A#12 (due 2/10)

Chapter 7 Closure (#124a&b, 127, 129, 130a, 131)

A#12 Handout (front)
A#12 Handout (back)

See all previous Chapter 7 notes in the Lectures sub-forum to help you out.

Below is a solution guide to the problems. It is recommended that you attempt them before watching the video, but nothing is stopping you from watching it. The important thing is that you know how it all works in the end.



EDIT:

As it turns out, I just witnessed an answer key which says there is a second point for D that can be found in problem #128 at (5,-. I disagree, because they tell us this quadrilateral is named ABCD, and the order of the letters matter (in other words, AC is a diagonal and not a side when it's ordered this way). I would just stick with the point (1,0). Watch the review video for more on what I'm talking about regarding naming quadrilaterals.

On problem 7-127, on the first statement. I'm not sure how this person could made such assumption with such little information.

Quinton R. wrote:

On problem 7-127, on the first statement. I'm not sure how this person could made such assumption with such little information.

Haha those are things that we just are told to be given, whether she knows it or somebody else knows it.

There are some certainties in life. That just must be one of them.

Sorry about me not actually doing the math on #129c, by the way. As I continually said, I was running out of time (before 2nd period classes came in) and had to rush right through it. If ya'll know the properties of a square, you can make the statement I made with that justification. If you're asked to prove that though, then I can see where we have issues.

wait, whats the triangle midsegment theorem ?   

Valerie H. wrote:

wait, whats the triangle midsegment theorem ?   

Not sure if you were absent on the day we did N#8.

I just finished writing the test, and midsegments pop up a lot in it (at least enough that you might want to know them). It means this...

A midsegment in a triangle is formed when connect two midpoints with a segment in a triangle, like so:




D and E are midpoints, which is why those portions are congruent to one another. The midsegment is DE, and it is drawn between CA and CB.

The Triangle Midsegment Theorem says that the midsegment connected between two sides of a triangle is exactly half of and parallel to the third side of the triangle. So, the length of DE will be half the length of AB, and it's also parallel to it.

We proved this in class by finding out the triangles were similar and finding the side ratios of the other sides, which is 1:2. That means the third sides in the similar triangles are also 1:2.

Watch the video first if there's anything else you need, otherwise feel free to ask any questions on here. Unfortunately, I'll have to see the questions when I wake up in the morning. I'll be getting to school at 6:00 AM, so feel free to drop by!