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Finding the maximum area of a right triangle inside a right triangle
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So I've encountered a question where it requires me to find the maximum area of a small right angled triangle inside a bigger one. The question stated the dimensions of the big triangle and approved the parallelism of 2 lines in the form below:
Everything in black is given by the question, and otherwise (red) is assumed by me.
I've assumed the line $overleftrightarroweb$ equal to $x$ and $overleftrightarrowdb$ equal to y. Then I proved that $Delta$ $deb$ is similar to $Delta$ $abc$ by sharing the same right angle and having the angle $<deb$ corresponding to angle $<acb$, thus:
$$fracy6 = fracx8$$, then $$ y = frac3x4 $$.
$$overleftrightarrowde$$ would be equal to $$frac5x4 $$
Now I had to find the height $$overleftrightarrowfe$$ in terms of $$x$$
Since $$overleftrightarrowde$$ is parallel to $$overleftrightarrowac$$ , angles $$<edf$$ and $$<afd$$ are alternate angles, thus they are equal to each other, and since $$<afd$$ and $$<acb$$ are corresponding angles, $$<edf$$ is equal to $$<acb$$, and both $$Delta abc, Delta fed$$ have right angles, then we can infer that $$<efd$$ is equal to $$<cab$$, so both triangles are similar.
Thus: $$fracn6 = fracfrac5x48$$ and $$n = frac15x16 $$.
Now we can find the area of the smaller triangle as a function of $x$.
$$f(x) = 0.5 × frac15x16 × frac5x4$$
But, if I were to take the derivative of that function to find a maximum value, I would end up with a minimum value at $x = 0$, which is utterly irrational. What mistake have I done here?
derivatives triangle area applications
asked Jul 22 at 10:58
Stephen Alexander
1315
add a comment |Â
up vote
0
down vote
favorite
So I've encountered a question where it requires me to find the maximum area of a small right angled triangle inside a bigger one. The question stated the dimensions of the big triangle and approved the parallelism of 2 lines in the form below:
Everything in black is given by the question, and otherwise (red) is assumed by me.
I've assumed the line $overleftrightarroweb$ equal to $x$ and $overleftrightarrowdb$ equal to y. Then I proved that $Delta$ $deb$ is similar to $Delta$ $abc$ by sharing the same right angle and having the angle $<deb$ corresponding to angle $<acb$, thus:
$$fracy6 = fracx8$$, then $$ y = frac3x4 $$.
$$overleftrightarrowde$$ would be equal to $$frac5x4 $$
Now I had to find the height $$overleftrightarrowfe$$ in terms of $$x$$
Since $$overleftrightarrowde$$ is parallel to $$overleftrightarrowac$$ , angles $$<edf$$ and $$<afd$$ are alternate angles, thus they are equal to each other, and since $$<afd$$ and $$<acb$$ are corresponding angles, $$<edf$$ is equal to $$<acb$$, and both $$Delta abc, Delta fed$$ have right angles, then we can infer that $$<efd$$ is equal to $$<cab$$, so both triangles are similar.
Thus: $$fracn6 = fracfrac5x48$$ and $$n = frac15x16 $$.
Now we can find the area of the smaller triangle as a function of $x$.
$$f(x) = 0.5 × frac15x16 × frac5x4$$
But, if I were to take the derivative of that function to find a maximum value, I would end up with a minimum value at $x = 0$, which is utterly irrational. What mistake have I done here?
derivatives triangle area applications
asked Jul 22 at 10:58
Stephen Alexander
1315
Pardon, I missed it, it's fde
– Stephen Alexander
Jul 22 at 12:59
I can see the triangle fde in your figure; but you have not clearly defined the set of admissible small triangles. As it stands the outer triangle does qualify, hence is trivially the largest.
– Christian Blatter
Jul 22 at 13:05
Nope, the question stated the maximum possible area of fde only.
– Stephen Alexander
Jul 22 at 13:44
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
So I've encountered a question where it requires me to find the maximum area of a small right angled triangle inside a bigger one. The question stated the dimensions of the big triangle and approved the parallelism of 2 lines in the form below:
Everything in black is given by the question, and otherwise (red) is assumed by me.
