Ok, here goes. I explained several of these in class so I will be brief on those.
#3. Do NOT set the expressions equal to each other. That would only happen in a rhombus.
#4. If you work with slopes first, you'll have to do two calculations. That will tell you parallelogram (?) and possibly rectangle. Regardless, you'll have to do a couple of distance calculations to see if you're dealing with a rhombus. Remember, rhombus + rectangle = square.
#5. You should have A = ....... and P = ......... In part c, the word evaluate means to find a value for A and P.
#6b. Kites?! What are they, I'd like to know. Do we take Wikipedia as a reliable source???? According to them, a kite may be a square. OK, so our textbook loses by Wiki rules.
#7. Focus on the angles that I numbered in class. Converse of AIA is useful.
#8. You already have a p-gram. To make it a rhombus, you only have to get two consecutive sides congruent. If you prove the overlapping triangles congruent, you've got CPCTC and you're almost done.
#9. Five steps, if you include the parallels in the Given. You could put Draw steps into the beginning of the proof which would add on two steps. It would be a professional geometer's way to go.
#10. Still in the A exercises... don't overthink it. I like an equidistance theorem to make this a slick proof.
#11. I have 9 steps because I just put in the fact that AE is NOT parallel to DC in the ending triple play. For a reason, just put Diagram. About that triple play, it includes one parallel statement, one non-parallel statement and one congruence statement. Wrap those up and you're done.
#12. Ah, one of my favorites. Here's a helpful hint... if I have to subtract two numbers from a third number but I only know the sum of those two numbers, can't I subtract the sum? In other words, I can subtract x and y from z even if I don't know what x and y each are. If I know their sum, it's all the same.
#13. No hints needed. Be observant.
#14. I worked on four slopes and two distances. You have other options. Keep in mind that slopes give a little more bang for the buck. They prove parallelograms and, if you're lucky, show perpendicular segments too. Show all work.
#15. Long proof. First focus on Triangle PQA. That will give you sides that you need in a transitive step. Then switch over to Triangle RAS. That will give you angles in a transitive step. Once you have the two angles at vertex S congruent, you're done. You'll need AIA and "If sides, then angles."
#16. Discussed in class. For the base, don't think horizontal.
#17. Draw the special quads and use their diagonals as mirrors.
#18. Took 10 steps for me. After I used the Given stating "Segments are drawn...", I followed up with simple parallel statements using the right letters. That makes the first Prove a breeze.
For the second part, do you remember which letter patterns we can use to describe alternate interior, corresponding, same-side interiors, ...? For example, SSI's take on a U-shape. Looking at letter patterns might help you. It is fine to say BC is parallel to MP, even though BC is an extension of the parallelogram. I figured that as a Restate (or Given would be fine).
#19. Discussed in class. Don't forget to finish up with the Perimeter.
#20. How can you figure out (x, y). One is already on the diagram, the other comes from the fact that the point is on a line.
#21. Area = pi r squared
#22. Answers will vary. But we all know what shape will result.
#23. After our circumcenter/centroid work, this is child's play, right? Midpoints are averages of x's and y's.
#24. Reflection of a point over an axis only changes the sign of one coordinate. So label D appropriately. Then set up an algebra equation by setting the slopes equal. Once you set up that equation, the answer might jump off your paper (or you can cross multiply).
#25. Probability of two things happening = (probability of first) X (probability of second) When you figure out the probability of choosing one true statement, assume SUCCESS. This means that there is one less true statement left in the pot. The first probability will have a denominator of 4. The second will have a denominator of 3.
#26. See #24 for reflection tips. I did two slopes and counted two distances.
#27. I proved two overlapping triangles congruent to get angles of one isosceles triangle. Then a quick subtraction property closes it up.
#28. Same old stuff.
#29. Imagine the whole figure shaded. What's its area? Subtract out the white part. And speaking of that white part, you know that x must be less than some quantity. So x squared will be less than that quantity squared. (And of course you might throw in there that x must be greater than 0.)
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