Confusion on Parallelogram [duplicate]












2












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This question already has an answer here:




  • Visualizing the Area of a Parallelogram

    2 answers




i know it's rude and pretentious and maybe stupid to ask such question but, i would like to know the reason why in geometry the area of a parallelogram follows this formula :



$ Base times Height $



and not this one :



$ Base times Side $



I saw this method which consists of transforming our figure into a rectangle by moving the black one to the right, and that seems logical to me, but why not adopt the second rule, all the horizontal points multiplied by the vertical ones ?



enter image description here



PS : I found the same Visualizing the Area of a Parallelogram, but was unconvinced by the answers










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marked as duplicate by mrtaurho, Cesareo, user21820, Xander Henderson, José Carlos Santos yesterday


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    @kkc, did you see the link ?
    $endgroup$
    – Bo Halim
    2 days ago






  • 4




    $begingroup$
    In mathematics, there is nothing rude and pretentious about asking questions
    $endgroup$
    – Taladris
    2 days ago










  • $begingroup$
    @Taladris thank you :)
    $endgroup$
    – Bo Halim
    yesterday
















2












$begingroup$



This question already has an answer here:




  • Visualizing the Area of a Parallelogram

    2 answers




i know it's rude and pretentious and maybe stupid to ask such question but, i would like to know the reason why in geometry the area of a parallelogram follows this formula :



$ Base times Height $



and not this one :



$ Base times Side $



I saw this method which consists of transforming our figure into a rectangle by moving the black one to the right, and that seems logical to me, but why not adopt the second rule, all the horizontal points multiplied by the vertical ones ?



enter image description here



PS : I found the same Visualizing the Area of a Parallelogram, but was unconvinced by the answers










share|cite|improve this question









New contributor




Bo Halim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







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marked as duplicate by mrtaurho, Cesareo, user21820, Xander Henderson, José Carlos Santos yesterday


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    @kkc, did you see the link ?
    $endgroup$
    – Bo Halim
    2 days ago






  • 4




    $begingroup$
    In mathematics, there is nothing rude and pretentious about asking questions
    $endgroup$
    – Taladris
    2 days ago










  • $begingroup$
    @Taladris thank you :)
    $endgroup$
    – Bo Halim
    yesterday














2












2








2





$begingroup$



This question already has an answer here:




  • Visualizing the Area of a Parallelogram

    2 answers




i know it's rude and pretentious and maybe stupid to ask such question but, i would like to know the reason why in geometry the area of a parallelogram follows this formula :



$ Base times Height $



and not this one :



$ Base times Side $



I saw this method which consists of transforming our figure into a rectangle by moving the black one to the right, and that seems logical to me, but why not adopt the second rule, all the horizontal points multiplied by the vertical ones ?



enter image description here



PS : I found the same Visualizing the Area of a Parallelogram, but was unconvinced by the answers










share|cite|improve this question









New contributor




Bo Halim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$





This question already has an answer here:




  • Visualizing the Area of a Parallelogram

    2 answers




i know it's rude and pretentious and maybe stupid to ask such question but, i would like to know the reason why in geometry the area of a parallelogram follows this formula :



$ Base times Height $



and not this one :



$ Base times Side $



I saw this method which consists of transforming our figure into a rectangle by moving the black one to the right, and that seems logical to me, but why not adopt the second rule, all the horizontal points multiplied by the vertical ones ?



enter image description here



PS : I found the same Visualizing the Area of a Parallelogram, but was unconvinced by the answers





This question already has an answer here:




  • Visualizing the Area of a Parallelogram

    2 answers








geometry






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New contributor




Bo Halim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Bo Halim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




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edited 2 days ago









dantopa

6,63342245




6,63342245






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Bo Halim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 2 days ago









Bo HalimBo Halim

1195




1195




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Bo Halim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





Bo Halim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Bo Halim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




marked as duplicate by mrtaurho, Cesareo, user21820, Xander Henderson, José Carlos Santos yesterday


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by mrtaurho, Cesareo, user21820, Xander Henderson, José Carlos Santos yesterday


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • $begingroup$
    @kkc, did you see the link ?
    $endgroup$
    – Bo Halim
    2 days ago






  • 4




    $begingroup$
    In mathematics, there is nothing rude and pretentious about asking questions
    $endgroup$
    – Taladris
    2 days ago










  • $begingroup$
    @Taladris thank you :)
    $endgroup$
    – Bo Halim
    yesterday


















  • $begingroup$
    @kkc, did you see the link ?
    $endgroup$
    – Bo Halim
    2 days ago






  • 4




    $begingroup$
    In mathematics, there is nothing rude and pretentious about asking questions
    $endgroup$
    – Taladris
    2 days ago










  • $begingroup$
    @Taladris thank you :)
    $endgroup$
    – Bo Halim
    yesterday
