I've assumed the line $overleftrightarroweb$ equal to $x$ and $overleftrightarrowdb$ equal to y. Then I proved that $Delta$ $deb$ is similar to $Delta$ $abc$ by sharing the same right angle and having the angle $<deb$ corresponding to angle $<acb$, thus:
$$fracy6 = fracx8$$, then $$ y = frac3x4 $$.
$$overleftrightarrowde$$ would be equal to $$frac5x4 $$
Now I had to find the height $$overleftrightarrowfe$$ in terms of $$x$$
Since $$overleftrightarrowde$$ is parallel to $$overleftrightarrowac$$ , angles $$<edf$$ and $$<afd$$ are alternate angles, thus they are equal to each other, and since $$<afd$$ and $$<acb$$ are corresponding angles, $$<edf$$ is equal to $$<acb$$, and both $$Delta abc, Delta fed$$ have right angles, then we can infer that $$<efd$$ is equal to $$<cab$$, so both triangles are similar.
Thus: $$fracn6 = fracfrac5x48$$ and $$n = frac15x16 $$.
Now we can find the area of the smaller triangle as a function of $x$.
$$f(x) = 0.5 × frac15x16 × frac5x4$$
But, if I were to take the derivative of that function to find a maximum value, I would end up with a minimum value at $x = 0$, which is utterly irrational. What mistake have I done here?
derivatives triangle area applications
asked Jul 22 at 10:58
Stephen Alexander
1315
So I've encountered a question where it requires me to find the maximum area of a small right angled triangle inside a bigger one. The question stated the dimensions of the big triangle and approved the parallelism of 2 lines in the form below:
Everything in black is given by the question, and otherwise (red) is assumed by me.
I've assumed the line $overleftrightarroweb$ equal to $x$ and $overleftrightarrowdb$ equal to y. Then I proved that $Delta$ $deb$ is similar to $Delta$ $abc$ by sharing the same right angle and having the angle $<deb$ corresponding to angle $<acb$, thus:
$$fracy6 = fracx8$$, then $$ y = frac3x4 $$.
$$overleftrightarrowde$$ would be equal to $$frac5x4 $$
Now I had to find the height $$overleftrightarrowfe$$ in terms of $$x$$
Since $$overleftrightarrowde$$ is parallel to $$overleftrightarrowac$$ , angles $$<edf$$ and $$<afd$$ are alternate angles, thus they are equal to each other, and since $$<afd$$ and $$<acb$$ are corresponding angles, $$<edf$$ is equal to $$<acb$$, and both $$Delta abc, Delta fed$$ have right angles, then we can infer that $$<efd$$ is equal to $$<cab$$, so both triangles are similar.
Thus: $$fracn6 = fracfrac5x48$$ and $$n = frac15x16 $$.
Now we can find the area of the smaller triangle as a function of $x$.
$$f(x) = 0.5 × frac15x16 × frac5x4$$
But, if I were to take the derivative of that function to find a maximum value, I would end up with a minimum value at $x = 0$, which is utterly irrational. What mistake have I done here?
derivatives triangle area applications
asked Jul 22 at 10:58
Stephen Alexander
1315
asked Jul 22 at 10:58
Stephen Alexander
1315
asked Jul 22 at 10:58
Stephen Alexander
1315
asked Jul 22 at 10:58
Stephen Alexander
1315
1315
Pardon, I missed it, it's fde
– Stephen Alexander
Jul 22 at 12:59
I can see the triangle fde in your figure; but you have not clearly defined the set of admissible small triangles. As it stands the outer triangle does qualify, hence is trivially the largest.
– Christian Blatter
Jul 22 at 13:05
Nope, the question stated the maximum possible area of fde only.
– Stephen Alexander
Jul 22 at 13:44
add a comment |Â
Pardon, I missed it, it's fde
– Stephen Alexander
Jul 22 at 12:59
I can see the triangle fde in your figure; but you have not clearly defined the set of admissible small triangles. As it stands the outer triangle does qualify, hence is trivially the largest.
– Christian Blatter
Jul 22 at 13:05
Nope, the question stated the maximum possible area of fde only.
