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

[mathspassion]

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...........

[Sriman Dutta]

[math] DE=BE-BD=\frac{BC}{2} -\sqrt{AB^2-AD^2}=\frac{BC}{2}-\sqrt{AB^2-(\frac{AC.AB}{BC})^2}[/math]

Edited December 24, 2016 by Sriman Dutta

[imatfaal]

Not sure if this is short enough for you

As triangles ABC and ADC are similar in a ratio of 5:4, DC is 4/5 of AC, EC is by definition 1/2 of BC; thus DE is (4/5)*4-(1/2)*5= 0.7

 

[mathspassion]

is it for 345 only or for 5 12 13 as well.....

 

my mean is to everyone can solve it (it is still too lengthy) , it can be too short if you know there is a new formula recently obtained.

 

study deep

[imatfaal]

 

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

[mathspassion]

ok wow gr8

[michel123456]

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

 

[imatfaal]

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

[michel123456]

 

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 December 27, 2016 by michel123456

[michel123456]

I couldn't post images the other day.

Edited December 29, 2016 by michel123456

[mathspassion]

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 .....................

[mathspassion]

bare links removed by moderator

Edited April 12, 2017 by imatfaal
removal of links

[imatfaal]

!

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[Commander]

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 !