SOLUTION: What is the length of the angle bisector to the short leg of a 9-12-15 right triangle? Answer in simplest radical form.

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Question 1177800: What is the length of the angle bisector to the short leg
of a 9-12-15 right triangle? Answer in simplest radical form.
Found 3 solutions by MathLover1, ikleyn, greenestamps:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Formula:

L+=+sqrt%28a%2Ab%2A%28a%2Bb%2Bc%29%28a%2Bb-c%29%29+%2F+%28a%2Bb%29

Where,
L+= Length of Angle Bisector
a+=+9
b+=+12
c+=+15

L+=+sqrt%289%2A12%2A%289%2B12%2B15%29%289%2B12-15%29%29+%2F+%289%2B12%29

L+=+sqrt%28108%2A36%2A6+%29%2F+21

L+=+sqrt%2823328%29+%2F+21 ....23328+=+2%2A%284%29%5E2%2A%2827%29%5E2

L+=+sqrt%282%2A%284%29%5E2%2A%2827%29%5E2%29+%2F+21+

L+=+%284%2A27sqrt%282%29%29+%2F+21+

L+=+%284%2Across%2827%299sqrt%282%29%29+%2F+cross%2821%297+

L+=+%284%2A9sqrt%282%29%29+%2F+7+

L+=+%2836sqrt%282%29%29%2F7


Answer by ikleyn(52866) About Me  (Show Source):
You can put this solution on YOUR website!
.

            The solution by  @MathLover1 is  INCORRECT.

            I came to bring the correct solution. 


Let the given triangle be ABC, with the right angle B;

     the leg AB = 12;

         leg BC =  9,

     and the hypotenuse AC = 15.


The bisector AD of the acute angle A divides the leg BC in two parts BD and DC.


Then the ratio   |BD|  to  |CD|  is equal to the ratio  |AB|  to  |AC|,  i.e.  12%2F15 = 4%2F5.


It implies that  |BD| = 4,  |BC| = 5.



Hence,  the length of the bisector AD is equal to  sqrt%2812%5E2%2B4%5E2%29 = 4%2Asqrt%283%5E2%2B1%5E1%29 = 4%2Asqrt%2810%29.    ANSWER

Solved     (C O R R E C T L Y).


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Looking into her activity in last several days, I start thinking that

@MathLover1 presents a real danger for any visitor to this forum, because she provides wrong solutions even to simplest Math problems.


See my notes to her posts in past two days

https://www.algebra.com/algebra/homework/Finance/Finance.faq.question.1177850.html

https://www.algebra.com/algebra/homework/Volume/Volume.faq.question.1177847.html

https://www.algebra.com/algebra/homework/Surface-area/Surface-area.faq.question.1177846.html

https://www.algebra.com/algebra/homework/Triangles/Triangles.faq.question.1177800.html

https://www.algebra.com/algebra/homework/Triangles/Triangles.faq.question.1177804.html



Answer by greenestamps(13206) About Me  (Show Source):
You can put this solution on YOUR website!


Only a comment regarding the other two responses you have received....

If you are in a job where you have to do this kind of calculation hundreds of times, then using an obscure and complicated formula as the first tutor TRIED to do makes sense -- program the formula into a computer or calculator and enter the input values to get the answer.

But to solve a single problem like this, it is absurd to use a difficult and obscure formula like that when the answer can be obtained with rather simple calculations, as the other tutor did.

If you just simply love to memorize formulas which magically give you the answers to problems, then go ahead and use the formula (but make sure it is correct!).

On the other hand, if you want to UNDERSTAND how to solve the problem, and you want to LEARN something from working the problem, then follow the path of the other tutor.