– Stephen Alexander
Jul 22 at 13:44
Pardon, I missed it, it's fde
– Stephen Alexander
Jul 22 at 12:59
Pardon, I missed it, it's fde
– Stephen Alexander
Jul 22 at 12:59
I can see the triangle fde in your figure; but you have not clearly defined the set of admissible small triangles. As it stands the outer triangle does qualify, hence is trivially the largest.
– Christian Blatter
Jul 22 at 13:05
I can see the triangle fde in your figure; but you have not clearly defined the set of admissible small triangles. As it stands the outer triangle does qualify, hence is trivially the largest.
– Christian Blatter
Jul 22 at 13:05
Nope, the question stated the maximum possible area of fde only.
– Stephen Alexander
Jul 22 at 13:44
Nope, the question stated the maximum possible area of fde only.
– Stephen Alexander
Jul 22 at 13:44
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
Hint:let $$ED=m$$ then we get:
$$A=frac12mn$$ where $$n=frac35(8-x)$$ and $$m=sqrtx^2+left(frac34xright)^2$$
We get then
$$A=frac38(8x-x^2)$$
And the maximum we get for $$x=4$$
answered Jul 22 at 11:24
Dr. Sonnhard Graubner
66.8k32659
I don't get it how did y become (3/5x) ? I mean, is my definition of y (y = 3x/4) incorrect? Also how did you get n? Can you elaborate please? Maybe I have some geometrical mistake
– Stephen Alexander
Jul 22 at 11:31
Yes it is $$fracyx=frac34$$ just a typo!
– Dr. Sonnhard Graubner
Jul 22 at 11:46
$$sin(alpha)=fracn8-x=frac6sqrt6^2+8^2$$ and $$alpha$$ denotes the angle at $$C$$
– Dr. Sonnhard Graubner
Jul 22 at 11:50
it is better now?
– Dr. Sonnhard Graubner
Jul 22 at 11:52
Note that the angle at $f$ is $fracpi2$
– Dr. Sonnhard Graubner
Jul 22 at 11:53
 |Â
show 1 more comment
up vote
1
down vote
$ overlinede = frac54x $ as you said,
since triangle $ cef$ and triangle $cab$ are similar triangles,
$ overlinece = 8-x $
$ 8-x:n=5:3 $
$5n = 24-3x$
$n = frac24-3x5$
$A = frac24-3x5timesfrac54xtimesfrac12 = frac24x-3x^24timesfrac12 =3x-frac38x^2$
$fractextdAtextdx = 3 - frac34x = 0$
$x = 4$
answered Jul 22 at 12:11
Pizzaroot
1056
Yeah, I see that you came with another method which works too, by using the similarity of two other triangles. But I have a question; why does my method give different answer? As you see, I used the theorem of triangles similarity twice, first time everything went ok, but the second time when I used it between triangles abc and fde, in order to fetch out "n", it gave me unsatisfying results. Utterly different. Why?
– Stephen Alexander
Jul 22 at 12:20
@StephenAlexander because $overlinedf$ and $overlinebc$ are not parallel
– Pizzaroot
Jul 22 at 12:27
Hmm so you are implying that adf and abc aren't similar triangles? Makes sense now. So let's say that if df was connecting the midpoints of ab and ac then it would be parallel with the third line right? Otherwise any line that comes above or below isn't parallel to the third line?
– Stephen Alexander
Jul 22 at 12:37
yeah but it wouldn't be the midpoints of ab and ac since the diagram only showed that ac and de are parallel and the angle def is right angle. df and bc could be parallel, but it wasn't mentioned in the question
– Pizzaroot
Jul 22 at 12:46
True. It doesn't even connect the midpoints too. Appreciated
– Stephen Alexander
Jul 22 at 12:53
add a comment |Â
up vote
1
down vote
Let $|BC|=a=8$cm, $|AB|=c=6$cm,
then $|AC|=b=10$cm.
beginalign
S_triangle FED(x)&=tfrac12|DE|cdot|FE|
\
&=tfrac12cdotfracxcosgammacdot(a-x)cosalpha
\
&=
tfrac12cdotfracxtfracabcdot(a-x)cdottfraccb
\
S_triangle FED(x)&=fracc2a,(ax-x^2)
.
endalign
beginalign
S'_triangle FED(x)&=
fracc2a,(a-2x)
,\
S''_triangle FED(x)&=
-fracca
<0quad forall x
,
endalign
hence $x=fraca2$ provides the maximum
beginalign
S_triangle FED(x)_max&=
S_triangle FED(tfraca2)
=fracac8
=6,mathrmcm^2
.
endalign
answered Jul 22 at 14:50
g.kov
5,5321717
Quick and neat method.