$begingroup$
@kkc, did you see the link ?
$endgroup$
– Bo Halim
2 days ago




$begingroup$
@kkc, did you see the link ?
$endgroup$
– Bo Halim
2 days ago




4




4




$begingroup$
In mathematics, there is nothing rude and pretentious about asking questions
$endgroup$
– Taladris
2 days ago




$begingroup$
In mathematics, there is nothing rude and pretentious about asking questions
$endgroup$
– Taladris
2 days ago












$begingroup$
@Taladris thank you :)
$endgroup$
– Bo Halim
yesterday




$begingroup$
@Taladris thank you :)
$endgroup$
– Bo Halim
yesterday










5 Answers
5






active

oldest

votes


















15












$begingroup$

Sometimes a figure is worth 1000 words:



parallelogram



Very long base and very long side and very small area.



Or...



...each of these parallelograms has the same base and side, but manifestly different areas:



enter image description here






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The best answer ever :)
    $endgroup$
    – Bo Halim
    yesterday






  • 1




    $begingroup$
    EVER? My goodness... thanks.
    $endgroup$
    – David G. Stork
    yesterday



















7












$begingroup$

You can't "adopt the second rule" because it gives the wrong answer. The parallelogram in your picture has all four sides of length $5$, There are lots of parallelograms like that, depending on the angle between adjacent sides. The largest is a square. If you make the acute angle smaller and smaller you find parallelograms that are nearly on a line, with tiny areas. But your formula would say they all have area $25$.






share|cite|improve this answer









$endgroup$





















    6












    $begingroup$

    You have to have some notion of what area must satisfy (otherwise, you could just say that area of everything is 0 and be done with it:). I think a principle with which you'll agree is that if one figure is a subset of another, it cannot have greater area.



    Well, in that case, imagine a world that looks like yours, where the area of any paralelogram is calculated according to formula you give---and look at the following picture:



    parallelogram inside rectangle



    As you see, by Bo's formula, the area of AECF would be 13*5=65, while the area of a containing rectangle (which is also a parallelogram) would be (13+3)*4=64<65. Contradiction.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      +1: This is an excellent refutation.
      $endgroup$
      – Cameron Buie
      yesterday



















    2












    $begingroup$

    As you point out, the first method is logical, as it uses a formula we already know to be true (the familiar area formula for a rectangle), and then transforms our figure without changing the area, so that we can apply the rule.



    To see why we can't simply multiply $5$ and $5$ to get the answer, draw the $5times 5$ square. Your parallelogram can't fill it by cut and paste. Alternatively, you can draw $25$ unit squares, and try to cut and paste them into the parallelogram. You'll only be able to make $20$ of them fit.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Hint:



      Drop a $perp$ to a point say $F$ from $C$, prove that $Delta AED cong Delta CFB$ and now consider the $text{ar}(parallel text{gm} ABCD)$ as the area of this rectangle because they're both equal and the area of the rectangle is very well known to be $text{base}cdottext{height}$.



      You may also want to see this fact as a consequence of the $2$-D analog of Cavalieri's Principle.






      share|cite|improve this answer











      $endgroup$




















        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        15












        $begingroup$

        Sometimes a figure is worth 1000 words:



        parallelogram



        Very long base and very long side and very small area.



        Or...



        ...each of these parallelograms has the same base and side, but manifestly different areas:



        enter image description here






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          The best answer ever :)
          $endgroup$
          – Bo Halim
          yesterday






        • 1




          $begingroup$
          EVER? My goodness... thanks.
          $endgroup$
          – David G. Stork
          yesterday
















        15












        $begingroup$

        Sometimes a figure is worth 1000 words:



        parallelogram



        Very long base and very long side and very small area.



        Or...



        ...each of these parallelograms has the same base and side, but manifestly different areas:



        enter image description here






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          The best answer ever :)
          $endgroup$
          – Bo Halim
          yesterday






        • 1




          $begingroup$
          EVER? My goodness... thanks.
          $endgroup$
          – David G. Stork
          yesterday














        15












        15








        15





        $begingroup$

        Sometimes a figure is worth 1000 words:



        parallelogram



        Very long base and very long side and very small area.



        Or...



        ...each of these parallelograms has the same base and side, but manifestly different areas:



        enter image description here






        share|cite|improve this answer











        $endgroup$



        Sometimes a figure is worth 1000 words:



        parallelogram



        Very long base and very long side and very small area.



        Or...



