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Area of parallelogram = Area of square. Shear transform

Visualizing the Area of a ParallelogramArea of stripe around cylinderProving the area of a square and the required axiomsarea of rotated squares on top of each otherArea of Square $neq$ the area of Rhombus created by stretched square?volume of a frustum in different wayParallelogram, Area and AnglesHow to prove area of the square and parallelogram the same?How can the area of a parallelogram be show to be equal to the determinant?Confusion on Parallelogram

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}

3

$begingroup$

Below the parallelogram is obtained from square by stretching the top side while fixing the bottom.

Since area of parallelogram is base x height, both square and parallelogram have the same area.

This is true no matter how far I stretch the top side.

In below figure it is easy to see why both areas are same.

But it's not that obvious in first two figures. Any help seeing why the area doesn't change in first figure?

geometry

share|cite|improve this question

asked 9 hours ago

rsadhvikarsadhvika

1,7991212 silver badges2929 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

Below the parallelogram is obtained from square by stretching the top side while fixing the bottom.

Since area of parallelogram is base x height, both square and parallelogram have the same area.

This is true no matter how far I stretch the top side.

In below figure it is easy to see why both areas are same.

But it's not that obvious in first two figures. Any help seeing why the area doesn't change in first figure?

geometry

share|cite|improve this question

asked 9 hours ago

rsadhvikarsadhvika

1,7991212 silver badges2929 bronze badges

$endgroup$

add a comment | 

$begingroup$

Below the parallelogram is obtained from square by stretching the top side while fixing the bottom.

Since area of parallelogram is base x height, both square and parallelogram have the same area.

This is true no matter how far I stretch the top side.

In below figure it is easy to see why both areas are same.

But it's not that obvious in first two figures. Any help seeing why the area doesn't change in first figure?

geometry

share|cite|improve this question

asked 9 hours ago

rsadhvikarsadhvika

1,7991212 silver badges2929 bronze badges

$endgroup$

share|cite|improve this question

asked 9 hours ago

rsadhvikarsadhvika

1,7991212 silver badges2929 bronze badges

share|cite|improve this question

asked 9 hours ago

rsadhvikarsadhvika

1,7991212 silver badges2929 bronze badges

asked 9 hours ago

rsadhvikarsadhvika

1,7991212 silver badges2929 bronze badges

asked 9 hours ago

rsadhvikarsadhvika

1,7991212 silver badges2929 bronze badges

4 Answers
4

active

oldest

votes

4

$begingroup$

Behold, $phantom{proof without words}$

share|cite|improve this answer

answered 5 hours ago

runway44runway44

68766 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah: that picture proof is pretty robust. Has all the elements to formulate a synthetic argument too.
  $endgroup$
  – rschwieb
  1 hour ago
  

  
  
  
  

add a comment | 

2

$begingroup$

It folows from Cavalieri's Principle, or else if you know that shear transforms have determinant $1$, and hence don't change area, that's another way to see it.

I don't understand why it isn't "easy to see" that all three figures have equal bases and heights... Maybe you should just take a look at computing the difference between the images of the endpoints of the top of the squares to convince yourself.

The general form of such a transformation is

$begin{bmatrix}1&a\0&1end{bmatrix}$ multiplying on the left of column vectors.)

Then you always have

$$[0,0]^Tmapsto [0,0]^T$$
$$[1,0]^Tmapsto [1,0]^T$$
$$[0,1]^Tmapsto [a,1]^T$$
$$[1,1]^Tmapsto [1+a,1]^T$$

From the first two you can see the length of the bottom horizontal line is $1$, and from the second two you can see the length of the top horizontal line is $1$. Obviously they also show the height was unchanged (since the $y$ coordinates were all preserved.)

