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mathmath333

  • one year ago

Prove

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} &\text{Prove that area of the figure}\hspace{.33em}\\~\\ &|x|+|y|=k ,\ k>0,\ k\in \mathbb{R}\hspace{.33em}\\~\\ &\text{is same as area of figure }\hspace{.33em}\\~\\ &|x+a|+|y+b|=k,\ \{a,b\}\in \mathbb{R}\hspace{.33em}\\~\\ \end{align}}\)

  2. ParthKohli
    • one year ago
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    Almost the same as the previous one - try to figure out the four lines.

  3. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} x+y=k\hspace{1.5em}\\~\\ x-y=k\hspace{1.5em}\\~\\ -x+y=k\hspace{1.5em}\\~\\ -x-y=k\hspace{1.5em}\\~\\ \end{align}}\)

  4. ParthKohli
    • one year ago
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    Yeah. Now graph them and see what you get.

  5. mathmath333
    • one year ago
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    square with area \(k^2\)

  6. ParthKohli
    • one year ago
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    Same process for the second case.

  7. mathmath333
    • one year ago
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    what about 'a' and 'b'

  8. ParthKohli
    • one year ago
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    don't worry about them. figure out the equations first.

  9. Here_to_Help15
    • one year ago
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    ! :D

  10. ParthKohli
    • one year ago
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    looks like the area is not \(k^2\) though.

  11. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} x+a-y-b=k\hspace{1.5em}\\~\\ x+a+y+b=k\hspace{1.5em}\\~\\ -x-a+y+b=k\hspace{1.5em}\\~\\ -x-a-y-b=k\hspace{1.5em}\\~\\ \end{align}}\)

  12. ParthKohli
    • one year ago
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    the diagonal of the square in the first case is \(2k\), so the side length is \(k\sqrt 2\)

  13. ParthKohli
    • one year ago
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    yeah, can you manage the rest now?

  14. mathmath333
    • one year ago
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    area of the first is \(2k^2\) ?

  15. ParthKohli
    • one year ago
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    that's correct.

  16. mathmath333
    • one year ago
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    and how to find 2nd one

  17. ParthKohli
    • one year ago
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    just find out the points where the lines meet and those would be the vertices of the square.

  18. Michele_Laino
    • one year ago
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    hint: if we make this traslation: \[\large \left\{ \begin{gathered} x + a = X \hfill \\ y + b = Y \hfill \\ \end{gathered} \right.\] where X,Y are the new coordinates, then the second equation can be rewritten as follows: \[\large \left| X \right| + \left| Y \right| = k\] and such equation has the same shape of the first one

  19. mathmath333
    • one year ago
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    would this be enough for proof

  20. Michele_Laino
    • one year ago
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    Yes I think so, since we have applied a traslation, and not a dilation, namely when we traslate a geometrical shape its area will be unchanged

  21. mathmath333
    • one year ago
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    ok thnx both

  22. Michele_Laino
    • one year ago
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    :)

  23. ParthKohli
    • one year ago
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    That is actually a nice way to see it, but it seems that he actually has to *prove* that translation has no effect on the shape, and in turn, has no effect on area.

  24. Michele_Laino
    • one year ago
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    I think that a proof of my statement above, can be this: when we traslate a rigid body into the euclidean space, its volume is unchanged, now we can imagine to build the geometrical shape involved into our exercise using rigid rods, so we can conclude that the area enclosed by our geometrical shape will be unchanged after a traslation

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