– Stephen Alexander
Jul 22 at 17:27
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint:let $$ED=m$$ then we get:
$$A=frac12mn$$ where $$n=frac35(8-x)$$ and $$m=sqrtx^2+left(frac34xright)^2$$
We get then
$$A=frac38(8x-x^2)$$
And the maximum we get for $$x=4$$
answered Jul 22 at 11:24
Dr. Sonnhard Graubner
66.8k32659
I don't get it how did y become (3/5x) ? I mean, is my definition of y (y = 3x/4) incorrect? Also how did you get n? Can you elaborate please? Maybe I have some geometrical mistake
– Stephen Alexander
Jul 22 at 11:31
Yes it is $$fracyx=frac34$$ just a typo!
– Dr. Sonnhard Graubner
Jul 22 at 11:46
$$sin(alpha)=fracn8-x=frac6sqrt6^2+8^2$$ and $$alpha$$ denotes the angle at $$C$$
– Dr. Sonnhard Graubner
Jul 22 at 11:50
it is better now?
– Dr. Sonnhard Graubner
Jul 22 at 11:52
Note that the angle at $f$ is $fracpi2$
– Dr. Sonnhard Graubner
Jul 22 at 11:53
 |Â
show 1 more comment
up vote
1
down vote
accepted
Hint:let $$ED=m$$ then we get:
$$A=frac12mn$$ where $$n=frac35(8-x)$$ and $$m=sqrtx^2+left(frac34xright)^2$$
We get then
$$A=frac38(8x-x^2)$$
And the maximum we get for $$x=4$$
answered Jul 22 at 11:24
Dr. Sonnhard Graubner
66.8k32659
I don't get it how did y become (3/5x) ? I mean, is my definition of y (y = 3x/4) incorrect? Also how did you get n? Can you elaborate please? Maybe I have some geometrical mistake
– Stephen Alexander
Jul 22 at 11:31
Yes it is $$fracyx=frac34$$ just a typo!
– Dr. Sonnhard Graubner
Jul 22 at 11:46
$$sin(alpha)=fracn8-x=frac6sqrt6^2+8^2$$ and $$alpha$$ denotes the angle at $$C$$
– Dr. Sonnhard Graubner
Jul 22 at 11:50
it is better now?
– Dr. Sonnhard Graubner
Jul 22 at 11:52
Note that the angle at $f$ is $fracpi2$
– Dr. Sonnhard Graubner
Jul 22 at 11:53
 |Â
show 1 more comment
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint:let $$ED=m$$ then we get:
$$A=frac12mn$$ where $$n=frac35(8-x)$$ and $$m=sqrtx^2+left(frac34xright)^2$$
We get then
$$A=frac38(8x-x^2)$$
And the maximum we get for $$x=4$$
answered Jul 22 at 11:24
Dr. Sonnhard Graubner
66.8k32659
Hint:let $$ED=m$$ then we get:
$$A=frac12mn$$ where $$n=frac35(8-x)$$ and $$m=sqrtx^2+left(frac34xright)^2$$
We get then
$$A=frac38(8x-x^2)$$
And the maximum we get for $$x=4$$
answered Jul 22 at 11:24
Dr. Sonnhard Graubner
66.8k32659
edited Jul 22 at 11:52
answered Jul 22 at 11:24
Dr. Sonnhard Graubner
66.8k32659
answered Jul 22 at 11:24
Dr. Sonnhard Graubner
66.8k32659
answered Jul 22 at 11:24
Dr. Sonnhard Graubner
66.8k32659
66.8k32659
I don't get it how did y become (3/5x) ? I mean, is my definition of y (y = 3x/4) incorrect? Also how did you get n? Can you elaborate please? Maybe I have some geometrical mistake
– Stephen Alexander
Jul 22 at 11:31
Yes it is $$fracyx=frac34$$ just a typo!