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

rschwiebrschwieb

111k1212 gold badges111111 silver badges260260 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's not clear to me OP doesn't see the figures have equal bases and heights. Presumably OP sees this, and understands the formula therefore dictates they have the same area, but doesn't understand why they ought to have the same area besides taking a formula for granted.
  $endgroup$
  – runway44
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @runway44 could be: although it could have been more clearly stated that a proof independent of the formula was requested.
  $endgroup$
  – rschwieb
  2 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Slice each figure by infinitely many infinitely thin horizontal layers. The area of each slice is the same as that of the corresponding slice in the original square: corresponding slices both have the same width and height and the ends can be neglected (in the given limit).

share|cite|improve this answer

edited 3 hours ago

answered 9 hours ago

David G. StorkDavid G. Stork

13.6k44 gold badges1919 silver badges3737 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Seems circular. You would like to say exactly the same thing about the thin slices as you would for the whole, but you don't have it yet (unless you assume the conclusion.)
  $endgroup$
  – rschwieb
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Agree with @rschwieb . You may want to put something in about the thin horizontal slices being just straight up rectangles instead of sheared rectangles on top of one another, so that they could've also been gotten in the original square, and as we make them infinitely small, the "empty space" due to having to shift over the ones getting stacked on top approaches 0?
  $endgroup$
  – Joshua Ronis
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not circular at all, though I clarified the answer to include this.
  $endgroup$
  – David G. Stork
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DavidG.Stork Taken at face value it was circular... but I see now you have expanded it to a sketch of Cavalieri’s principle, yes.
  $endgroup$
  – rschwieb
  2 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

In your first two figures, note that
$$text{area}(EBGH)=text{area}(EBCH)+text{area}(HCG)$$
and
$$text{area}(EBGH)=text{area}(EFGH)+text{area}(BEF).$$
But the triangles $HCG$ and $BEF$ are congruent, so have the same area.
Subtracting that gives
$$text{area}(EBCH)+text{area}(BCG)=text{area}(EFGH).$$
Come to think about it, this works just as well in the third figure.

share|cite|improve this answer

edited 13 mins ago

answered 9 hours ago

Lord Shark the UnknownLord Shark the Unknown

116k1111 gold badges6767 silver badges148148 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't understand the letters you're using. For instance, I don't see a point $P$ in the picture, and $B,C,G$ are collinear points. But a tribagle sounds quite nice. :) I suspect your proof is almost the same as mine though.
  $endgroup$
  – runway44
  5 hours ago
  
  
  

  
  
  
  

add a comment | 

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4

$begingroup$

Behold, $phantom{proof without words}$

share|cite|improve this answer

answered 5 hours ago

runway44runway44

68766 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah: that picture proof is pretty robust. Has all the elements to formulate a synthetic argument too.
  $endgroup$
  – rschwieb
  1 hour ago
  

  
  
  
  

add a comment | 

4

$begingroup$

Behold, $phantom{proof without words}$

share|cite|improve this answer

answered 5 hours ago

runway44runway44

68766 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah: that picture proof is pretty robust. Has all the elements to formulate a synthetic argument too.
  $endgroup$
  – rschwieb
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

Behold, $phantom{proof without words}$

share|cite|improve this answer

answered 5 hours ago

runway44runway44

68766 bronze badges

$endgroup$

share|cite|improve this answer

answered 5 hours ago

runway44runway44

68766 bronze badges

answered 5 hours ago

runway44runway44

68766 bronze badges

answered 5 hours ago

runway44runway44

68766 bronze badges

2

$begingroup$

It folows from Cavalieri's Principle, or else if you know that shear transforms have determinant $1$, and hence don't change area, that's another way to see it.

I don't understand why it isn't "easy to see" that all three figures have equal bases and heights... Maybe you should just take a look at computing the difference between the images of the endpoints of the top of the squares to convince yourself.

The general form of such a transformation is

$begin{bmatrix}1&a\0&1end{bmatrix}$ multiplying on the left of column vectors.)

Then you always have

$$[0,0]^Tmapsto [0,0]^T$$
$$[1,0]^Tmapsto [1,0]^T$$
$$[0,1]^Tmapsto [a,1]^T$$
$$[1,1]^Tmapsto [1+a,1]^T$$

From the first two you can see the length of the bottom horizontal line is $1$, and from the second two you can see the length of the top horizontal line is $1$. Obviously they also show the height was unchanged (since the $y$ coordinates were all preserved.)

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

rschwiebrschwieb

111k1212 gold badges111111 silver badges260260 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's not clear to me OP doesn't see the figures have equal bases and heights. Presumably OP sees this, and understands the formula therefore dictates they have the same area, but doesn't understand why they ought to have the same area besides taking a formula for granted.
  $endgroup$
  – runway44
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @runway44 could be: although it could have been more clearly stated that a proof independent of the formula was requested.
  $endgroup$
  – rschwieb
  2 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

It folows from Cavalieri's Principle, or else if you know that shear transforms have determinant $1$, and hence don't change area, that's another way to see it.

