Decide whether each triangle below lies in the Euclidean plane, on a sphere (of radius 1), or in the hyperbolic plane

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A triangle has sides of length $a$, $b$, $c$ (given correct to $2$dp) and a right angle opposite $c$. Decide whether each triangle below lies in the Euclidean plane, on a sphere (of radius $1$), or in the hyperbolic plane



  1. $a=b=1.00, c=1.27$

  2. $a=0.50, b=1.20, c=1.34$

  3. $a=1.0, b=2.40, c=2.60$

  4. $a=b=3.00, c=0.20$

In each case, answer "Euclidean", "spherical" or "hyperbolic".



How to do this without calculator? I have an exam tomorrow please help.







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    up vote
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    A triangle has sides of length $a$, $b$, $c$ (given correct to $2$dp) and a right angle opposite $c$. Decide whether each triangle below lies in the Euclidean plane, on a sphere (of radius $1$), or in the hyperbolic plane



    1. $a=b=1.00, c=1.27$

    2. $a=0.50, b=1.20, c=1.34$

    3. $a=1.0, b=2.40, c=2.60$

    4. $a=b=3.00, c=0.20$

    In each case, answer "Euclidean", "spherical" or "hyperbolic".



    How to do this without calculator? I have an exam tomorrow please help.







    share|cite|improve this question























      up vote
      -1
      down vote

      favorite









      up vote
      -1
      down vote

      favorite











      A triangle has sides of length $a$, $b$, $c$ (given correct to $2$dp) and a right angle opposite $c$. Decide whether each triangle below lies in the Euclidean plane, on a sphere (of radius $1$), or in the hyperbolic plane



      1. $a=b=1.00, c=1.27$

      2. $a=0.50, b=1.20, c=1.34$

      3. $a=1.0, b=2.40, c=2.60$

      4. $a=b=3.00, c=0.20$

      In each case, answer "Euclidean", "spherical" or "hyperbolic".



      How to do this without calculator? I have an exam tomorrow please help.







      share|cite|improve this question













      A triangle has sides of length $a$, $b$, $c$ (given correct to $2$dp) and a right angle opposite $c$. Decide whether each triangle below lies in the Euclidean plane, on a sphere (of radius $1$), or in the hyperbolic plane



      1. $a=b=1.00, c=1.27$

      2. $a=0.50, b=1.20, c=1.34$

      3. $a=1.0, b=2.40, c=2.60$

      4. $a=b=3.00, c=0.20$

      In each case, answer "Euclidean", "spherical" or "hyperbolic".



      How to do this without calculator? I have an exam tomorrow please help.









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      edited Aug 7 at 18:12
























      asked Aug 7 at 17:46









      mathsnerd22

      12




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          The Pythagorean theorem says $a^2+b^2=c^2$ for the Euclidean case, the left hand side would be larger than the right one in spherical geomatry and smaller for the hyperbolic one. Thus you would have to calculate simply $sqrta^2+b^2$ and compare that to $c$.



          That root would become $1.41$, $ 1.3$, $ 2.6$, $ 4.23$ respectively.



          You even could do all this without calculator. Because of $sqrt2=1.41$ respectively $13^2=169$. The remainder is just a multiplication by $2$ respectively $3$.



          --- rk






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          • ah thanks that makes sense! :)
            – mathsnerd22
            Aug 7 at 19:06










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          up vote
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          The Pythagorean theorem says $a^2+b^2=c^2$ for the Euclidean case, the left hand side would be larger than the right one in spherical geomatry and smaller for the hyperbolic one. Thus you would have to calculate simply $sqrta^2+b^2$ and compare that to $c$.



          That root would become $1.41$, $ 1.3$, $ 2.6$, $ 4.23$ respectively.



          You even could do all this without calculator. Because of $sqrt2=1.41$ respectively $13^2=169$. The remainder is just a multiplication by $2$ respectively $3$.



          --- rk






          share|cite|improve this answer























          • ah thanks that makes sense! :)
            – mathsnerd22
            Aug 7 at 19:06














          up vote
          0
          down vote













          The Pythagorean theorem says $a^2+b^2=c^2$ for the Euclidean case, the left hand side would be larger than the right one in spherical geomatry and smaller for the hyperbolic one. Thus you would have to calculate simply $sqrta^2+b^2$ and compare that to $c$.



          That root would become $1.41$, $ 1.3$, $ 2.6$, $ 4.23$ respectively.



          You even could do all this without calculator. Because of $sqrt2=1.41$ respectively $13^2=169$. The remainder is just a multiplication by $2$ respectively $3$.



          --- rk






          share|cite|improve this answer























          • ah thanks that makes sense! :)
            – mathsnerd22
            Aug 7 at 19:06












          up vote
          0
          down vote










          up vote
          0
          down vote









          The Pythagorean theorem says $a^2+b^2=c^2$ for the Euclidean case, the left hand side would be larger than the right one in spherical geomatry and smaller for the hyperbolic one. Thus you would have to calculate simply $sqrta^2+b^2$ and compare that to $c$.



          That root would become $1.41$, $ 1.3$, $ 2.6$, $ 4.23$ respectively.



          You even could do all this without calculator. Because of $sqrt2=1.41$ respectively $13^2=169$. The remainder is just a multiplication by $2$ respectively $3$.



          --- rk






          share|cite|improve this answer















          The Pythagorean theorem says $a^2+b^2=c^2$ for the Euclidean case, the left hand side would be larger than the right one in spherical geomatry and smaller for the hyperbolic one. Thus you would have to calculate simply $sqrta^2+b^2$ and compare that to $c$.



          That root would become $1.41$, $ 1.3$, $ 2.6$, $ 4.23$ respectively.



          You even could do all this without calculator. Because of $sqrt2=1.41$ respectively $13^2=169$. The remainder is just a multiplication by $2$ respectively $3$.



          --- rk







          share|cite|improve this answer















          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 7 at 18:52


























          answered Aug 7 at 18:45









          Dr. Richard Klitzing

          7566




          7566











          • ah thanks that makes sense! :)
            – mathsnerd22
            Aug 7 at 19:06
















          • ah thanks that makes sense! :)
            – mathsnerd22
            Aug 7 at 19:06















          ah thanks that makes sense! :)
          – mathsnerd22
          Aug 7 at 19:06




          ah thanks that makes sense! :)
          – mathsnerd22
          Aug 7 at 19:06












           

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