– Dr. Sonnhard Graubner
Jul 22 at 11:46
$$sin(alpha)=fracn8-x=frac6sqrt6^2+8^2$$ and $$alpha$$ denotes the angle at $$C$$
– Dr. Sonnhard Graubner
Jul 22 at 11:50
it is better now?
– Dr. Sonnhard Graubner
Jul 22 at 11:52
Note that the angle at $f$ is $fracpi2$
– Dr. Sonnhard Graubner
Jul 22 at 11:53
 |Â
show 1 more comment
I don't get it how did y become (3/5x) ? I mean, is my definition of y (y = 3x/4) incorrect? Also how did you get n? Can you elaborate please? Maybe I have some geometrical mistake
– Stephen Alexander
Jul 22 at 11:31
Yes it is $$fracyx=frac34$$ just a typo!
– Dr. Sonnhard Graubner
Jul 22 at 11:46
$$sin(alpha)=fracn8-x=frac6sqrt6^2+8^2$$ and $$alpha$$ denotes the angle at $$C$$
– Dr. Sonnhard Graubner
Jul 22 at 11:50
it is better now?
– Dr. Sonnhard Graubner
Jul 22 at 11:52
Note that the angle at $f$ is $fracpi2$
– Dr. Sonnhard Graubner
Jul 22 at 11:53
I don't get it how did y become (3/5x) ? I mean, is my definition of y (y = 3x/4) incorrect? Also how did you get n? Can you elaborate please? Maybe I have some geometrical mistake
– Stephen Alexander
Jul 22 at 11:31
I don't get it how did y become (3/5x) ? I mean, is my definition of y (y = 3x/4) incorrect? Also how did you get n? Can you elaborate please? Maybe I have some geometrical mistake
– Stephen Alexander
Jul 22 at 11:31
Yes it is $$fracyx=frac34$$ just a typo!
– Dr. Sonnhard Graubner
Jul 22 at 11:46
Yes it is $$fracyx=frac34$$ just a typo!
– Dr. Sonnhard Graubner
Jul 22 at 11:46
$$sin(alpha)=fracn8-x=frac6sqrt6^2+8^2$$ and $$alpha$$ denotes the angle at $$C$$
– Dr. Sonnhard Graubner
Jul 22 at 11:50
$$sin(alpha)=fracn8-x=frac6sqrt6^2+8^2$$ and $$alpha$$ denotes the angle at $$C$$
– Dr. Sonnhard Graubner
Jul 22 at 11:50
it is better now?
– Dr. Sonnhard Graubner
Jul 22 at 11:52
it is better now?
– Dr. Sonnhard Graubner
Jul 22 at 11:52
Note that the angle at $f$ is $fracpi2$
– Dr. Sonnhard Graubner
Jul 22 at 11:53
Note that the angle at $f$ is $fracpi2$
– Dr. Sonnhard Graubner
Jul 22 at 11:53
 |Â
show 1 more comment
up vote
1
down vote
$ overlinede = frac54x $ as you said,
since triangle $ cef$ and triangle $cab$ are similar triangles,
$ overlinece = 8-x $
$ 8-x:n=5:3 $
$5n = 24-3x$
$n = frac24-3x5$
$A = frac24-3x5timesfrac54xtimesfrac12 = frac24x-3x^24timesfrac12 =3x-frac38x^2$
$fractextdAtextdx = 3 - frac34x = 0$
$x = 4$
answered Jul 22 at 12:11
Pizzaroot
1056
Yeah, I see that you came with another method which works too, by using the similarity of two other triangles. But I have a question; why does my method give different answer? As you see, I used the theorem of triangles similarity twice, first time everything went ok, but the second time when I used it between triangles abc and fde, in order to fetch out "n", it gave me unsatisfying results. Utterly different. Why?
– Stephen Alexander
Jul 22 at 12:20
@StephenAlexander because $overlinedf$ and $overlinebc$ are not parallel
– Pizzaroot
Jul 22 at 12:27
Hmm so you are implying that adf and abc aren't similar triangles? Makes sense now. So let's say that if df was connecting the midpoints of ab and ac then it would be parallel with the third line right? Otherwise any line that comes above or below isn't parallel to the third line?
