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Old 09-09-2012, 06:37 AM   #1
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2 Math Questions.

I know the answers, but I don't know how to get to them.

1. What is the perimeter of the six sided figure?
http://www.abload.de/img/unbenanntfxj6q.png

I know that the missing sides somehow must add up to 12 and 10. But why?

2. In the figure angle ABC has measure 140. What is the value of r+s+t?
http://www.abload.de/img/unbenannt22vkh8.png

The three triangles have B in common. And I somehow need a system of equations to solve this, but I keep getting 150 and it has to be 130.

Thx in advance, guys.

Last edited by DuffMan1991; 09-09-2012 at 06:47 AM.
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Old 09-09-2012, 06:49 AM   #2
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1. Use the standard strategy "move the sides of the figure around" to form a 10 by 12 rectangle which has perimeter 2(10) + 2(12) = 20 + 24 = 44.

Note: When you move the sides of a figure the perimeter is unchanged. Be careful though - the area may change!
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Old 09-09-2012, 06:52 AM   #3
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2. Geometric solution: If we label the bottom 2 vertices D and E, we have r = 90 - ABD, s = 90 - DBE, and t = 90 - EBC. So r+s+t = 270 - (ABD+DBE+EBC) = 270-140=130.

Solution by picking numbers: Let ABD=60, DBE=40, and EBC=40 (note that they add to 140). Then, using the fact that a triangle has 180 degrees and each triangle is a right triangle, r=30, s=50, and t=50. So r+s+t=130.
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Old 09-09-2012, 06:59 AM   #4
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1) The perimeter of this shape equals the perimeter of the retangular 10x12, so it is: (10+12)x2 =44
2) Take the three small angles of ABC is B1, B2, B3
So: B1 + B2 + B3 =140
(90-r)+(90-s)+(90-t) =140
Then we have r+s+t=180-140=40
Hope this help :3
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Old 09-11-2012, 08:22 AM   #5
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Thx for the explanations so far. I have another one for you. I somehow fail to process the highlighted phrasing.

Edit: I think I got it. Do they mean the midpoints of the triangle sides?

http://www.abload.de/img/unbenanntdffffkeel7.png

Last edited by DuffMan1991; 09-11-2012 at 08:29 AM.
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Old 09-11-2012, 09:50 AM   #6
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OK here's a last one:
For all x, let the function f be defined by f(x)=a(x-h)^2+k, where a, h and k are constants. If a and k are positive, which of the following CANNOT be true?
A) f(10) = 1
B) f(0) = -5
C) f(0) = 5
D) f(1) = -h
E) f(-1) = h
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Old 09-11-2012, 10:29 AM   #7
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^^

I wasn't sure for a second either, but I'm guessing the fact that only one of the answers isn't using the midpoints of the triangle edges rules, that one out.

I would say it was C, although I find the wording confusing as well. They say "drawn on the triangular faces", which to me says that they don't go through the triangle (so it rules out the cross answers) and the points are on the edges, so it's not A.

I could be completely wrong though.
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Old 09-11-2012, 10:34 AM   #8
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I skimmed that last question, but I would say B.

If A is positive and K is positive, the answer must be positive. (x-h)^2 will always be positive and so will +K

-h and h can be true, as you have no idea about what h is.

Once again, I'm not expert and I'm not putting too much thought in to these, I would gladly be corrected.
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Old 09-11-2012, 11:08 AM   #9
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@Duffman1991, it's B. All the other choices *could* be true, but B cannot be true since a(x-h)^2 >= 0, and k > 0.
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Old 09-21-2012, 05:41 AM   #10
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Here's another tricky one for you:

Each student in a group of 30 students studies German, Italian, or both. The total number of students studying German is three more than the total number of students studying Italian. If the number of students that study both subjects is the same as the number of students that study exactly one subject, how many students in the group study only Italian?
A: 6
B: 9
C: 15
D: 21
E: 24

The answer is 6.
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Old 09-21-2012, 06:26 AM   #11
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Solution by starting with choice (C): If we take (C) as our first guess, then 15 students study only Italian, and 18 study only German, so 15+18=33 study both. This is already way too big. So let's try (A) next. Then 6 study Italian and 9 study German. So 6+9=15 study both. All together we have 30 so that (A) is the answer.

Quickest Solution: Since there are 30 students, and the number that study both subjects equals the number that study only one subject, there are 15 students that study only one subject. So we want to split up 15 as a sum of two numbers such that one is 3 more than the other - we have 9 + 6 = 15 (we can get this using "informal" or "formal" algebra). So 6 students study Italian only, choice (A).

Notes: (1) Here is the "formal" algebra in case you need it: Let x be the number of students that study Italian only. Then the number of students that study German only is x+3. So x + (x+3) = 15. Thus, 2x + 3 = 15. So 2x = 12, and x = 6.

(2) To do this with a Venn diagram, you would draw 2 overlapping circles labelling the first G and the second I. The intersection gets 15 right away, and then we need to put a total of 15 in the other two regions such that region G has 3 more than region I (by G, I actually mean "G only," and similarly for I) . This should clearly be 9 and 6, respectively.

(3) To solve this algebraically, we can let x be the number of students that study German only, y the number of students that study both, and z the number of students that study Italian only. Then we get the system of equations:

x+y+z=30
x+z=y
x+y=y+z+3

This system simplifies to:

x+y+z=30
x-y+z=0
x-z=3

You can solve this system pretty quickly by hand using the elimination method, or you can do Gauss Jordan Reduction pretty quickly in your graphing calculator by inputting a matrix and using the rref( feature.
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Old 09-21-2012, 09:53 AM   #12
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@Duffman1991,

Suppose that k students study only Italian. It follows that k+3 students study only German*, so therefore 2k+3 students study one subject, and 2k+3 students study both. Therefore 2(2k+3) = 30 --> k = 6. So six students study Italian, A.

*It might look like I'm confusing the initial condition with the "total number of students studying..." but it's correct because I just subtracted the size of the intersection of the sets. e.g. if k students study only Italian and m students study both, then k+m study Italian and k+m+3 study German, subtracting m leaves k+3 studying only German.
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Old 09-26-2012, 03:23 PM   #13
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Thank you very much for your help! Here's another one. Either CB is wrong here, or, more likely, I'm missing something very simple which I'll bang my head in the wall about after you enlighten me.
http://www.abload.de/img/unbenannthhl28.png

This is how I went about it:
The perimeter of ABC is 18. So one side of the large ABC triangle: 18/3=6
Then for the ratio, we need the simple equation: 6x3=4X. So the longer part of the ratio is 4,5. Smilarly, the shorter part of the ratio is 1,5. Then, we use the Pythagorean theorem to find one side of the enclosed triangle: 4,5^2=1,5^2+X^2.
Result:3sqrt2.
Then 3(3sqrt2)= 9sqrt2. CB says that the answer is B, 6sqrt3.

Last edited by DuffMan1991; 09-26-2012 at 03:32 PM.
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Old 09-26-2012, 03:45 PM   #14
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I think you made some calculation mistake...sqrt(2) doesn't go well with equilateral triangles.

Clearly, AB = 6 and BD = 2. Because BDE is a 30-60-90 triangle, it follows that DE = 2sqrt(3), so the perimeter of triangle DEF is 3(2sqrt(3)) = 6sqrt(3), B.
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Old 09-26-2012, 04:57 PM   #15
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ok thanks.
Apparently, I calculated the ratio part incorrectly. -.- I hate these small mistakes.
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