particular case of Pythagoras' theorem
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According to Pythagoras' theorem the hypotenuse of a rectangle triangle is calculated:
$$c=sqrt{a^2+b^2}$$
There is a similar rectangle triangle that fulfills:
$$C^2=A+B$$
For it:
1) We find the relation of the legs:
$r={aover b}$
2) We raise r squared and calculate:
$d={1over r^2+1}$
3) We obtain a first triangle similar to d, e, f with less leg equal to d, and e, greater leg:
$e=dr$
4) We obtain a second triangle also similar g, h, i where:
$g=e$ and $h=1−d$
5) Now we add the minor and major legs of d, e, f and h, i, j to obtain the ABC triangle:
$A=d+e$ and $B=h+e$
6) This ABC triangle complies with:
$C=sqrt{A^2+B^2}=sqrt{A+B}$
This I have checked in all tests, but I could not make a demonstration.
Note that the calculated C hypotenuse is equal to the sum of the sines of the angles of the triangle not straight.
geometry
asked Nov 8 at 16:17
user167434
113
|
show 5 more comments
up vote
1
down vote
favorite
According to Pythagoras' theorem the hypotenuse of a rectangle triangle is calculated:
$$c=sqrt{a^2+b^2}$$
There is a similar rectangle triangle that fulfills:
$$C^2=A+B$$
For it:
1) We find the relation of the legs:
$r={aover b}$
2) We raise r squared and calculate:
$d={1over r^2+1}$
3) We obtain a first triangle similar to d, e, f with less leg equal to d, and e, greater leg:
$e=dr$
4) We obtain a second triangle also similar g, h, i where:
$g=e$ and $h=1−d$
5) Now we add the minor and major legs of d, e, f and h, i, j to obtain the ABC triangle:
$A=d+e$ and $B=h+e$
6) This ABC triangle complies with:
$C=sqrt{A^2+B^2}=sqrt{A+B}$
This I have checked in all tests, but I could not make a demonstration.
Note that the calculated C hypotenuse is equal to the sum of the sines of the angles of the triangle not straight.
geometry
asked Nov 8 at 16:17
user167434
113
Please take a look at math.meta.stackexchange.com/questions/5020/… to learn how to format your text.
– KM101
Nov 8 at 16:19
2
Pythagoras' theorem is $c^2=a^2+b^2$, so $c=sqrt{a^2+b^2}$, not $c^2$
– Aleksa
Nov 8 at 16:22
This is confusing: $C^2=sqrt{A+B}$ .... $A, B$ are not squared?
– user376343
Nov 8 at 16:31
It's hard to understand what you mean, since you've got Pythagoras's theorem wrong. Please try to restate your question.
– saulspatz
Nov 8 at 16:31
1
So they need to be squared. But above all, what is the question??
– NickD
Nov 9 at 15:45
|
show 5 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
According to Pythagoras' theorem the hypotenuse of a rectangle triangle is calculated:
$$c=sqrt{a^2+b^2}$$
There is a similar rectangle triangle that fulfills:
$$C^2=A+B$$
For it:
1) We find the relation of the legs:
$r={aover b}$
2) We raise r squared and calculate:
$d={1over r^2+1}$
3) We obtain a first triangle similar to d, e, f with less leg equal to d, and e, greater leg:
$e=dr$
4) We obtain a second triangle also similar g, h, i where:
$g=e$ and $h=1−d$
5) Now we add the minor and major legs of d, e, f and h, i, j to obtain the ABC triangle:
$A=d+e$ and $B=h+e$
6) This ABC triangle complies with:
$C=sqrt{A^2+B^2}=sqrt{A+B}$
This I have checked in all tests, but I could not make a demonstration.
Note that the calculated C hypotenuse is equal to the sum of the sines of the angles of the triangle not straight.
geometry
asked Nov 8 at 16:17
user167434
113
According to Pythagoras' theorem the hypotenuse of a rectangle triangle is calculated:
$$c=sqrt{a^2+b^2}$$
There is a similar rectangle triangle that fulfills:
$$C^2=A+B$$
For it:
1) We find the relation of the legs:
$r={aover b}$
2) We raise r squared and calculate:
$d={1over r^2+1}$
3) We obtain a first triangle similar to d, e, f with less leg equal to d, and e, greater leg:
$e=dr$
4) We obtain a second triangle also similar g, h, i where:
$g=e$ and $h=1−d$
5) Now we add the minor and major legs of d, e, f and h, i, j to obtain the ABC triangle:
$A=d+e$ and $B=h+e$
6) This ABC triangle complies with:
$C=sqrt{A^2+B^2}=sqrt{A+B}$
This I have checked in all tests, but I could not make a demonstration.
Note that the calculated C hypotenuse is equal to the sum of the sines of the angles of the triangle not straight.
geometry
geometry
asked Nov 8 at 16:17
user167434
113
asked Nov 8 at 16:17
user167434
113
edited Nov 14 at 12:27
asked Nov 8 at 16:17
user167434
113
asked Nov 8 at 16:17
user167434
113
asked Nov 8 at 16:17
user167434
113
113
Please take a look at math.meta.stackexchange.com/questions/5020/… to learn how to format your text.
