What's the ratio of triangles made by diagonals of a trapezoid/trapezium?












3












$begingroup$


Trapezoid



In the above image, what will be the ratio of areas of triangle $A$ and $B$?



From Googling, I've found that:



$operatorname{Ar}(A) = dfrac{a^2h}{2(a+b)}$



and



$operatorname{Ar}(B) = dfrac{b^2h}{2(a+b)}$



but how do I get these formulas from the classic formula of $dfrac{rm base times height}2$?!



Basically, how were the formulas in this image:



this image http://www.geometryexpressions.com/explorations/04-Examples/02-Example_Book/02-Quadrilaterals/Example%20036-Areas_Of_Triangles_In_A_Trapezoid_files/image004.gif



figured out?






geometry triangle







share|cite|improve this question











$endgroup$












  • $begingroup$
    Oh, and reference link: Geometry Expressions
    $endgroup$
    – Nommy G
    Jun 26 '13 at 16:56


















3












$begingroup$


Trapezoid



In the above image, what will be the ratio of areas of triangle $A$ and $B$?



From Googling, I've found that:



$operatorname{Ar}(A) = dfrac{a^2h}{2(a+b)}$



and



$operatorname{Ar}(B) = dfrac{b^2h}{2(a+b)}$



but how do I get these formulas from the classic formula of $dfrac{rm base times height}2$?!



Basically, how were the formulas in this image:



this image http://www.geometryexpressions.com/explorations/04-Examples/02-Example_Book/02-Quadrilaterals/Example%20036-Areas_Of_Triangles_In_A_Trapezoid_files/image004.gif



figured out?






geometry triangle







share|cite|improve this question











$endgroup$












  • $begingroup$
    Oh, and reference link: Geometry Expressions
    $endgroup$
    – Nommy G
    Jun 26 '13 at 16:56
















3












3








3





$begingroup$


Trapezoid



In the above image, what will be the ratio of areas of triangle $A$ and $B$?



From Googling, I've found that:



$operatorname{Ar}(A) = dfrac{a^2h}{2(a+b)}$



and



$operatorname{Ar}(B) = dfrac{b^2h}{2(a+b)}$



but how do I get these formulas from the classic formula of $dfrac{rm base times height}2$?!



Basically, how were the formulas in this image:



this image http://www.geometryexpressions.com/explorations/04-Examples/02-Example_Book/02-Quadrilaterals/Example%20036-Areas_Of_Triangles_In_A_Trapezoid_files/image004.gif



figured out?






geometry triangle







share|cite|improve this question











$endgroup$




Trapezoid



In the above image, what will be the ratio of areas of triangle $A$ and $B$?



From Googling, I've found that:



$operatorname{Ar}(A) = dfrac{a^2h}{2(a+b)}$



and



$operatorname{Ar}(B) = dfrac{b^2h}{2(a+b)}$



but how do I get these formulas from the classic formula of $dfrac{rm base times height}2$?!



Basically, how were the formulas in this image:



this image http://www.geometryexpressions.com/explorations/04-Examples/02-Example_Book/02-Quadrilaterals/Example%20036-Areas_Of_Triangles_In_A_Trapezoid_files/image004.gif



figured out?






geometry triangle




geometry triangle






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 23 '18 at 21:12









Glorfindel

3,41981830




3,41981830










asked Jun 26 '13 at 16:56









Nommy GNommy G

1813




1813












  • $begingroup$
    Oh, and reference link: Geometry Expressions
    $endgroup$
    – Nommy G
    Jun 26 '13 at 16:56




















  • $begingroup$
    Oh, and reference link: Geometry Expressions
    $endgroup$
    – Nommy G
    Jun 26 '13 at 16:56


















$begingroup$
Oh, and reference link: Geometry Expressions
$endgroup$
– Nommy G
Jun 26 '13 at 16:56






$begingroup$
Oh, and reference link: Geometry Expressions
$endgroup$
– Nommy G
Jun 26 '13 at 16:56












1 Answer
1






active

oldest

votes


















6












$begingroup$

Well, what is the height of the triangle $Delta a$? It is similar to the triangle $Delta b$ (why?), so and the heights $h_a$ of $Delta a$ and $h_b$ of $Delta b$ are related in the following way:



