Confusion on ParallelogramVisualizing 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?
Freedom of speech and where it applies
We have a love-hate relationship
ArcGIS not connecting to PostgreSQL db with all upper-case name
How do I implement a file system driver driver in Linux?
Java - What do constructor type arguments mean when placed *before* the type?
Find last 3 digits of this monster number
How do I extrude a face to a single vertex
Using a siddur to Daven from in a seforim store
How to align and center standalone amsmath equations?
Could the E-bike drivetrain wear down till needing replacement after 400 km?
Journal losing indexing services
How should I respond when I lied about my education and the company finds out through background check?
Frequency of inspection at vegan restaurants
What's the difference between 違法 and 不法?
THT: What is a squared annular “ring”?
What is the grammatical term for “‑ed” words like these?
Did US corporations pay demonstrators in the German demonstrations against article 13?
A social experiment. What is the worst that can happen?
Has Darkwing Duck ever met Scrooge McDuck?
Drawing ramified coverings with tikz
Will adding a BY-SA image to a blog post make the entire post BY-SA?
API Access HTML/Javascript
What does this horizontal bar at the first measure mean?
Is it possible to use .desktop files to open local pdf files on specific pages with a browser?
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?
$begingroup$
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 1 hour 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.
$endgroup$
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 1 hour 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 1 hour 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 1 hour 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 1 hour 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 1 hour ago
dantopa
6,63342245
edited 1 hour ago
dantopa
6,63342245
edited 1 hour 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
$endgroup$
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 |
$begingroup$
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 |
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.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
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
Required, but never shown
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.
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
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
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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