I don't understand why it isn't "easy to see" that all three figures have equal bases and heights... Maybe you should just take a look at computing the difference between the images of the endpoints of the top of the squares to convince yourself.

The general form of such a transformation is

$begin{bmatrix}1&a\0&1end{bmatrix}$ multiplying on the left of column vectors.)

Then you always have

$$[0,0]^Tmapsto [0,0]^T$$
$$[1,0]^Tmapsto [1,0]^T$$
$$[0,1]^Tmapsto [a,1]^T$$
$$[1,1]^Tmapsto [1+a,1]^T$$

From the first two you can see the length of the bottom horizontal line is $1$, and from the second two you can see the length of the top horizontal line is $1$. Obviously they also show the height was unchanged (since the $y$ coordinates were all preserved.)

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

rschwiebrschwieb

111k1212 gold badges111111 silver badges260260 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's not clear to me OP doesn't see the figures have equal bases and heights. Presumably OP sees this, and understands the formula therefore dictates they have the same area, but doesn't understand why they ought to have the same area besides taking a formula for granted.
  $endgroup$
  – runway44
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @runway44 could be: although it could have been more clearly stated that a proof independent of the formula was requested.
  $endgroup$
  – rschwieb
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

It folows from Cavalieri's Principle, or else if you know that shear transforms have determinant $1$, and hence don't change area, that's another way to see it.

I don't understand why it isn't "easy to see" that all three figures have equal bases and heights... Maybe you should just take a look at computing the difference between the images of the endpoints of the top of the squares to convince yourself.

The general form of such a transformation is

$begin{bmatrix}1&a\0&1end{bmatrix}$ multiplying on the left of column vectors.)

Then you always have

$$[0,0]^Tmapsto [0,0]^T$$
$$[1,0]^Tmapsto [1,0]^T$$
$$[0,1]^Tmapsto [a,1]^T$$
$$[1,1]^Tmapsto [1+a,1]^T$$

From the first two you can see the length of the bottom horizontal line is $1$, and from the second two you can see the length of the top horizontal line is $1$. Obviously they also show the height was unchanged (since the $y$ coordinates were all preserved.)

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

rschwiebrschwieb

111k1212 gold badges111111 silver badges260260 bronze badges

$endgroup$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

rschwiebrschwieb

111k1212 gold badges111111 silver badges260260 bronze badges

answered 9 hours ago

rschwiebrschwieb

111k1212 gold badges111111 silver badges260260 bronze badges

answered 9 hours ago

rschwiebrschwieb

111k1212 gold badges111111 silver badges260260 bronze badges

2

$begingroup$

Slice each figure by infinitely many infinitely thin horizontal layers. The area of each slice is the same as that of the corresponding slice in the original square: corresponding slices both have the same width and height and the ends can be neglected (in the given limit).

share|cite|improve this answer

edited 3 hours ago

answered 9 hours ago

David G. StorkDavid G. Stork

13.6k44 gold badges1919 silver badges3737 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Seems circular. You would like to say exactly the same thing about the thin slices as you would for the whole, but you don't have it yet (unless you assume the conclusion.)
  $endgroup$
  – rschwieb
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Agree with @rschwieb . You may want to put something in about the thin horizontal slices being just straight up rectangles instead of sheared rectangles on top of one another, so that they could've also been gotten in the original square, and as we make them infinitely small, the "empty space" due to having to shift over the ones getting stacked on top approaches 0?
  $endgroup$
  – Joshua Ronis
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not circular at all, though I clarified the answer to include this.
  $endgroup$
  – David G. Stork
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DavidG.Stork Taken at face value it was circular... but I see now you have expanded it to a sketch of Cavalieri’s principle, yes.
  $endgroup$
  – rschwieb
  2 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Slice each figure by infinitely many infinitely thin horizontal layers. The area of each slice is the same as that of the corresponding slice in the original square: corresponding slices both have the same width and height and the ends can be neglected (in the given limit).

