what is wrong with following right triangle's hypotenuse's square reasoning [closed]
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In the image given in the link, the square of the hypotenuse is 2ab. But it can't be true. I can't be true. I can't seem to find the reason. any help would be appreciated.
triangle
asked Dec 8 '18 at 19:14
Nasir AzizNasir Aziz
31
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closed as unclear what you're asking by Scientifica, Lord Shark the Unknown, Cesareo, José Carlos Santos, amWhy Dec 9 '18 at 16:53
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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$begingroup$
In the image given in the link, the square of the hypotenuse is 2ab. But it can't be true. I can't be true. I can't seem to find the reason. any help would be appreciated.
triangle
asked Dec 8 '18 at 19:14
Nasir AzizNasir Aziz
31
$endgroup$
closed as unclear what you're asking by Scientifica, Lord Shark the Unknown, Cesareo, José Carlos Santos, amWhy Dec 9 '18 at 16:53
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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Is that figure meant to be a square? If so...the diagonals of a square are equal so we'd need $a=b$. If it's not meant to be a square then the area of the figure is not $c^2$.
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– lulu
Dec 8 '18 at 19:17
add a comment |
$begingroup$
In the image given in the link, the square of the hypotenuse is 2ab. But it can't be true. I can't be true. I can't seem to find the reason. any help would be appreciated.
triangle
asked Dec 8 '18 at 19:14
Nasir AzizNasir Aziz
31
$endgroup$
In the image given in the link, the square of the hypotenuse is 2ab. But it can't be true. I can't be true. I can't seem to find the reason. any help would be appreciated.
triangle
triangle
asked Dec 8 '18 at 19:14
Nasir AzizNasir Aziz
31
asked Dec 8 '18 at 19:14
Nasir AzizNasir Aziz
31
asked Dec 8 '18 at 19:14
Nasir AzizNasir Aziz
31
asked Dec 8 '18 at 19:14
Nasir AzizNasir Aziz
31
asked Dec 8 '18 at 19:14
Nasir AzizNasir Aziz
31
31
closed as unclear what you're asking by Scientifica, Lord Shark the Unknown, Cesareo, José Carlos Santos, amWhy Dec 9 '18 at 16:53
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Scientifica, Lord Shark the Unknown, Cesareo, José Carlos Santos, amWhy Dec 9 '18 at 16:53
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Is that figure meant to be a square? If so...the diagonals of a square are equal so we'd need $a=b$. If it's not meant to be a square then the area of the figure is not $c^2$.
$endgroup$
– lulu
Dec 8 '18 at 19:17
add a comment |
$begingroup$
Is that figure meant to be a square? If so...the diagonals of a square are equal so we'd need $a=b$. If it's not meant to be a square then the area of the figure is not $c^2$.
$endgroup$
– lulu
Dec 8 '18 at 19:17
$begingroup$
Is that figure meant to be a square? If so...the diagonals of a square are equal so we'd need $a=b$. If it's not meant to be a square then the area of the figure is not $c^2$.
$endgroup$
– lulu
Dec 8 '18 at 19:17
$begingroup$
Is that figure meant to be a square? If so...the diagonals of a square are equal so we'd need $a=b$. If it's not meant to be a square then the area of the figure is not $c^2$.
$endgroup$
– lulu
Dec 8 '18 at 19:17
add a comment |
1 Answer
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Since you talk about hypotenuse, it means that the diagonals of the quadrilateral are perpendicular. Using $$c^2=a^2+b^2$$ and $$c^2=2ab$$ you get $$a^2+b^2-2ab=0$$ or $$(a-b)^2=0$$This is valid for $a=b$
answered Dec 8 '18 at 19:19
AndreiAndrei
11.8k21026
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$begingroup$
i was trying to prove pythagoras theorem
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– Nasir Aziz
Dec 8 '18 at 19:50
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since you talk about hypotenuse, it means that the diagonals of the quadrilateral are perpendicular. Using $$c^2=a^2+b^2$$ and $$c^2=2ab$$ you get $$a^2+b^2-2ab=0$$ or $$(a-b)^2=0$$This is valid for $a=b$
answered Dec 8 '18 at 19:19
AndreiAndrei
11.8k21026
$endgroup$
$begingroup$
i was trying to prove pythagoras theorem
$endgroup$
– Nasir Aziz
Dec 8 '18 at 19:50
add a comment |
$begingroup$
Since you talk about hypotenuse, it means that the diagonals of the quadrilateral are perpendicular. Using $$c^2=a^2+b^2$$ and $$c^2=2ab$$ you get $$a^2+b^2-2ab=0$$ or $$(a-b)^2=0$$This is valid for $a=b$
answered Dec 8 '18 at 19:19
AndreiAndrei
11.8k21026
$endgroup$
$begingroup$
i was trying to prove pythagoras theorem
$endgroup$
– Nasir Aziz
Dec 8 '18 at 19:50
add a comment |
$begingroup$
Since you talk about hypotenuse, it means that the diagonals of the quadrilateral are perpendicular. Using $$c^2=a^2+b^2$$ and $$c^2=2ab$$ you get $$a^2+b^2-2ab=0$$ or $$(a-b)^2=0$$This is valid for $a=b$
answered Dec 8 '18 at 19:19
AndreiAndrei
11.8k21026
$endgroup$
Since you talk about hypotenuse, it means that the diagonals of the quadrilateral are perpendicular. Using $$c^2=a^2+b^2$$ and $$c^2=2ab$$ you get $$a^2+b^2-2ab=0$$ or $$(a-b)^2=0$$This is valid for $a=b$
answered Dec 8 '18 at 19:19
AndreiAndrei
11.8k21026
answered Dec 8 '18 at 19:19
AndreiAndrei
11.8k21026
answered Dec 8 '18 at 19:19
AndreiAndrei
11.8k21026
answered Dec 8 '18 at 19:19
AndreiAndrei
11.8k21026
11.8k21026
$begingroup$
i was trying to prove pythagoras theorem
$endgroup$
– Nasir Aziz
Dec 8 '18 at 19:50
add a comment |
$begingroup$
i was trying to prove pythagoras theorem
$endgroup$
– Nasir Aziz
Dec 8 '18 at 19:50
$begingroup$
i was trying to prove pythagoras theorem
$endgroup$
– Nasir Aziz
Dec 8 '18 at 19:50
$begingroup$
i was trying to prove pythagoras theorem
$endgroup$
– Nasir Aziz
Dec 8 '18 at 19:50
add a comment |
$begingroup$
Is that figure meant to be a square? If so...the diagonals of a square are equal so we'd need $a=b$. If it's not meant to be a square then the area of the figure is not $c^2$.
$endgroup$
– lulu
Dec 8 '18 at 19:17