Calendar Project

[Saber Knight]

Let me start by saying I'M REALLY SORRY TO BE BUGGING EVERYONE WITH THIS! I don't normally like to do this, but our teacher even said that our textbook would not teach us how to find the answers, and that we needed to use outside sources to be able to do this project.

So, you may have seen me mention something about the Calendar Project of doom. Well, it's 31 math problems, many of which are just like "WHAT THE CRAP?! HOW DO I DO THIS?!" even at this stage in time. It's due January 20th, and I still have a bunch of problems to go, and I'm really not certain I'm going to finish in time.

I really need your help.

If you can't help, I understand completly. After all, this is some pretty tricky stuff. I showed it to my friend's sister who's a senior (and a top student at that), and the few that I showed her she really didn't know how to do.

Alright, so I'm going to post all of the problems I haven't done. Any help would be appriciated SOOO MUCH! Maybe I'll even come up with a reward, or something. I could totally take spriting requests for Pokemon or FE sprites (but, Saber, you'd do those free. Shaddup.).

Well, all things aside, PLEASE! Just take a look, see if you understand anything. If so, I really need your help. This project has me super worried. Even if you're not sure, a hint, even moral support, anything will help! Who knows, a tiny idea may set me off on a trian of thought, and I'll figure out the rest (it happened before).

So, here come the problems of doom. Under each problem, it tells how my teacher wants us to write the final answer.


2. These two concentric circles contain two arcs with lengths s and t, defined by the right triangle with base r and height y. List s, t, and y, in order from smallest to largest, and justify your answer.

Must be written as an Inequality.

4.
Must be written as a decimal

8 The triangle inequality guarantees that the sum of the lengths of two sides of a triangle is greater than the length of the third. As a consequence, if x and y are the legs of a right triangle, with x (less than or equal to, the symbol didn't copy over) y , and z is the hypotenuse, then x + y > z, so x > z-y. Under what circumstances will x > 2(z - y) be true?
Well, she just wrote "Answer the question" as the format for the answer.

10 The circular table in the diagram is pushed against two perpendicular walls. The point P on the circumference of the table is a distance 2 from one wall and a distance 9 from the other. What is the radius of the table?

She just wrote "Exact" as format for the answer.

11 A square is divided into three pieces of equal area by using two parallel cuts, as shown. The distance between the parallel lines is 10 inches. What is the area of the square?

Sq units

13 Can two angle bisectors in a triangle be perpendicular?
I know the answer is "no", but she wants a proof for the answer, and I don't know how to prove it. Anyone know a postulate or theorem or whatever that would help to prove this?

14 Consider two mirrors placed at a right angle to each other. A person standing at ponig A (x, y) shines a laser pointer so that the light hits both mirrors and then a person at point B (a, b). What is the total distance that the light travels, in terms of a, b, x, and y? Assume x, y, a, and b are positive.

Not sure exactly what happened to that image she gave us, but whatever.
The answer needs to be expressed as "exact with variables"

20 Pam has an unusual dog run in her yard. A fifty-foot rope is tied at each end to 2 pegs that are fourteen feet apart. The dog is tethered to the rope, but the tether is loose and slides freely along the rope between the pegs. Pam laid tree bark over the area of the yard that the dog can reach. What is the area of the region that the dog can reach?

Exact feet

21 ABCD is a piece of paper 1 foot square, and M is the midpoint of AB. The vertex at C is folded up to coincide with M, as shown. Show that all three right triangles – BEM , AMG , and DFG - have side lengths in a 3:4:5 ratio.

A proof

23 A dodecahedron has twelve faces, all of which are regular pentagons. Three edges meet at each vertex of the dodecahedron. An interior diagonal is a segment connecting two vertices such that the segment is not an edge or along a face of the dodecahedron. How many interior diagonals does a regular dodecahedron have?
Must be expressed as a "quantity"

25 Find the area of the shaded region below, given that ABCDEF is a regular hexagon with side length 6.

I HAVE NO IDEA FOR THIS ONE, and neither does anyone in class as far as I know. Anyone?
Must be expressed as "exact"

28 Consider acute triangle ABC with area of 20 square units and sides AB = 7 and AC = 10 . Find the exact length of the remaining side.

"exact"

29 A triangular trough has sides meeting at angles of Ө degrees. A ball of radius R is placed in the trough. In terms of R, what is the radius of the largest ball that will just fit in the trough beneath the bigger ball?

"exact"

Whew...so there they are.

Help...me.....

[Blue Fire Cyndaquil]

I may be in 7th grade, but I'm two years ahead in math. I'll give it a shot.
I'll edit later with my answers.

[Josiah]

My brother's friend Jaun (the "math genius" of their group) and I came up with the following. Not 100% on most of them though.

2) s<y<t
"s" is less than "t" because it is the corresponding arc of the smaller circle. The hypotenuse of the right triangle pictured is smaller than the arc "t". "y" is smaller than "t" because it is the leg of that right triangle, and both legs are smaller than the hypotenuse.

8) *Jaun* My best answer is: "x" cannot be equal to "y"
Not sure how he came up with that.

11) 900 sq units
Each part of the square in equal area. The center part is a parrallelagram with a height of 10. A parrallelagram can be can be turned into a rectangle with indeticle height and width. Therefor, the square is equal to three rectangles, each with a height of 10. So total height of the square is 30 In a square, width = height. So 30 x 30 = 900.

20) *Jaun* 105 sq ft ......... *Josiah* 336 sq ft (I think Jaun wrote a number wrong when he was working it.)
The dog can only make triangles where one side is always 14 ft and the other two sides must add up to 50. Create a triangle with lengths at 14, 25, 25. Divide it down the center into two right triangles. Find the height using a^2 + b^2 = c^2. Using the height, find the area of the big triangle 1/2 base x height. Multiply by 2 since the dog can create an identicle triangle going in the other direction.

23) 5 per face. So 60 interior diagonals total.
Just look at an image of a dodecahedron and count.

28) 5.84
Area of a triangle is 1/2 base x height. Therefor, the height must equal 4. Divide into two right triangles. The one on the left will have have lengths of 7, 4 and y. Solve for y. Subtract y from 10. You now know that the other right triangle has lengths of 4, (10-y), and z. Solve for z and that's your answer.

29) R/2
We were both fairly certain that's the correct answer, but not sure how to prove it. I think it has something to do with a formula about arcs in triangles.

[Saber Knight]

THANK YOU! You are amazing!

Though....I hate to say this, but I really need more than just the answer. I have to type up a whole thing on how the problem was done, just the answer isn't good enough (though it does get you partial credit, which is always good).
The grading rubric is (for each problem):

3: The answer is correct with a very complete explanation documenting the rationale for the answer.
2: Answer is only slightly incorrect; reasoning is strong or answer is correct, but reasoning is weak.
1: Incorrect answer; some good documentation.
0: Nothing provided or an unacceptable attempt made.

Um, so yeah. You don't have to put every single detail, just some breif stuff on how you did it (though #11 still confuses me a lot, even after my friend tried to explain it to me (in the span of about 4 minutes, cuz I had to leave)) and with that I might be able to figure out the rest. Thank you so much for your help already!

[Siranae]

Wow, I remember doing this stuff, but I can't remember how to do it! Man, it makes me wonder if all that time in Geometry was wasted. I do know that some of these things were mentioned on note cards. Maybe they're in the basement somewhere...
I'll look. Keep going Saber!

[Josiah]

I corrected(?) one of the answers and added explantions for the others.

[Saber Knight]

Thank you so much!

This is really helping me not freak out too much about the project. I may stay sane yet (Silly Saber, you went insane long ago! I said shaddup!)