$

$SQ = 7.5cm$

Is this correct?
• Sep 23rd 2010, 07:12 AM
Wilmer
NO. AB=6. Try again.
• Sep 23rd 2010, 07:35 AM
umangarora