– Stephen Alexander
Jul 22 at 12:37
yeah but it wouldn't be the midpoints of ab and ac since the diagram only showed that ac and de are parallel and the angle def is right angle. df and bc could be parallel, but it wasn't mentioned in the question
– Pizzaroot
Jul 22 at 12:46
True. It doesn't even connect the midpoints too. Appreciated
– Stephen Alexander
Jul 22 at 12:53
add a comment |Â
up vote
1
down vote
$ overlinede = frac54x $ as you said,
since triangle $ cef$ and triangle $cab$ are similar triangles,
$ overlinece = 8-x $
$ 8-x:n=5:3 $
$5n = 24-3x$
$n = frac24-3x5$
$A = frac24-3x5timesfrac54xtimesfrac12 = frac24x-3x^24timesfrac12 =3x-frac38x^2$
$fractextdAtextdx = 3 - frac34x = 0$
$x = 4$
answered Jul 22 at 12:11
Pizzaroot
1056
Yeah, I see that you came with another method which works too, by using the similarity of two other triangles. But I have a question; why does my method give different answer? As you see, I used the theorem of triangles similarity twice, first time everything went ok, but the second time when I used it between triangles abc and fde, in order to fetch out "n", it gave me unsatisfying results. Utterly different. Why?
– Stephen Alexander
Jul 22 at 12:20
@StephenAlexander because $overlinedf$ and $overlinebc$ are not parallel
– Pizzaroot
Jul 22 at 12:27
Hmm so you are implying that adf and abc aren't similar triangles? Makes sense now. So let's say that if df was connecting the midpoints of ab and ac then it would be parallel with the third line right? Otherwise any line that comes above or below isn't parallel to the third line?
– Stephen Alexander
Jul 22 at 12:37
yeah but it wouldn't be the midpoints of ab and ac since the diagram only showed that ac and de are parallel and the angle def is right angle. df and bc could be parallel, but it wasn't mentioned in the question
– Pizzaroot
Jul 22 at 12:46
True. It doesn't even connect the midpoints too. Appreciated
– Stephen Alexander
Jul 22 at 12:53
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$ overlinede = frac54x $ as you said,
since triangle $ cef$ and triangle $cab$ are similar triangles,
$ overlinece = 8-x $
$ 8-x:n=5:3 $
$5n = 24-3x$
$n = frac24-3x5$
$A = frac24-3x5timesfrac54xtimesfrac12 = frac24x-3x^24timesfrac12 =3x-frac38x^2$
$fractextdAtextdx = 3 - frac34x = 0$
$x = 4$
answered Jul 22 at 12:11
Pizzaroot
1056
$ overlinede = frac54x $ as you said,
since triangle $ cef$ and triangle $cab$ are similar triangles,
$ overlinece = 8-x $
$ 8-x:n=5:3 $
$5n = 24-3x$
$n = frac24-3x5$
$A = frac24-3x5timesfrac54xtimesfrac12 = frac24x-3x^24timesfrac12 =3x-frac38x^2$
$fractextdAtextdx = 3 - frac34x = 0$
$x = 4$
answered Jul 22 at 12:11
Pizzaroot
1056
answered Jul 22 at 12:11
Pizzaroot
1056
answered Jul 22 at 12:11
Pizzaroot
1056
answered Jul 22 at 12:11
Pizzaroot
1056
1056
Yeah, I see that you came with another method which works too, by using the similarity of two other triangles. But I have a question; why does my method give different answer? As you see, I used the theorem of triangles similarity twice, first time everything went ok, but the second time when I used it between triangles abc and fde, in order to fetch out "n", it gave me unsatisfying results. Utterly different. Why?
– Stephen Alexander
Jul 22 at 12:20
@StephenAlexander because $overlinedf$ and $overlinebc$ are not parallel
– Pizzaroot
Jul 22 at 12:27
Hmm so you are implying that adf and abc aren't similar triangles? Makes sense now. So let's say that if df was connecting the midpoints of ab and ac then it would be parallel with the third line right? Otherwise any line that comes above or below isn't parallel to the third line?