        ...each of these parallelograms has the same base and side, but manifestly different areas:



        enter image description here







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 days ago

























        answered 2 days ago









        David G. StorkDavid G. Stork

        11.5k41533




        11.5k41533












        • $begingroup$
          The best answer ever :)
          $endgroup$
          – Bo Halim
          yesterday






        • 1




          $begingroup$
          EVER? My goodness... thanks.
          $endgroup$
          – David G. Stork
          yesterday


















        • $begingroup$
          The best answer ever :)
          $endgroup$
          – Bo Halim
          yesterday






        • 1




          $begingroup$
          EVER? My goodness... thanks.
          $endgroup$
          – David G. Stork
          yesterday
















        $begingroup$
        The best answer ever :)
        $endgroup$
        – Bo Halim
        yesterday




        $begingroup$
        The best answer ever :)
        $endgroup$
        – Bo Halim
        yesterday




        1




        1




        $begingroup$
        EVER? My goodness... thanks.
        $endgroup$
        – David G. Stork
        yesterday




        $begingroup$
        EVER? My goodness... thanks.
        $endgroup$
        – David G. Stork
        yesterday











        7












        $begingroup$

        You can't "adopt the second rule" because it gives the wrong answer. The parallelogram in your picture has all four sides of length $5$, There are lots of parallelograms like that, depending on the angle between adjacent sides. The largest is a square. If you make the acute angle smaller and smaller you find parallelograms that are nearly on a line, with tiny areas. But your formula would say they all have area $25$.






        share|cite|improve this answer









        $endgroup$


















          7












          $begingroup$

          You can't "adopt the second rule" because it gives the wrong answer. The parallelogram in your picture has all four sides of length $5$, There are lots of parallelograms like that, depending on the angle between adjacent sides. The largest is a square. If you make the acute angle smaller and smaller you find parallelograms that are nearly on a line, with tiny areas. But your formula would say they all have area $25$.






          share|cite|improve this answer









          $endgroup$
















            7












            7








            7





            $begingroup$

            You can't "adopt the second rule" because it gives the wrong answer. The parallelogram in your picture has all four sides of length $5$, There are lots of parallelograms like that, depending on the angle between adjacent sides. The largest is a square. If you make the acute angle smaller and smaller you find parallelograms that are nearly on a line, with tiny areas. But your formula would say they all have area $25$.






            share|cite|improve this answer









            $endgroup$



            You can't "adopt the second rule" because it gives the wrong answer. The parallelogram in your picture has all four sides of length $5$, There are lots of parallelograms like that, depending on the angle between adjacent sides. The largest is a square. If you make the acute angle smaller and smaller you find parallelograms that are nearly on a line, with tiny areas. But your formula would say they all have area $25$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 days ago









            Ethan BolkerEthan Bolker

            45.4k553120




            45.4k553120























                6












                $begingroup$

                You have to have some notion of what area must satisfy (otherwise, you could just say that area of everything is 0 and be done with it:). I think a principle with which you'll agree is that if one figure is a subset of another, it cannot have greater area.



                Well, in that case, imagine a world that looks like yours, where the area of any paralelogram is calculated according to formula you give---and look at the following picture:



                parallelogram inside rectangle



                As you see, by Bo's formula, the area of AECF would be 13*5=65, while the area of a containing rectangle (which is also a parallelogram) would be (13+3)*4=64<65. Contradiction.






                share|cite|improve this answer











                $endgroup$









                • 1




                  $begingroup$
                  +1: This is an excellent refutation.
                  $endgroup$
                  – Cameron Buie
                  yesterday
















                6












                $begingroup$

                You have to have some notion of what area must satisfy (otherwise, you could just say that area of everything is 0 and be done with it:). I think a principle with which you'll agree is that if one figure is a subset of another, it cannot have greater area.



                Well, in that case, imagine a world that looks like yours, where the area of any paralelogram is calculated according to formula you give---and look at the following picture:



                parallelogram inside rectangle



                As you see, by Bo's formula, the area of AECF would be 13*5=65, while the area of a containing rectangle (which is also a parallelogram) would be (13+3)*4=64<65. Contradiction.






                share|cite|improve this answer











                $endgroup$









                • 1




                  $begingroup$
                  +1: This is an excellent refutation.
                  $endgroup$
                  – Cameron Buie
                  yesterday














                6












                6








                6





                $begingroup$

                You have to have some notion of what area must satisfy (otherwise, you could just say that area of everything is 0 and be done with it:). I think a principle with which you'll agree is that if one figure is a subset of another, it cannot have greater area.



                Well, in that case, imagine a world that looks like yours, where the area of any paralelogram is calculated according to formula you give---and look at the following picture:



                parallelogram inside rectangle



                As you see, by Bo's formula, the area of AECF would be 13*5=65, while the area of a containing rectangle (which is also a parallelogram) would be (13+3)*4=64<65. Contradiction.






                share|cite|improve this answer











                $endgroup$



                You have to have some notion of what area must satisfy (otherwise, you could just say that area of everything is 0 and be done with it:). I think a principle with which you'll agree is that if one figure is a subset of another, it cannot have greater area.