First, we have $h = h_a + h_b$. This is obvious if you draw them on your figure. Secondly, from similarity, we have that
$$
frac{h_b}{h_a} = frac{b}{a}.
$$
By manipulating the last equation, and with a little help from the first to rid ourselves of $h_b$, we get
$$
frac{h-h_a}{h_a} = frac{b}{a} \\
frac{h}{h_a} - 1 = frac{b}{a} \\
frac{h}{h_a} = 1 + frac{b}{a} \\
h_a = frac{h}{1 + frac{b}{a}} = frac{ha}{a + b}
$$
This means that the area of $Delta a$ is equal to
$$
frac{1}{2}cdot a cdot h_a = frac{1}{2}cdot a cdot frac{ha}{a + b} = frac{a^2h}{2(a+b)}
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much! Great explanation! :)
    $endgroup$
    – Nommy G
    Jun 26 '13 at 17:22










  • $begingroup$
    One thing, though, how is $1 + b/a$ = $a + b$?
    $endgroup$
    – Nommy G
    Jun 26 '13 at 17:23










  • $begingroup$
    It isn't. The $a$ in the denominator was 'thrown' up.
    $endgroup$
    – ohad
    Jun 26 '13 at 17:25










  • $begingroup$
    @ohad Now I feel dumb... Thanks for explaining this!
    $endgroup$
    – Nommy G
    Jun 26 '13 at 17:26











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f430159%2fwhat39s-the-ratio-of-triangles-made-by-diagonals-of-a-trapezoid-trapezium%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

Well, what is the height of the triangle $Delta a$? It is similar to the triangle $Delta b$ (why?), so and the heights $h_a$ of $Delta a$ and $h_b$ of $Delta b$ are related in the following way:



First, we have $h = h_a + h_b$. This is obvious if you draw them on your figure. Secondly, from similarity, we have that
$$
frac{h_b}{h_a} = frac{b}{a}.
$$
By manipulating the last equation, and with a little help from the first to rid ourselves of $h_b$, we get
$$
frac{h-h_a}{h_a} = frac{b}{a} \\
frac{h}{h_a} - 1 = frac{b}{a} \\
frac{h}{h_a} = 1 + frac{b}{a} \\
h_a = frac{h}{1 + frac{b}{a}} = frac{ha}{a + b}
$$
This means that the area of $Delta a$ is equal to
$$
frac{1}{2}cdot a cdot h_a = frac{1}{2}cdot a cdot frac{ha}{a + b} = frac{a^2h}{2(a+b)}
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much! Great explanation! :)
    $endgroup$
    – Nommy G
    Jun 26 '13 at 17:22










  • $begingroup$
    One thing, though, how is $1 + b/a$ = $a + b$?
    $endgroup$
    – Nommy G
    Jun 26 '13 at 17:23










  • $begingroup$
    It isn't. The $a$ in the denominator was 'thrown' up.
    $endgroup$
    – ohad
    Jun 26 '13 at 17:25










  • $begingroup$
    @ohad Now I feel dumb... Thanks for explaining this!
    $endgroup$
    – Nommy G
    Jun 26 '13 at 17:26
















6












$begingroup$

Well, what is the height of the triangle $Delta a$? It is similar to the triangle $Delta b$ (why?), so and the heights $h_a$ of $Delta a$ and $h_b$ of $Delta b$ are related in the following way:



First, we have $h = h_a + h_b$. This is obvious if you draw them on your figure. Secondly, from similarity, we have that
$$
frac{h_b}{h_a} = frac{b}{a}.
$$
By manipulating the last equation, and with a little help from the first to rid ourselves of $h_b$, we get
$$
frac{h-h_a}{h_a} = frac{b}{a} \\
frac{h}{h_a} - 1 = frac{b}{a} \\
frac{h}{h_a} = 1 + frac{b}{a} \\
h_a = frac{h}{1 + frac{b}{a}} = frac{ha}{a + b}
$$
This means that the area of $Delta a$ is equal to
$$
frac{1}{2}cdot a cdot h_a = frac{1}{2}cdot a cdot frac{ha}{a + b} = frac{a^2h}{2(a+b)}
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much! Great explanation! :)
    $endgroup$
    – Nommy G
    Jun 26 '13 at 17:22










  • $begingroup$
    One thing, though, how is $1 + b/a$ = $a + b$?
    $endgroup$
    – Nommy G
    Jun 26 '13 at 17:23










  • $begingroup$
    It isn't. The $a$ in the denominator was 'thrown' up.
    $endgroup$
    – ohad
    Jun 26 '13 at 17:25










  • $begingroup$
    @ohad Now I feel dumb... Thanks for explaining this!
    $endgroup$
    – Nommy G
    Jun 26 '13 at 17:26














6












6








6





$begingroup$

Well, what is the height of the triangle $Delta a$? It is similar to the triangle $Delta b$ (why?), so and the heights $h_a$ of $Delta a$ and $h_b$ of $Delta b$ are related in the following way:



First, we have $h = h_a + h_b$. This is obvious if you draw them on your figure. Secondly, from similarity, we have that
$$
frac{h_b}{h_a} = frac{b}{a}.
$$
By manipulating the last equation, and with a little help from the first to rid ourselves of $h_b$, we get
$$
frac{h-h_a}{h_a} = frac{b}{a} \\
frac{h}{h_a} - 1 = frac{b}{a} \\
frac{h}{h_a} = 1 + frac{b}{a} \\
h_a = frac{h}{1 + frac{b}{a}} = frac{ha}{a + b}
$$
This means that the area of $Delta a$ is equal to
$$
frac{1}{2}cdot a cdot h_a = frac{1}{2}cdot a cdot frac{ha}{a + b} = frac{a^2h}{2(a+b)}
$$






share|cite|improve this answer









$endgroup$



Well, what is the height of the triangle $Delta a$? It is similar to the triangle $Delta b$ (why?), so and the heights $h_a$ of $Delta a$ and $h_b$ of $Delta b$ are related in the following way:



First, we have $h = h_a + h_b$. This is obvious if you draw them on your figure. Secondly, from similarity, we have that
$$
frac{h_b}{h_a} = frac{b}{a}.
$$
By manipulating the last equation, and with a little help from the first to rid ourselves of $h_b$, we get
$$
frac{h-h_a}{h_a} = frac{b}{a} \\
frac{h}{h_a} - 1 = frac{b}{a} \\
frac{h}{h_a} = 1 + frac{b}{a} \\
h_a = frac{h}{1 + frac{b}{a}} = frac{ha}{a + b}
$$
This means that the area of $Delta a$ is equal to
$$
frac{1}{2}cdot a cdot h_a = frac{1}{2}cdot a cdot frac{ha}{a + b} = frac{a^2h}{2(a+b)}
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jun 26 '13 at 17:18









ArthurArthur

116k7116198




116k7116198












  • $begingroup$
    Thank you very much! Great explanation! :)
    $endgroup$
    – Nommy G
    Jun 26 '13 at 17:22










  • $begingroup$
    One thing, though, how is $1 + b/a$ = $a + b$?
    $endgroup$
    – Nommy G
    Jun 26 '13 at 17:23










  • $begingroup$
    It isn't. The $a$ in the denominator was 'thrown' up.
    $endgroup$
    – ohad
    Jun 26 '13 at 17:25










  • $begingroup$
    @ohad Now I feel dumb... Thanks for explaining this!
    $endgroup$
    – Nommy G
    Jun 26 '13 at 17:26


















  • $begingroup$
    Thank you very much! Great explanation! :)
    $endgroup$
    – Nommy G
    Jun 26 '13 at 17:22










  • $begingroup$
    One thing, though, how is $1 + b/a$ = $a + b$?
    $endgroup$
    – Nommy G
    Jun 26 '13 at 17:23










  • $begingroup$
    It isn't. The $a$ in the denominator was 'thrown' up.
    $endgroup$
    – ohad
    Jun 26 '13 at 17:25










  • $begingroup$
    @ohad Now I feel dumb... Thanks for explaining this!
    $endgroup$
    – Nommy G
    Jun 26 '13 at 17:26
















$begingroup$
Thank you very much! Great explanation! :)
$endgroup$
– Nommy G
Jun 26 '13 at 17:22




$begingroup$
Thank you very much! Great explanation! :)
$endgroup$
– Nommy G
Jun 26 '13 at 17:22












$begingroup$
One thing, though, how is $1 + b/a$ = $a + b$?
$endgroup$
– Nommy G
Jun 26 '13 at 17:23




$begingroup$
One thing, though, how is $1 + b/a$ = $a + b$?
$endgroup$
– Nommy G
Jun 26 '13 at 17:23












$begingroup$
It isn't. The $a$ in the denominator was 'thrown' up.
$endgroup$
– ohad
Jun 26 '13 at 17:25




$begingroup$
It isn't. The $a$ in the denominator was 'thrown' up.
$endgroup$
– ohad
Jun 26 '13 at 17:25












$begingroup$
@ohad Now I feel dumb... Thanks for explaining this!
$endgroup$
– Nommy G
Jun 26 '13 at 17:26




$begingroup$
@ohad Now I feel dumb... Thanks for explaining this!
$endgroup$
– Nommy G
Jun 26 '13 at 17:26


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f430159%2fwhat39s-the-ratio-of-triangles-made-by-diagonals-of-a-trapezoid-trapezium%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Ellipse (mathématiques)

Image
Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
Read more

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
Read more