– KM101
Nov 8 at 16:19
2
Pythagoras' theorem is $c^2=a^2+b^2$, so $c=sqrt{a^2+b^2}$, not $c^2$
– Aleksa
Nov 8 at 16:22
This is confusing: $C^2=sqrt{A+B}$ .... $A, B$ are not squared?
– user376343
Nov 8 at 16:31
It's hard to understand what you mean, since you've got Pythagoras's theorem wrong. Please try to restate your question.
– saulspatz
Nov 8 at 16:31
1
So they need to be squared. But above all, what is the question??
– NickD
Nov 9 at 15:45
|
show 5 more comments
Please take a look at math.meta.stackexchange.com/questions/5020/… to learn how to format your text.
– KM101
Nov 8 at 16:19
2
Pythagoras' theorem is $c^2=a^2+b^2$, so $c=sqrt{a^2+b^2}$, not $c^2$
– Aleksa
Nov 8 at 16:22
This is confusing: $C^2=sqrt{A+B}$ .... $A, B$ are not squared?
– user376343
Nov 8 at 16:31
It's hard to understand what you mean, since you've got Pythagoras's theorem wrong. Please try to restate your question.
– saulspatz
Nov 8 at 16:31
1
So they need to be squared. But above all, what is the question??
– NickD
Nov 9 at 15:45
Please take a look at math.meta.stackexchange.com/questions/5020/… to learn how to format your text.
– KM101
Nov 8 at 16:19
Please take a look at math.meta.stackexchange.com/questions/5020/… to learn how to format your text.
– KM101
Nov 8 at 16:19
2
2
Pythagoras' theorem is $c^2=a^2+b^2$, so $c=sqrt{a^2+b^2}$, not $c^2$
– Aleksa
Nov 8 at 16:22
Pythagoras' theorem is $c^2=a^2+b^2$, so $c=sqrt{a^2+b^2}$, not $c^2$
– Aleksa
Nov 8 at 16:22
This is confusing: $C^2=sqrt{A+B}$ .... $A, B$ are not squared?
– user376343
Nov 8 at 16:31
This is confusing: $C^2=sqrt{A+B}$ .... $A, B$ are not squared?
– user376343
Nov 8 at 16:31
It's hard to understand what you mean, since you've got Pythagoras's theorem wrong. Please try to restate your question.
– saulspatz
Nov 8 at 16:31
It's hard to understand what you mean, since you've got Pythagoras's theorem wrong. Please try to restate your question.
– saulspatz
Nov 8 at 16:31
1
1
So they need to be squared. But above all, what is the question??
– NickD
Nov 9 at 15:45
So they need to be squared. But above all, what is the question??
– NickD
Nov 9 at 15:45
|
show 5 more comments
1 Answer
1
active
oldest
votes
up vote
0
down vote
I believe the question can be simplified to find particular cases of $$sqrt{A^2+B^2}=sqrt{A+B}$$ We can square both sides and move $A$ and $B^2$ to their other sides, getting $$A^2-A=B-B^2$$ If we replace $B-B^2$ with $C$ then we can create the quadratic equation $$A^2-A-C=0$$ and using the quadratic formula and replacing $C$ we obtain $$A=frac{1pmsqrt{1+4(B-B^2)}}2$$
answered Nov 14 at 12:30
Kyky
399113
Consider A = 0.84 and b = 1.12 (values obtained for the relation of legs $4over3$) $$A^2-A = -0.1344$$ and $$ B^2-B = 0.1344$$ Then the equality $$A^2-A=B^2-B $$ is not met. On the other hand, if you do it using $$A^2-A = B-B^2$$
– user167434
Nov 15 at 10:09
@user167434 Thanks for spotting that error there. Yes, it should be $A^2-A=B-B^2$ than the other way around.
– Kyky
Nov 16 at 13:41
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I believe the question can be simplified to find particular cases of $$sqrt{A^2+B^2}=sqrt{A+B}$$ We can square both sides and move $A$ and $B^2$ to their other sides, getting $$A^2-A=B-B^2$$ If we replace $B-B^2$ with $C$ then we can create the quadratic equation $$A^2-A-C=0$$ and using the quadratic formula and replacing $C$ we obtain $$A=frac{1pmsqrt{1+4(B-B^2)}}2$$
answered Nov 14 at 12:30
Kyky
399113
Consider A = 0.84 and b = 1.12 (values obtained for the relation of legs $4over3$) $$A^2-A = -0.1344$$ and $$ B^2-B = 0.1344$$ Then the equality $$A^2-A=B^2-B $$ is not met. On the other hand, if you do it using $$A^2-A = B-B^2$$
– user167434
Nov 15 at 10:09
@user167434 Thanks for spotting that error there. Yes, it should be $A^2-A=B-B^2$ than the other way around.