– Stephen Alexander
Jul 22 at 12:37
yeah but it wouldn't be the midpoints of ab and ac since the diagram only showed that ac and de are parallel and the angle def is right angle. df and bc could be parallel, but it wasn't mentioned in the question
– Pizzaroot
Jul 22 at 12:46
True. It doesn't even connect the midpoints too. Appreciated
– Stephen Alexander
Jul 22 at 12:53
add a comment |Â
Yeah, I see that you came with another method which works too, by using the similarity of two other triangles. But I have a question; why does my method give different answer? As you see, I used the theorem of triangles similarity twice, first time everything went ok, but the second time when I used it between triangles abc and fde, in order to fetch out "n", it gave me unsatisfying results. Utterly different. Why?
– Stephen Alexander
Jul 22 at 12:20
@StephenAlexander because $overlinedf$ and $overlinebc$ are not parallel
– Pizzaroot
Jul 22 at 12:27
Hmm so you are implying that adf and abc aren't similar triangles? Makes sense now. So let's say that if df was connecting the midpoints of ab and ac then it would be parallel with the third line right? Otherwise any line that comes above or below isn't parallel to the third line?
– Stephen Alexander
Jul 22 at 12:37
yeah but it wouldn't be the midpoints of ab and ac since the diagram only showed that ac and de are parallel and the angle def is right angle. df and bc could be parallel, but it wasn't mentioned in the question
– Pizzaroot
Jul 22 at 12:46
True. It doesn't even connect the midpoints too. Appreciated
– Stephen Alexander
Jul 22 at 12:53
Yeah, I see that you came with another method which works too, by using the similarity of two other triangles. But I have a question; why does my method give different answer? As you see, I used the theorem of triangles similarity twice, first time everything went ok, but the second time when I used it between triangles abc and fde, in order to fetch out "n", it gave me unsatisfying results. Utterly different. Why?
– Stephen Alexander
Jul 22 at 12:20
Yeah, I see that you came with another method which works too, by using the similarity of two other triangles. But I have a question; why does my method give different answer? As you see, I used the theorem of triangles similarity twice, first time everything went ok, but the second time when I used it between triangles abc and fde, in order to fetch out "n", it gave me unsatisfying results. Utterly different. Why?
– Stephen Alexander
Jul 22 at 12:20
@StephenAlexander because $overlinedf$ and $overlinebc$ are not parallel
– Pizzaroot
Jul 22 at 12:27
@StephenAlexander because $overlinedf$ and $overlinebc$ are not parallel
– Pizzaroot
Jul 22 at 12:27
Hmm so you are implying that adf and abc aren't similar triangles? Makes sense now. So let's say that if df was connecting the midpoints of ab and ac then it would be parallel with the third line right? Otherwise any line that comes above or below isn't parallel to the third line?
– Stephen Alexander
Jul 22 at 12:37
Hmm so you are implying that adf and abc aren't similar triangles? Makes sense now. So let's say that if df was connecting the midpoints of ab and ac then it would be parallel with the third line right? Otherwise any line that comes above or below isn't parallel to the third line?
– Stephen Alexander
Jul 22 at 12:37
yeah but it wouldn't be the midpoints of ab and ac since the diagram only showed that ac and de are parallel and the angle def is right angle. df and bc could be parallel, but it wasn't mentioned in the question
– Pizzaroot
Jul 22 at 12:46
yeah but it wouldn't be the midpoints of ab and ac since the diagram only showed that ac and de are parallel and the angle def is right angle. df and bc could be parallel, but it wasn't mentioned in the question
– Pizzaroot
Jul 22 at 12:46
True. It doesn't even connect the midpoints too. Appreciated
– Stephen Alexander
Jul 22 at 12:53
True. It doesn't even connect the midpoints too. Appreciated
– Stephen Alexander
Jul 22 at 12:53
add a comment |Â
up vote
1
down vote
Let $|BC|=a=8$cm, $|AB|=c=6$cm,
then $|AC|=b=10$cm.
beginalign
S_triangle FED(x)&=tfrac12|DE|cdot|FE|
\
&=tfrac12cdotfracxcosgammacdot(a-x)cosalpha
\
&=
tfrac12cdotfracxtfracabcdot(a-x)cdottfraccb
\
S_triangle FED(x)&=fracc2a,(ax-x^2)
.
endalign
beginalign
S'_triangle FED(x)&=
fracc2a,(a-2x)
,\
S''_triangle FED(x)&=
-fracca
<0quad forall x
,
endalign
hence $x=fraca2$ provides the maximum
beginalign
S_triangle FED(x)_max&=
S_triangle FED(tfraca2)
=fracac8
=6,mathrmcm^2
.
endalign
answered Jul 22 at 14:50
g.kov
5,5321717
Quick and neat method.
