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Confusion on Parallelogram [duplicate]

Visualizing the Area of a ParallelogramHow do I find the surface area of an angled conic base?Visualizing the Area of a ParallelogramArea of stripe around cylinderArea of Square $neq$ the area of Rhombus created by stretched square?How can the area of a parallelogram be show to be equal to the determinant?Finding the Longer Diagonal of a ParallelogramGRE: Can area of this parallelogram be known?Area of a regular hexagon via area of trianglesFinding the Relation of the Areas in Two Similar Right TrianglesDid Euclid prove the formula for the area of a triangle?

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 ?

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

geometry

share|cite|improve this question

edited yesterday

dantopa

6,63342245

asked yesterday

Bo HalimBo Halim

1195

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.

$endgroup$

marked as duplicate by mrtaurho, Cesareo, user21820, Xander Henderson, José Carlos Santos 14 hours ago

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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Taladris thank you :)
  $endgroup$
  – Bo Halim
  19 hours ago
  

  
  
  
  

add a comment | 

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 ?

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

geometry

share|cite|improve this question

edited yesterday

dantopa

6,63342245

asked yesterday

Bo HalimBo Halim

1195

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.

$endgroup$

marked as duplicate by mrtaurho, Cesareo, user21820, Xander Henderson, José Carlos Santos 14 hours ago

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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Taladris thank you :)
  $endgroup$
  – Bo Halim
  19 hours ago
  

  
  
  
  

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

share|cite|improve this question

edited yesterday

dantopa

6,63342245

asked yesterday

Bo HalimBo Halim

1195

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.

$endgroup$

share|cite|improve this question

edited yesterday

dantopa

6,63342245

asked yesterday

Bo HalimBo Halim

1195

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

edited yesterday

dantopa

6,63342245

asked yesterday

Bo HalimBo Halim

1195

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 yesterday

dantopa

6,63342245

edited yesterday

dantopa

6,63342245

asked yesterday

Bo HalimBo Halim

1195

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 yesterday

Bo HalimBo Halim

1195

5 Answers
5

active

oldest

votes

15

$begingroup$

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:

share|cite|improve this answer

edited yesterday

answered yesterday

David G. StorkDavid G. Stork

11.4k41533

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The best answer ever :)
  $endgroup$
  – Bo Halim
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  EVER? My goodness... thanks.
  $endgroup$
  – David G. Stork
  10 hours ago
  

  
  
  
  

add a comment | 

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

answered yesterday

Ethan BolkerEthan Bolker

45.4k553120

$endgroup$

add a comment | 

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:

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

edited 17 hours ago

answered 22 hours ago

VekyVeky

24019

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +1: This is an excellent refutation.
  $endgroup$
  – Cameron Buie
  16 hours ago
  

  
  
  
  

add a comment | 

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

answered yesterday

Cameron BuieCameron Buie

86.2k772161

$endgroup$

add a comment | 

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

edited 22 hours ago

answered 22 hours ago

Paras KhoslaParas Khosla

2,663323

$endgroup$

add a comment | 

15

$begingroup$

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:

share|cite|improve this answer

edited yesterday

answered yesterday

David G. StorkDavid G. Stork

11.4k41533

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The best answer ever :)
  $endgroup$
  – Bo Halim
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  EVER? My goodness... thanks.
  $endgroup$
  – David G. Stork
  10 hours ago
  

  
  
  
  

add a comment | 

15

$begingroup$

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:

share|cite|improve this answer

edited yesterday

answered yesterday

David G. StorkDavid G. Stork

11.4k41533

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The best answer ever :)
  $endgroup$
  – Bo Halim
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  EVER? My goodness... thanks.
  $endgroup$
  – David G. Stork
  10 hours ago
  

  
  
  
  

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:

share|cite|improve this answer

edited yesterday

answered yesterday

David G. StorkDavid G. Stork

11.4k41533

$endgroup$

share|cite|improve this answer

edited yesterday

answered yesterday

David G. StorkDavid G. Stork

11.4k41533

answered yesterday

David G. StorkDavid G. Stork

11.4k41533

answered yesterday

David G. StorkDavid G. Stork

11.4k41533

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

answered yesterday

Ethan BolkerEthan Bolker

45.4k553120

$endgroup$

add a comment | 

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

answered yesterday

Ethan BolkerEthan Bolker

45.4k553120

$endgroup$

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 $25$.

share|cite|improve this answer

answered yesterday

Ethan BolkerEthan Bolker

45.4k553120

$endgroup$

share|cite|improve this answer

answered yesterday

Ethan BolkerEthan Bolker

45.4k553120

answered yesterday

Ethan BolkerEthan Bolker

45.4k553120

answered yesterday

Ethan BolkerEthan Bolker

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:

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

edited 17 hours ago

answered 22 hours ago

VekyVeky

24019

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +1: This is an excellent refutation.
  $endgroup$
  – Cameron Buie
  16 hours ago
  

  
  
  
  

add a comment | 

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:

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

edited 17 hours ago

answered 22 hours ago

VekyVeky

24019

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +1: This is an excellent refutation.
  $endgroup$
  – Cameron Buie
  16 hours ago
  

  
  
  
  

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.

share|cite|improve this answer

edited 17 hours ago

answered 22 hours ago

VekyVeky

24019

$endgroup$

share|cite|improve this answer

edited 17 hours ago

answered 22 hours ago

VekyVeky

24019

answered 22 hours ago

VekyVeky

24019

answered 22 hours ago

VekyVeky

24019

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

answered yesterday

Cameron BuieCameron Buie

86.2k772161

$endgroup$

add a comment | 

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

answered yesterday

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.

share|cite|improve this answer

answered yesterday

Cameron BuieCameron Buie

86.2k772161

$endgroup$

share|cite|improve this answer

answered yesterday

Cameron BuieCameron Buie

86.2k772161

answered yesterday

Cameron BuieCameron Buie

86.2k772161

answered yesterday

Cameron BuieCameron Buie

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

edited 22 hours ago

answered 22 hours ago

Paras KhoslaParas Khosla

2,663323

$endgroup$

add a comment | 

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

edited 22 hours ago

answered 22 hours ago

Paras KhoslaParas Khosla

2,663323

$endgroup$

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.

share|cite|improve this answer

edited 22 hours ago

answered 22 hours ago

Paras KhoslaParas Khosla

2,663323

$endgroup$

share|cite|improve this answer

edited 22 hours ago

answered 22 hours ago

Paras KhoslaParas Khosla

2,663323

answered 22 hours ago

Paras KhoslaParas Khosla

2,663323

answered 22 hours ago

Paras KhoslaParas Khosla

2,663323