Confusion on Parallelogram
$begingroup$
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 ?
PS : I found the same Visualizing the Area of a Parallelogram, but was unconvinced by the answers
geometry
edited 59 mins ago
dantopa
6,63342245
asked 6 hours ago
Bo HalimBo Halim
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.
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add a comment |
$begingroup$
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 ?
PS : I found the same Visualizing the Area of a Parallelogram, but was unconvinced by the answers
geometry
edited 59 mins ago
dantopa
6,63342245
asked 6 hours ago
Bo HalimBo Halim
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.
$endgroup$
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
add a comment |
$begingroup$
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 ?
PS : I found the same Visualizing the Area of a Parallelogram, but was unconvinced by the answers
geometry
edited 59 mins ago
dantopa
6,63342245
asked 6 hours ago
Bo HalimBo Halim
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.
$endgroup$
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 ?
PS : I found the same Visualizing the Area of a Parallelogram, but was unconvinced by the answers
geometry
geometry
edited 59 mins ago
dantopa
6,63342245
asked 6 hours ago
Bo HalimBo Halim
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.
edited 59 mins ago
dantopa
6,63342245
asked 6 hours ago
Bo HalimBo Halim
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.
edited 59 mins ago
dantopa
6,63342245
edited 59 mins ago
dantopa
6,63342245
edited 59 mins ago
dantopa
6,63342245
6,63342245
asked 6 hours ago
Bo HalimBo Halim
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.
asked 6 hours ago
Bo HalimBo Halim
1094
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
add a comment |
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
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Sometimes a figure is worth 1000 words:
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:
answered 6 hours ago
David G. StorkDavid G. Stork
11.3k41432
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add a comment |
$begingroup$
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 $25$.
answered 6 hours ago
Ethan BolkerEthan Bolker
45.3k553120
$endgroup$
add a comment |
$begingroup$
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.
answered 6 hours ago
Cameron BuieCameron Buie
86.2k772161
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Sometimes a figure is worth 1000 words:
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:
answered 6 hours ago
David G. StorkDavid G. Stork
11.3k41432
$endgroup$
add a comment |
$begingroup$
Sometimes a figure is worth 1000 words:
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:
answered 6 hours ago
David G. StorkDavid G. Stork
11.3k41432
$endgroup$
add a comment |
$begingroup$
Sometimes a figure is worth 1000 words:
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:
answered 6 hours ago
David G. StorkDavid G. Stork
11.3k41432
$endgroup$
Sometimes a figure is worth 1000 words:
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:
answered 6 hours ago
David G. StorkDavid G. Stork
11.3k41432
edited 5 hours ago
answered 6 hours ago
David G. StorkDavid G. Stork
11.3k41432
answered 6 hours ago
David G. StorkDavid G. Stork
11.3k41432
answered 6 hours ago
David G. StorkDavid G. Stork
11.3k41432
11.3k41432
add a comment |
add a comment |
$begingroup$
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 $25$.
answered 6 hours ago
Ethan BolkerEthan Bolker
45.3k553120
$endgroup$
add a comment |
$begingroup$
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 $25$.
answered 6 hours ago
Ethan BolkerEthan Bolker
45.3k553120
$endgroup$
add a comment |
$begingroup$
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 $25$.
answered 6 hours ago
Ethan BolkerEthan Bolker
45.3k553120
$endgroup$
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 $25$.
answered 6 hours ago
Ethan BolkerEthan Bolker
45.3k553120
answered 6 hours ago
Ethan BolkerEthan Bolker
45.3k553120
answered 6 hours ago
Ethan BolkerEthan Bolker
45.3k553120
answered 6 hours ago
Ethan BolkerEthan Bolker
45.3k553120
45.3k553120
add a comment |
add a comment |
$begingroup$
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.
answered 6 hours ago
Cameron BuieCameron Buie
86.2k772161
$endgroup$
add a comment |
$begingroup$
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.
answered 6 hours ago
Cameron BuieCameron Buie
86.2k772161
$endgroup$
add a comment |
$begingroup$
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.
answered 6 hours ago
Cameron BuieCameron Buie
86.2k772161
$endgroup$
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.
answered 6 hours ago
Cameron BuieCameron Buie
86.2k772161
answered 6 hours ago
Cameron BuieCameron Buie
86.2k772161
answered 6 hours ago
Cameron BuieCameron Buie
86.2k772161
answered 6 hours ago
Cameron BuieCameron Buie
86.2k772161
86.2k772161
add a comment |
add a comment |
Bo Halim is a new contributor. Be nice, and check out our Code of Conduct.
Bo Halim is a new contributor. Be nice, and check out our Code of Conduct.
Bo Halim is a new contributor. Be nice, and check out our Code of Conduct.
Bo Halim is a new contributor. Be nice, and check out our Code of Conduct.
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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