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Thread: I proved my teacher wrong, please look.

  1. #1

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    Hyperactive Member DerekM's Avatar
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    I proved my teacher wrong, please look.

    Alright, I just want some back-up on this matter, just to know I am not crazy. I have already done my math correctly, and I know I am correct, but I just want to see if any of you will also see reason to what I saw. I will tell you a little about myself so that you can understand my story. I am currently in high school in America (grade 10), I have an IQ above 127, and most of the stuff my mind does I have no clue how it does it. In my Geometry class today, I encountered one of those moments where my mind seems to process things instantly without me even thinking about it and made me realize that something was not right about a math problem we were given. My teacher drew a parallelogram on the board, gave us the measurement of each side, and the height (I must remind you that this parallelogram did not have any 90 degree angles). On the top side of the parallelogram, there was a triangle, this triangle at the top formed a 90 degree angle. This triangle's hypotenuse was the parallelogram's top side, and my teacher wanted us to find the length of one of the sides of the triangle (but not the hypotenuse). There are multiple ways you could find the answer to this problem, but the thing is, with the measurements she gave us, it made it impossible for the triangle to have a 90 degree angle. You may be confused at this point, and that is why I will provide several diagrams. I will list the different mathematical equations I used to disprove my teacher, and how it is correct.
    Equations:
    Code:
    
    Distance between two coordinate points on a grid = Square root((X2-X1)2 + (Y2-Y1)2)
    The Pythagorean theorem: A2 + B2 = C2 (Leg2 + Leg2 = Hypotenuse2)
    Inverse Pythagorean Theorem: Hypotenuse2 - Leg2 = Leg2
    
    I will also be using common math such as subtraction, and also some of the rules of geometry.
    Here is my shape:

    ***Please note that I only added the letters on the angle's to make this easier to explain***
    Here is what the prompt was:
    Find the measure of line segment EC.
    We were also given that angle E was 90 degrees. According to my math, it would be impossible for angle E to equal 90 degrees. If you are confused, here is my reason.

    Using the Inverse Pythagorean theorem on the triangle to find the measure of line segment AF (102 - 82 = ?2) You get the measurement of 6.
    62 = 36
    82 = 64
    of course, 102 = 100
    36 + 64 = *100*
    We can now infer that the rectangle within the parallelogram is a perfect square:
    14 - 6 = 8
    Base = 8
    height = 8

    Visual:

    In the above image, the pink square represents the perfect square left over on the parallelogram if the triangles are taken off. (Sorry, I had forgotten to add in the measurements)
    Code:
    
    In a perfect square, the diagonals form two 45 degree angles when they split the square in half.(I'm talking about a single diagonal)
    So, if you make a line from D to B, you will then notice that you can form a triangle. This newly formed triangle is congruent to triangle DCE: ADB is congruent to DCE
    
    Here's another picture:

    Here's another:

    And the final of this set!:

    The last image I will include so some of you may get an idea of the shape:

    (The large black dot represents the origin of the grid, also known as (0,0))
    So, now, the proof that it can't be 90.
    If you take the 45 degrees from and FDB, and then you use some common knowledge and think "Well, if 90 - 45 = 45, and angle FAD couldn't be 45 degrees because the triangle ADF would have to be congruent to triangle FDB, which it's not." So, there you have it, angle ADB is not equal to 90, since triangle DCE is congruent to triangle ABD, then angle E could not possibly be 90 degrees. My teacher had the wrong measurements.
    Since the problem was to find the measure of the line segment I have labeled to be EC, the answer is the same as the measure of segment DB. We can find the measure of DB with the distance formula. I'll tell you, I already did the math, the distance is 11.3 rounded off:
    sqrt((0-8)2 + (8 - 0)2)
    The answer the teacher gave us to the problem (which was amazingly incorrect) was actually 11.2 (which was also rounded off). So, in conclusion. The teachers of America are not very good at what they do because obviously, they can't even find the correct measurement of a line! Please tell me what you think.

  2. #2
    Only Slightly Obsessive jemidiah's Avatar
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    Re: I proved my teacher wrong, please look.

    I believe the teacher is right. You've assumed that ADB and DEC are congruent, which is the same as assuming DB and EC are parallel, which is also the same as assuming that AD and DE have the same length. Other than the image looking a bit suggestive, I don't see any reason for assuming this. If you allow the length of DE to vary, you can indeed get a right angle on the triangle while keeping the height, width, and altitude of the parallelogram at the given values. My calculation shows the length of EC to be 11.2 (exactly) in this case.

    I agree a little that "the teachers of America are not very good at what they do" if you're talking about high school. However, generalizing this example that far is a logical fallacy, which is ironically what you're accusing other people of. (I'd also say that a *ton* is asked of teachers in general, often for terrible salaries.) I hope you plan on going to college and continuing your education; you would probably find a challenging school to be a great environment.
    The time you enjoy wasting is not wasted time.
    Bertrand Russell

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  3. #3

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    Hyperactive Member DerekM's Avatar
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    Re: I proved my teacher wrong, please look.

    Quote Originally Posted by jemidiah View Post
    I believe the teacher is right. You've assumed that ADB and DEC are congruent, which is the same as assuming DB and EC are parallel, which is also the same as assuming that AD and DE have the same length. Other than the image looking a bit suggestive, I don't see any reason for assuming this. If you allow the length of DE to vary, you can indeed get a right angle on the triangle while keeping the height, width, and altitude of the parallelogram at the given values. My calculation shows the length of EC to be 11.2 (exactly) in this case.

    I agree a little that "the teachers of America are not very good at what they do" if you're talking about high school. However, generalizing this example that far is a logical fallacy, which is ironically what you're accusing other people of. (I'd also say that a *ton* is asked of teachers in general, often for terrible salaries.) I hope you plan on going to college and continuing your education; you would probably find a challenging school to be a great environment.
    Yeah, I didn't realize my grave mistake until I had a dream about it, but I don't care right now.
    Last edited by Hack; Feb 6th, 2010 at 08:07 AM. Reason: Cleaned up some comments.

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    Re: I proved my teacher wrong, please look.

    ha,ha. I think your a bit full of yourself arent you

  5. #5
    Junior Member Shadow-GK's Avatar
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    Re: I proved my teacher wrong, please look.

    yeah, you did the problem backwards. you determined that its not 90 by saying they are congruent, where you had to use 90 to find if they are congruent and then eventually find the sides you need. well, that happens. that's why on the diagrams it is usually said "NOT DRAWN TO SCALE!"

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