– Stephen Alexander
Jul 22 at 17:27
add a comment |Â
up vote
1
down vote
Let $|BC|=a=8$cm, $|AB|=c=6$cm,
then $|AC|=b=10$cm.
beginalign
S_triangle FED(x)&=tfrac12|DE|cdot|FE|
\
&=tfrac12cdotfracxcosgammacdot(a-x)cosalpha
\
&=
tfrac12cdotfracxtfracabcdot(a-x)cdottfraccb
\
S_triangle FED(x)&=fracc2a,(ax-x^2)
.
endalign
beginalign
S'_triangle FED(x)&=
fracc2a,(a-2x)
,\
S''_triangle FED(x)&=
-fracca
<0quad forall x
,
endalign
hence $x=fraca2$ provides the maximum
beginalign
S_triangle FED(x)_max&=
S_triangle FED(tfraca2)
=fracac8
=6,mathrmcm^2
.
endalign
answered Jul 22 at 14:50
g.kov
5,5321717
Quick and neat method.
– Stephen Alexander
Jul 22 at 17:27
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $|BC|=a=8$cm, $|AB|=c=6$cm,
then $|AC|=b=10$cm.
beginalign
S_triangle FED(x)&=tfrac12|DE|cdot|FE|
\
&=tfrac12cdotfracxcosgammacdot(a-x)cosalpha
\
&=
tfrac12cdotfracxtfracabcdot(a-x)cdottfraccb
\
S_triangle FED(x)&=fracc2a,(ax-x^2)
.
endalign
beginalign
S'_triangle FED(x)&=
fracc2a,(a-2x)
,\
S''_triangle FED(x)&=
-fracca
<0quad forall x
,
endalign
hence $x=fraca2$ provides the maximum
beginalign
S_triangle FED(x)_max&=
S_triangle FED(tfraca2)
=fracac8
=6,mathrmcm^2
.
endalign
answered Jul 22 at 14:50
g.kov
5,5321717
Let $|BC|=a=8$cm, $|AB|=c=6$cm,
then $|AC|=b=10$cm.
beginalign
S_triangle FED(x)&=tfrac12|DE|cdot|FE|
\
&=tfrac12cdotfracxcosgammacdot(a-x)cosalpha
\
&=
tfrac12cdotfracxtfracabcdot(a-x)cdottfraccb
\
S_triangle FED(x)&=fracc2a,(ax-x^2)
.
endalign
beginalign
S'_triangle FED(x)&=
fracc2a,(a-2x)
,\
S''_triangle FED(x)&=
-fracca
<0quad forall x
,
endalign
hence $x=fraca2$ provides the maximum
beginalign
S_triangle FED(x)_max&=
S_triangle FED(tfraca2)
=fracac8
=6,mathrmcm^2
.
endalign
answered Jul 22 at 14:50
g.kov
5,5321717
answered Jul 22 at 14:50
g.kov
5,5321717
answered Jul 22 at 14:50
g.kov
5,5321717
answered Jul 22 at 14:50
g.kov
5,5321717
5,5321717
Quick and neat method.
– Stephen Alexander
Jul 22 at 17:27
add a comment |Â
Quick and neat method.
– Stephen Alexander
Jul 22 at 17:27
Quick and neat method.
– Stephen Alexander
Jul 22 at 17:27
Quick and neat method.
– Stephen Alexander
Jul 22 at 17:27
add a comment |Â
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Pardon, I missed it, it's fde
– Stephen Alexander
Jul 22 at 12:59
I can see the triangle fde in your figure; but you have not clearly defined the set of admissible small triangles. As it stands the outer triangle does qualify, hence is trivially the largest.
– Christian Blatter
Jul 22 at 13:05
Nope, the question stated the maximum possible area of fde only.
– Stephen Alexander
Jul 22 at 13:44