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











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1 Answer
1






active

oldest

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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


















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