                Well, in that case, imagine a world that looks like yours, where the area of any paralelogram is calculated according to formula you give---and look at the following picture:



                parallelogram inside rectangle



                As you see, by Bo's formula, the area of AECF would be 13*5=65, while the area of a containing rectangle (which is also a parallelogram) would be (13+3)*4=64<65. Contradiction.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited yesterday

























                answered 2 days ago









                VekyVeky

                24019




                24019








                • 1




                  $begingroup$
                  +1: This is an excellent refutation.
                  $endgroup$
                  – Cameron Buie
                  yesterday














                • 1




                  $begingroup$
                  +1: This is an excellent refutation.
                  $endgroup$
                  – Cameron Buie
                  yesterday








                1




                1




                $begingroup$
                +1: This is an excellent refutation.
                $endgroup$
                – Cameron Buie
                yesterday




                $begingroup$
                +1: This is an excellent refutation.
                $endgroup$
                – Cameron Buie
                yesterday











                2












                $begingroup$

                As you point out, the first method is logical, as it uses a formula we already know to be true (the familiar area formula for a rectangle), and then transforms our figure without changing the area, so that we can apply the rule.



                To see why we can't simply multiply $5$ and $5$ to get the answer, draw the $5times 5$ square. Your parallelogram can't fill it by cut and paste. Alternatively, you can draw $25$ unit squares, and try to cut and paste them into the parallelogram. You'll only be able to make $20$ of them fit.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  As you point out, the first method is logical, as it uses a formula we already know to be true (the familiar area formula for a rectangle), and then transforms our figure without changing the area, so that we can apply the rule.



                  To see why we can't simply multiply $5$ and $5$ to get the answer, draw the $5times 5$ square. Your parallelogram can't fill it by cut and paste. Alternatively, you can draw $25$ unit squares, and try to cut and paste them into the parallelogram. You'll only be able to make $20$ of them fit.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    As you point out, the first method is logical, as it uses a formula we already know to be true (the familiar area formula for a rectangle), and then transforms our figure without changing the area, so that we can apply the rule.



                    To see why we can't simply multiply $5$ and $5$ to get the answer, draw the $5times 5$ square. Your parallelogram can't fill it by cut and paste. Alternatively, you can draw $25$ unit squares, and try to cut and paste them into the parallelogram. You'll only be able to make $20$ of them fit.






                    share|cite|improve this answer









                    $endgroup$



                    As you point out, the first method is logical, as it uses a formula we already know to be true (the familiar area formula for a rectangle), and then transforms our figure without changing the area, so that we can apply the rule.



                    To see why we can't simply multiply $5$ and $5$ to get the answer, draw the $5times 5$ square. Your parallelogram can't fill it by cut and paste. Alternatively, you can draw $25$ unit squares, and try to cut and paste them into the parallelogram. You'll only be able to make $20$ of them fit.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 days ago









                    Cameron BuieCameron Buie

                    86.2k772161




                    86.2k772161























                        1












                        $begingroup$

                        Hint:



                        Drop a $perp$ to a point say $F$ from $C$, prove that $Delta AED cong Delta CFB$ and now consider the $text{ar}(parallel text{gm} ABCD)$ as the area of this rectangle because they're both equal and the area of the rectangle is very well known to be $text{base}cdottext{height}$.



                        You may also want to see this fact as a consequence of the $2$-D analog of Cavalieri's Principle.






                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$

                          Hint:



                          Drop a $perp$ to a point say $F$ from $C$, prove that $Delta AED cong Delta CFB$ and now consider the $text{ar}(parallel text{gm} ABCD)$ as the area of this rectangle because they're both equal and the area of the rectangle is very well known to be $text{base}cdottext{height}$.



                          You may also want to see this fact as a consequence of the $2$-D analog of Cavalieri's Principle.






                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Hint:



                            Drop a $perp$ to a point say $F$ from $C$, prove that $Delta AED cong Delta CFB$ and now consider the $text{ar}(parallel text{gm} ABCD)$ as the area of this rectangle because they're both equal and the area of the rectangle is very well known to be $text{base}cdottext{height}$.



                            You may also want to see this fact as a consequence of the $2$-D analog of Cavalieri's Principle.






                            share|cite|improve this answer











                            $endgroup$



                            Hint:



                            Drop a $perp$ to a point say $F$ from $C$, prove that $Delta AED cong Delta CFB$ and now consider the $text{ar}(parallel text{gm} ABCD)$ as the area of this rectangle because they're both equal and the area of the rectangle is very well known to be $text{base}cdottext{height}$.



                            You may also want to see this fact as a consequence of the $2$-D analog of Cavalieri's Principle.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 2 days ago

























                            answered 2 days ago









                            Paras KhoslaParas Khosla

                            2,676323




                            2,676323















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