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?
How should I respond when I lied about my education and the company finds out through background check?
Does the Mind Blank spell prevent the target from being frightened?
Have I saved too much for retirement so far?
How can "mimic phobia" be cured or prevented?
Global amount of publications over time
Greatest common substring
What (else) happened July 1st 1858 in London?
What does this horizontal bar at the first measure mean?
How much character growth crosses the line into breaking the character
Proving a function is onto where f(x)=|x|.
Diode in opposite direction?
Why does Async/Await work properly when the loop is inside the async function and not the other way around?
ArcGIS not connecting to PostgreSQL db with all upper-case name
Translation of Scottish 16th century church stained glass
Is camera lens focus an exact point or a range?
How do I repair my stair bannister?
Longest common substring in linear time
Is there a word to describe the feeling of being transfixed out of horror?
How can Trident be so inexpensive? Will it orbit Triton or just do a (slow) flyby?
Fly on a jet pack vs fly with a jet pack?
Why has "pence" been used in this sentence, not "pences"?
Is a model fitted to data or is data fitted to a model?
Should I install hardwood flooring or cabinets first?
Two-sided logarithm inequality
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
New contributor
$endgroup$
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
New contributor
$endgroup$
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
New contributor
$endgroup$
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
New contributor
New contributor
edited 53 mins ago
dantopa
6,63342245
6,63342245
New contributor
asked 6 hours ago
Bo HalimBo Halim
1094
1094
New contributor
New contributor
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:
$endgroup$
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 |
$begingroup$
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 |
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:
$endgroup$
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:
$endgroup$
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:
$endgroup$
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 5 hours ago
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 $25$.
$endgroup$
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 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.
$endgroup$
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 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