Is the Pythagorean Theorem Applied to Determine the Length of AB?

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Discussion Overview

The discussion revolves around determining the length of segment AB using the Pythagorean Theorem and related mathematical concepts. Participants explore various methods, including the distance formula and geometric reasoning, while addressing a specific geometric configuration.

Discussion Character

  • Homework-related, Mathematical reasoning, Debate/contested

Main Points Raised

  • Some participants propose that the length of AB is 6, but they do not provide detailed reasoning initially.
  • One participant suggests using the distance formula to find AB, indicating that there are "pleasant cancellations" in the calculations.
  • Another participant calculates the hypotenuse of a triangle with legs 4 + sqrt(2) and 4 - sqrt(2) but arrives at sqrt(34).
  • There is a discussion about the legs of triangles formed by splitting a square, with one participant asserting that each leg would be sqrt(2), which leads to a correction of an earlier mistake.
  • A participant provides a detailed calculation involving coordinates and the distance formula, arriving at a length of 6, while acknowledging the difficulty of typing mathematical notation.
  • One participant questions whether the figures with sides 2 and 4 are squares, which could affect the calculations.

Areas of Agreement / Disagreement

Participants express differing views on the length of AB, with some asserting it is 6 and others calculating it as sqrt(34). The discussion remains unresolved regarding the correct length and the assumptions about the geometric figures involved.

Contextual Notes

There are limitations in the clarity of the geometric configuration and the assumptions about the shapes involved, which may affect the calculations presented.

Ismael Nunez
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Anyone want to take a crack at it? My class has been discussing it: Find the length of AB:http://t4.rbxcdn.com/84e25f3830d66e6bbaeaba48e35c0781
 
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Six.
 
Misha Kuznetsov said:
Six.
Will you please elaborate?
 
6?
 
Jerry Friedman said:
6?
Ok, great, but how did you get there?
 
I am probably wrong, but I got sqrt(34).
 
Find the coordinates of point A and use the distance formula to get AB. There are pleasant cancellations.
 
I just found the hypotenuse of a triangle with legs 4+sqrt(2) and 4-sqrt(2).
 
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Jerry Friedman said:
Find the coordinates of point A and use the distance formula to get AB. There are pleasant cancellations.
I did, and I still got sqrt(34)... One question though, the square at the top left corner... If split vertically, to get 4 triangles, wouldn't the legs of one of those triangles be one?
 
  • #10
No, each leg would be sqrt(2) .
 
  • #11
Misha Kuznetsov said:
No, each leg would be sqrt(2) .
Alright, I see my mistake, I checked my work again. Thanks.
 
  • #12
Yep, no problem. :smile:
 
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  • #13
First, we choose the axes so as to simplify our calculation. Let us drop a perpendicular from A to the base and consider the the y-axis.

The location of A is (0, 4 +√²) and B is (4 - √2, 0)
Distance is √x²+y²

As we know, (a - b)² + (a + b)² = 2( a² + b²)

Which here is 2( 16 +2) = 36, the square root of it is 6.

I'm sorry if I skipped some steps. It's very difficult to type in mathematical notation.
 
  • #14
Was it given that the figures with sides 2 and 4 are squares?

If so, 6.
 
Last edited:

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