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







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.
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  • 2




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










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










  • $begingroup$
    In mathematics, there is nothing rude and pretentious about asking questions
    $endgroup$
    – Taladris
    28 mins 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
    7 hours ago










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










  • $begingroup$
    In mathematics, there is nothing rude and pretentious about asking questions
    $endgroup$
    – Taladris
    28 mins 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 2 hours 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 7 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
    7 hours ago










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










  • $begingroup$
    In mathematics, there is nothing rude and pretentious about asking questions
    $endgroup$
    – Taladris
    28 mins ago














  • 2




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










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










  • $begingroup$
    In mathematics, there is nothing rude and pretentious about asking questions
    $endgroup$
    – Taladris
    28 mins ago








2




2




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




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












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




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












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




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










4 Answers
4






active

oldest

votes


















7












$begingroup$

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











$endgroup$





















    4












    $begingroup$

    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









      $endgroup$





















        0












        $begingroup$

        Hint:



        Drop a $perp$ to $F$ 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 $text{base}cdottext{height}$.






        share|cite









        $endgroup$













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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          7












          $begingroup$

          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











          $endgroup$


















            7












            $begingroup$

            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











            $endgroup$
















              7












              7








              7





              $begingroup$

              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 6 hours ago

























              answered 7 hours ago









              David G. StorkDavid G. Stork

              11.3k41432




              11.3k41432























                  4












                  $begingroup$

                  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









                  $endgroup$


















                    4












                    $begingroup$

                    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









                    $endgroup$
















                      4












                      4








                      4





                      $begingroup$

                      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 7 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 7 hours ago









                              Cameron BuieCameron Buie

                              86.2k772161




                              86.2k772161























                                  0












                                  $begingroup$

                                  Hint:



                                  Drop a $perp$ to $F$ 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 $text{base}cdottext{height}$.






                                  share|cite









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    Hint:



                                    Drop a $perp$ to $F$ 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 $text{base}cdottext{height}$.






                                    share|cite









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      Hint:



                                      Drop a $perp$ to $F$ 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 $text{base}cdottext{height}$.






                                      share|cite









                                      $endgroup$



                                      Hint:



                                      Drop a $perp$ to $F$ 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 $text{base}cdottext{height}$.







                                      share|cite












                                      share|cite



                                      share|cite










                                      answered 1 min ago









                                      Paras KhoslaParas Khosla

                                      2,643323




                                      2,643323






















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