Question

In △JKL, KM is an altitude
What is the length, in units, of JK? Image posted.
5 radical 10.
10.
10 radical 5.
15.

Answer 1

KL^2 = 25*5 => KL = 5radical5
JL^2 = JK^2 + KL^2
JK = Square root of(JL^2 - KL^2)

\[\sqrt((25)^2 - (5\sqrt5)^2)) = 5\sqrt10\]

Answer 2

sorry but i think that not is right this answer

Answer 3

if JM = 20 so JK need to be greater than 20 so not can be from those answers just
10sqrt5

Answer 4

yes - i rechecked it too
correct answer is sqrt500 = sqrt ( 5 * 100) = 10 sqrt5

Answer 5

Nevertheless, I cannot really find any mistake in my reasoning. Knowing that:

The square of the length of a cathetus equals the product of the lengths of its orthographic projection on the hypotenuse times the length of this.

Follow up my few calcs up there and they seem to be right. Find the error, so I may learn something as well? ;)

Answer 6

\[\frac{20}{h}=\frac{h}{5} \]h =10\[\text{JK}=20\text{ Csc}\left[\text{ArcTan}\left[\frac{20}{10}\right]\right]=10 \sqrt{5} \]

Answer 7

Still, what's the error in my reasoning, which is different from yours?

Answer 8

to be honest i've never heard of the theorem you quoted in your second sentence.
but i just used pythagoras theorem and i'm sure 10sqrt5 is correct.
rob's method using similar triangles and trig ratios is also valid.
your mistake is on the last line

sqrt(25^2 - (5sqrt5)^2 ) = sqrt (625 - 125) = sqrt 500 = sqrt(100*5) = 10sqrt5 not 5sqrt10

Answer 9

just a simple error at the end!!! it happens

Answer 10

Oh well, I'm happy I didn't do any error reasoning-wise :) since I did everything on my mind and not on the sheet I did a calc error in the very last step, thank you! :)

Answer 11

yw