Wednesday, February 25, 2009

Class on 2/24/09

Do problems 1, 4, 5, 12 & 15 on page 235 of your textbook. You may use this post to discuss any of the questions.

8 comments:

Anonymous said...

In the homework journal I assume?

Anonymous said...

can anyone push me in the right direction for number 1?

Anonymous said...

nevermind

Anonymous said...

can anyone help me with 12?

Anonymous said...

Virginia, I started by frist looking at triangles CDA and BDA in relation to each other and then triangle DBA in relation to triangle ABC. Hope this helps.

Anonymous said...

For number 15, I figured out a, but b is making me confused. Can someone help me in figuring out how triangles BPS and QCR are similar, if that is the path you are supposed to take.

Anonymous said...

Jack, I have your homework journals so you can't possibly do it in there. Remember, this is classwork and so do it in your notebook.

As for other questions, I shall leave it upon all of you to figure things out and help each other out.

Cheers

Anonymous said...

well since AB and AC are transversals of PQ and BC you know that angle APQ=ABC and AQP=ACB and angles A PSB, and QRC are all 90 degreesso with this info you can prove that each little triangle is equal to triangle APQ. In triangle QRC, angle QRC = angle A, and angle ACB = angle AQP, so we know that angle CQR must equal angle APQ which equals angle ABC and by the same method angle BPS must equal ACB. So you know that all three angle in the little triangle are equal which is more than enough for ~ity or similarity. But i couldn't figure out the (PS)squared stuff. How this helps just a little bit Victoria