# Thread: Similar triangles - calculating the length

1. ## Similar triangles - calculating the length

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.

2. 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

3. 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"?

4. 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:

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

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

$\displaystyle SQ = 7.5cm$

Is this correct?

5. NO. AB=6. Try again.

6. No. ur answer is wrong,Pythagoras..
AQ/QC=1/2
So, BP/PC=1/2
Let BP be x, x+ 2x=15 => BP=5.
SQBP is a parallelogram.
so BP=SQ=5cm.

7. P.s.
Similar triangles to PCQ are
ASQ and ABC