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See and Solve in one line...................


mathspassion

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Between intersection of altitude and bisector on hypotenuse BC of any right-triangle ABC is (AC x AC/BC) - BC/2 where AC is longer leg

 

 

That's one line and general - dunno what new formula you might be referring. Perhaps you could stop being so elliptic and ask in a more straight forward manner

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In a Right angle triangle ABC

 

AB=3,AC=4 & BC=5

at BC,E is middle point and D is altitude AD

 

Find distance DE=?

 

it is not a difficult question eveyone who know maths can solve easly but question is solve it in one line...........

If I follow your setup D is in the middle of AB, then the distance DE is

half AC=2

 

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If I follow your setup D is in the middle of AB, then the distance DE is

half AC=2

 

 

The altitude of a triangle divides the angle and forms a perpendicular to the side opposite the angle; ie D must lie on BC (side opposite A) and line AD must be perpendicular to line BC *

 

* I think for the other two side/angles of a right triangle the altitude is the same as the side itself - but for the hypoteneuse / right angle this is not the case

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The altitude of a triangle divides the angle and forms a perpendicular to the side opposite the angle; ie D must lie on BC (side opposite A) and line AD must be perpendicular to line BC *

 

* I think for the other two side/angles of a right triangle the altitude is the same as the side itself - but for the hypoteneuse / right angle this is not the case

Thank you.

I missed a sketch.. And now that I put it on paper I realize your post #3 is correct. +1

Edited by michel123456
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  • 3 months later...

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sir figure is right AB=3,AC=4 AND BC=5

 

as per figure find DE=?

 

only two seconds question if you know the particular theorum which recently obtained by .....................

 

I get DE = [ AC2 - AB2 ] / 2 x BC as a General Solution and therefore [42 - 32 ] /2 x5 = 7/10 = 0.7 in this case [the same result found by imatfaal !

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