– Kyky
Nov 16 at 13:41
add a comment |
up vote
0
down vote
I believe the question can be simplified to find particular cases of $$sqrt{A^2+B^2}=sqrt{A+B}$$ We can square both sides and move $A$ and $B^2$ to their other sides, getting $$A^2-A=B-B^2$$ If we replace $B-B^2$ with $C$ then we can create the quadratic equation $$A^2-A-C=0$$ and using the quadratic formula and replacing $C$ we obtain $$A=frac{1pmsqrt{1+4(B-B^2)}}2$$
answered Nov 14 at 12:30
Kyky
399113
Consider A = 0.84 and b = 1.12 (values obtained for the relation of legs $4over3$) $$A^2-A = -0.1344$$ and $$ B^2-B = 0.1344$$ Then the equality $$A^2-A=B^2-B $$ is not met. On the other hand, if you do it using $$A^2-A = B-B^2$$
– user167434
Nov 15 at 10:09
@user167434 Thanks for spotting that error there. Yes, it should be $A^2-A=B-B^2$ than the other way around.
– Kyky
Nov 16 at 13:41
add a comment |
up vote
0
down vote
up vote
0
down vote
I believe the question can be simplified to find particular cases of $$sqrt{A^2+B^2}=sqrt{A+B}$$ We can square both sides and move $A$ and $B^2$ to their other sides, getting $$A^2-A=B-B^2$$ If we replace $B-B^2$ with $C$ then we can create the quadratic equation $$A^2-A-C=0$$ and using the quadratic formula and replacing $C$ we obtain $$A=frac{1pmsqrt{1+4(B-B^2)}}2$$
answered Nov 14 at 12:30
Kyky
399113
I believe the question can be simplified to find particular cases of $$sqrt{A^2+B^2}=sqrt{A+B}$$ We can square both sides and move $A$ and $B^2$ to their other sides, getting $$A^2-A=B-B^2$$ If we replace $B-B^2$ with $C$ then we can create the quadratic equation $$A^2-A-C=0$$ and using the quadratic formula and replacing $C$ we obtain $$A=frac{1pmsqrt{1+4(B-B^2)}}2$$
answered Nov 14 at 12:30
Kyky
399113
edited Nov 16 at 13:43
answered Nov 14 at 12:30
Kyky
399113
answered Nov 14 at 12:30
Kyky
399113
answered Nov 14 at 12:30
Kyky
399113
399113
Consider A = 0.84 and b = 1.12 (values obtained for the relation of legs $4over3$) $$A^2-A = -0.1344$$ and $$ B^2-B = 0.1344$$ Then the equality $$A^2-A=B^2-B $$ is not met. On the other hand, if you do it using $$A^2-A = B-B^2$$
– user167434
Nov 15 at 10:09
@user167434 Thanks for spotting that error there. Yes, it should be $A^2-A=B-B^2$ than the other way around.
– Kyky
Nov 16 at 13:41
add a comment |
Consider A = 0.84 and b = 1.12 (values obtained for the relation of legs $4over3$) $$A^2-A = -0.1344$$ and $$ B^2-B = 0.1344$$ Then the equality $$A^2-A=B^2-B $$ is not met. On the other hand, if you do it using $$A^2-A = B-B^2$$
– user167434
Nov 15 at 10:09
@user167434 Thanks for spotting that error there. Yes, it should be $A^2-A=B-B^2$ than the other way around.
– Kyky
Nov 16 at 13:41
Consider A = 0.84 and b = 1.12 (values obtained for the relation of legs $4over3$) $$A^2-A = -0.1344$$ and $$ B^2-B = 0.1344$$ Then the equality $$A^2-A=B^2-B $$ is not met. On the other hand, if you do it using $$A^2-A = B-B^2$$
– user167434
Nov 15 at 10:09
Consider A = 0.84 and b = 1.12 (values obtained for the relation of legs $4over3$) $$A^2-A = -0.1344$$ and $$ B^2-B = 0.1344$$ Then the equality $$A^2-A=B^2-B $$ is not met. On the other hand, if you do it using $$A^2-A = B-B^2$$
– user167434
Nov 15 at 10:09
@user167434 Thanks for spotting that error there. Yes, it should be $A^2-A=B-B^2$ than the other way around.
– Kyky
Nov 16 at 13:41
@user167434 Thanks for spotting that error there. Yes, it should be $A^2-A=B-B^2$ than the other way around.
– Kyky
Nov 16 at 13:41
add a comment |
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Please take a look at math.meta.stackexchange.com/questions/5020/… to learn how to format your text.
– KM101
Nov 8 at 16:19
2
Pythagoras' theorem is $c^2=a^2+b^2$, so $c=sqrt{a^2+b^2}$, not $c^2$
– Aleksa
Nov 8 at 16:22
This is confusing: $C^2=sqrt{A+B}$ .... $A, B$ are not squared?
– user376343
Nov 8 at 16:31
It's hard to understand what you mean, since you've got Pythagoras's theorem wrong. Please try to restate your question.
– saulspatz
Nov 8 at 16:31
1
So they need to be squared. But above all, what is the question??
– NickD
Nov 9 at 15:45