Ttykuu

Should I stop contributing to retirement accounts?

Translation of Scottish 16th century church stained glass

Varistor? Purpose and principle

Proof of Lemma: Every nonzero integer can be written as a product of primes

THT: What is a squared annular “ring”?

Is there a word to describe the feeling of being transfixed out of horror?

Java - What do constructor type arguments mean when placed *before* the type?

Why does Async/Await work properly when the loop is inside the async function and not the other way around?

Melting point of aspirin, contradicting sources

How do I extrude a face to a single vertex

Longest common substring in linear time

Does the Mind Blank spell prevent the target from being frightened?

Query about absorption line spectra

Flux received by a negative charge

Create all possible words using a set or letters

List of people who lose a child in תנ"ך

Constructing Group Divisible Designs - Algorithms?

Drawing ramified coverings with tikz

Can somebody explain Brexit in a few child-proof sentences?

Find last 3 digits of this monster number

Visiting the UK as unmarried couple

How to color a curve

Has Darkwing Duck ever met Scrooge McDuck?

ArcGIS not connecting to PostgreSQL db with all upper-case name

Confusion on Parallelogram

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?

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 ?

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

geometry

share|cite|improve this question

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 | 

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 ?

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

geometry

share|cite|improve this question

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

share|cite|improve this question

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.

$endgroup$

share|cite|improve this question

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.

share|cite|improve this question

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

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

3 Answers
3

active

oldest

votes

6

$begingroup$

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 5 hours ago

answered 6 hours ago

David G. StorkDavid G. Stork

11.3k41432

$endgroup$

add a comment | 

3

$begingroup$

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 6 hours ago

Ethan BolkerEthan Bolker

45.3k553120

$endgroup$

add a comment | 

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

answered 6 hours ago

Cameron BuieCameron Buie

86.2k772161

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

Bo Halim is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');

var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);

;
$window.on('scroll', onScroll);

);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3161100%2fconfusion-on-parallelogram%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

6

$begingroup$

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 5 hours ago

answered 6 hours ago

David G. StorkDavid G. Stork

11.3k41432

$endgroup$

add a comment | 

6

$begingroup$

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 5 hours ago

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:

share|cite|improve this answer

edited 5 hours ago

answered 6 hours ago

David G. StorkDavid G. Stork

11.3k41432

$endgroup$

share|cite|improve this answer

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

3

$begingroup$

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 6 hours ago

Ethan BolkerEthan Bolker

45.3k553120

$endgroup$

add a comment | 

3

$begingroup$

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 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 $25$.

share|cite|improve this answer

answered 6 hours ago

Ethan BolkerEthan Bolker

45.3k553120

$endgroup$

share|cite|improve this answer

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

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

answered 6 hours ago

Cameron BuieCameron Buie

86.2k772161

$endgroup$

add a comment | 

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

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.

share|cite|improve this answer

answered 6 hours ago

Cameron BuieCameron Buie

86.2k772161

$endgroup$

share|cite|improve this answer

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