Confusion on Parallelogram












1












$begingroup$


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






geometry







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  • 2




    $begingroup$
    Possible duplicate of Visualizing the Area of a Parallelogram
    $endgroup$
    – kkc
    6 hours ago










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
















1












$begingroup$


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






geometry







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.







$endgroup$








  • 2




    $begingroup$
    Possible duplicate of Visualizing the Area of a Parallelogram
    $endgroup$
    – kkc
    6 hours ago










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














1












1








1





$begingroup$


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






geometry







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.







$endgroup$




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






geometry




geometry






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









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




share|cite|improve this question








edited 1 hour ago









dantopa

6,63342245




6,63342245






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.









asked 6 hours ago









Bo HalimBo Halim

1094




1094




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.





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.








  • 2




    $begingroup$
    Possible duplicate of Visualizing the Area of a Parallelogram
    $endgroup$
    – kkc
    6 hours ago










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














  • 2




    $begingroup$
    Possible duplicate of Visualizing the Area of a Parallelogram
    $endgroup$
    – kkc
    6 hours ago










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








2




2




$begingroup$
Possible duplicate of Visualizing the Area of a Parallelogram
$endgroup$
– kkc
6 hours ago




$begingroup$
Possible duplicate of Visualizing the Area of a Parallelogram
$endgroup$
– kkc
6 hours ago












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




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










3 Answers
3






active

oldest

votes


















6












$begingroup$

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











$endgroup$





















    3












    $begingroup$

    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









    $endgroup$





















      1












      $begingroup$

      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









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        6












        $begingroup$

        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











        $endgroup$


















          6












          $begingroup$

          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











          $endgroup$
















            6












            6








            6





            $begingroup$

            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 5 hours ago

























            answered 6 hours ago









            David G. StorkDavid G. Stork

            11.3k41432




            11.3k41432























                3












                $begingroup$

                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









                $endgroup$


















                  3












                  $begingroup$

                  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









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    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 6 hours ago









                    Ethan BolkerEthan Bolker

                    45.3k553120




                    45.3k553120























                        1












                        $begingroup$

                        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












                          $begingroup$

                          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












                            1








                            1





                            $begingroup$

                            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 6 hours ago









                            Cameron BuieCameron Buie

                            86.2k772161




                            86.2k772161






















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