share|cite|improve this answer

edited 3 hours ago

answered 9 hours ago

David G. StorkDavid G. Stork

13.6k44 gold badges1919 silver badges3737 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Seems circular. You would like to say exactly the same thing about the thin slices as you would for the whole, but you don't have it yet (unless you assume the conclusion.)
  $endgroup$
  – rschwieb
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Agree with @rschwieb . You may want to put something in about the thin horizontal slices being just straight up rectangles instead of sheared rectangles on top of one another, so that they could've also been gotten in the original square, and as we make them infinitely small, the "empty space" due to having to shift over the ones getting stacked on top approaches 0?
  $endgroup$
  – Joshua Ronis
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not circular at all, though I clarified the answer to include this.
  $endgroup$
  – David G. Stork
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DavidG.Stork Taken at face value it was circular... but I see now you have expanded it to a sketch of Cavalieri’s principle, yes.
  $endgroup$
  – rschwieb
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Slice each figure by infinitely many infinitely thin horizontal layers. The area of each slice is the same as that of the corresponding slice in the original square: corresponding slices both have the same width and height and the ends can be neglected (in the given limit).

share|cite|improve this answer

edited 3 hours ago

answered 9 hours ago

David G. StorkDavid G. Stork

13.6k44 gold badges1919 silver badges3737 bronze badges

$endgroup$

share|cite|improve this answer

edited 3 hours ago

answered 9 hours ago

David G. StorkDavid G. Stork

13.6k44 gold badges1919 silver badges3737 bronze badges

answered 9 hours ago

David G. StorkDavid G. Stork

13.6k44 gold badges1919 silver badges3737 bronze badges

answered 9 hours ago

David G. StorkDavid G. Stork

13.6k44 gold badges1919 silver badges3737 bronze badges

1

$begingroup$

In your first two figures, note that
$$text{area}(EBGH)=text{area}(EBCH)+text{area}(HCG)$$
and
$$text{area}(EBGH)=text{area}(EFGH)+text{area}(BEF).$$
But the triangles $HCG$ and $BEF$ are congruent, so have the same area.
Subtracting that gives
$$text{area}(EBCH)+text{area}(BCG)=text{area}(EFGH).$$
Come to think about it, this works just as well in the third figure.

share|cite|improve this answer

edited 13 mins ago

answered 9 hours ago

Lord Shark the UnknownLord Shark the Unknown

116k1111 gold badges6767 silver badges148148 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't understand the letters you're using. For instance, I don't see a point $P$ in the picture, and $B,C,G$ are collinear points. But a tribagle sounds quite nice. :) I suspect your proof is almost the same as mine though.
  $endgroup$
  – runway44
  5 hours ago
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

In your first two figures, note that
$$text{area}(EBGH)=text{area}(EBCH)+text{area}(HCG)$$
and
$$text{area}(EBGH)=text{area}(EFGH)+text{area}(BEF).$$
But the triangles $HCG$ and $BEF$ are congruent, so have the same area.
Subtracting that gives
$$text{area}(EBCH)+text{area}(BCG)=text{area}(EFGH).$$
Come to think about it, this works just as well in the third figure.

share|cite|improve this answer

edited 13 mins ago

answered 9 hours ago

Lord Shark the UnknownLord Shark the Unknown

116k1111 gold badges6767 silver badges148148 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't understand the letters you're using. For instance, I don't see a point $P$ in the picture, and $B,C,G$ are collinear points. But a tribagle sounds quite nice. :) I suspect your proof is almost the same as mine though.
  $endgroup$
  – runway44
  5 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

In your first two figures, note that
$$text{area}(EBGH)=text{area}(EBCH)+text{area}(HCG)$$
and
$$text{area}(EBGH)=text{area}(EFGH)+text{area}(BEF).$$
But the triangles $HCG$ and $BEF$ are congruent, so have the same area.
Subtracting that gives
$$text{area}(EBCH)+text{area}(BCG)=text{area}(EFGH).$$
Come to think about it, this works just as well in the third figure.

share|cite|improve this answer

edited 13 mins ago

answered 9 hours ago

Lord Shark the UnknownLord Shark the Unknown

116k1111 gold badges6767 silver badges148148 bronze badges

$endgroup$

share|cite|improve this answer

edited 13 mins ago

answered 9 hours ago

Lord Shark the UnknownLord Shark the Unknown

116k1111 gold badges6767 silver badges148148 bronze badges

answered 9 hours ago

Lord Shark the UnknownLord Shark the Unknown

116k1111 gold badges6767 silver badges148148 bronze badges

answered 9 hours ago

Lord Shark the UnknownLord Shark the Unknown

116k1111 gold badges6767 silver badges148148 bronze badges