Confusion on Parallelogram [duplicate]
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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 ?
PS : I found the same Visualizing the Area of a Parallelogram, but was unconvinced by the answers
geometry
New contributor
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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.
add a comment |
$begingroup$
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 ?
PS : I found the same Visualizing the Area of a Parallelogram, but was unconvinced by the answers
geometry
New contributor
$endgroup$
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.
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@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
add a comment |
$begingroup$
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 ?
PS : I found the same Visualizing the Area of a Parallelogram, but was unconvinced by the answers
geometry
New contributor
$endgroup$
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 ?
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
geometry
New contributor
New contributor
edited 2 days ago
dantopa
6,63342245
6,63342245
New contributor
asked 2 days ago
Bo HalimBo Halim
1195
1195
New contributor
New contributor
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
add a comment |
$begingroup$
@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
add a comment |
5 Answers
5
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:
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$begingroup$
The best answer ever :)
$endgroup$
– Bo Halim
yesterday
1
$begingroup$
EVER? My goodness... thanks.
$endgroup$
– David G. Stork
yesterday
add a comment |
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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 $25$.
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add a comment |
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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:
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.
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1
$begingroup$
+1: This is an excellent refutation.
$endgroup$
– Cameron Buie
yesterday
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.
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add a comment |
$begingroup$
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.
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add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
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:
$endgroup$
$begingroup$
The best answer ever :)
$endgroup$
– Bo Halim
yesterday
1
$begingroup$
EVER? My goodness... thanks.
$endgroup$
– David G. Stork
yesterday
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:
$endgroup$
$begingroup$
The best answer ever :)
$endgroup$
– Bo Halim
yesterday
1
$begingroup$
EVER? My goodness... thanks.
$endgroup$
– David G. Stork
yesterday
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:
$endgroup$
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:
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
add a comment |
$begingroup$
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
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 $25$.
$endgroup$
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 $25$.
$endgroup$
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 $25$.
$endgroup$
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 2 days ago
Ethan BolkerEthan Bolker
45.4k553120
45.4k553120
add a comment |
add a comment |
$begingroup$
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:
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.
$endgroup$
1
$begingroup$
+1: This is an excellent refutation.
$endgroup$
– Cameron Buie
yesterday
add a comment |
$begingroup$
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:
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.
$endgroup$
1
$begingroup$
+1: This is an excellent refutation.
$endgroup$
– Cameron Buie
yesterday
add a comment |
$begingroup$
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:
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.
$endgroup$
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:
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.
edited yesterday
answered 2 days ago
VekyVeky
24019
24019
1
$begingroup$
+1: This is an excellent refutation.
$endgroup$
– Cameron Buie
yesterday
add a comment |
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
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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 2 days ago
Cameron BuieCameron Buie
86.2k772161
86.2k772161
add a comment |
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
edited 2 days ago
answered 2 days ago
Paras KhoslaParas Khosla
2,676323
2,676323
add a comment |
add a comment |
$begingroup$
@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