Start by doing exercises 4-10 on page 224 and then from Exercises [A] on pages 225-226 do problems 1-4, 7 and 9. Next, read pages 226-228 and make sure to understand the proofs for the theorems regarding similar triangles discussed on those pages. You will be required to take notes on this reading. In your notes please try and re-write, in paragraph form, the proofs for the theorems.
[Note: For those students that are absent on Friday, Feb 13th, you can download the handout we will have done in class from the class conference folder and you will be expected to complete the handout and then do the homework that is assigned in this post.]
Subscribe to:
Post Comments (Atom)
15 comments:
Here's my answer to number 7 like you requested Mr. Kerai.
Prompt: The sides of a hexagon are 2, 3, 2, 4, 5, and 6 inches. Find the perimeter of a similar hexagon, two of whose sides are 3 inches.
{- Hexagon A already has two sides which are 2 inches, and so the three inch sides of similar hexagon B are corresponding parts. The ratio of similitude between the hexagon B and A is 3:2, as that is the ratio between a pair of corresponding sides from the hexagons. If x inches is the perimeter of hexagon B and 16 inches is the perimeter of hexagon A, then x:16 = 3:2. Once one uses cross multiplication, we find that 2x = 3(16) => 2x = 48. We then divide two by both sides to isolate x, and we find that x = 24, thus, the perimeter of hexagon B is 24 inches.
Thank you Scott, your a gem!
I think the perimeter of hexagon A might be 22 unless I made a math mistake. 2+2+3+4+5+6=22
Mr. Kerai I believe that is Whitney STOTT, not SCOTT Newton
Oh Gosh! I'm sorry, thank you Whitney and not Scott, but you're both gems! I'm getting old I guess. But really, it couldn't hurt to just use your first names...
(sorry this is in the wrong section, there was not a post for tonights hw.)For number 12 on p. 235, I am at a loss at where to start looking for similar triangles. I know that in each triangle there is a right angle, but I feel like I am missing something obvious to say they are similar. Any suggestions?
Wait, did I assign number 12 on page 235?
Anyway, even if I didn't assign it, since you're on that question then pay attention to triangles ADC and ABD. Because of the two angle corollary you'll see that those triangles are both similar to ABC. Now get some proportions going with corresponding sides, especially the ones included in the equation you have to prove.
Mr. Kerai, your right. I was looking at the wrong number 12.
I'm confused on where to start on 12 on 234. Maybe I was doing the wrong one... Can anyone help me?
I know that triangles QDC and QAB are similar, but I'm not sure where to go from there.
Looking at the right number 12 on p. 234, I see the triangle QDC, and that AB cuts QD and QC proportionally. I am thinking that the distance from DC to AB is a perpendicular line from the mentioned lines and measuring 1.5 in. (given). Am not sure how this will give me the distance from p and q to AB. Suggestions?
Any ideas for # 13 on p.229?
Okay, so if you drop an altitude from Q to AB then you can get the length of the altitude from the ratio 3:2 from AB:DC. Because 1.5 is the portion of that altitude between DC and AB and the remaining portion can be obtained by the proportion of 3:2 = x/1.5 where x is that remaining portion. Then add x to 1.5 to get the distance of Q from AB. Then for the distance of P, start by make a line through P that is perpendicular to both AB and DC (which it has to be since they are parallel). Then try some things from there...
Sam, for number 13 on page 229, remember that QB and PB represent the light from the tree reflected up to the boy's eye level. A property of reflection the angle the light makes with the mirror and the dotted vertical line is the same as the angle the reflected ray of light makes with the dotted vertical line. This is also know as the angle of incidence. This means that angle QBC = angle PBA.
Post a Comment