I'm self taught (still in the beginning ...sad i know)
not sure if it's a mistake by the author i think it's impossible to evaluate JK
since i assume i will always end up having two variables.
i couldn't solve it just thought about it and figured that one of the segments of the hypotenuse should be known so we use the proportion relationship of the segments and the altitude.
and two more things...
1-is there any easier way to draw figures and notate with mathematical symbols rather than playing around with Photoshop.
2- i hope I'm not offending anybody with my English , I'm not aware of my weaknesses on the subject .
unfortunately i couldn't solve it.
i ended up having a chaos !
i used bagatrix algebra and it produced six pages of steps then ended up using quadratic equation ???
i am about to study intermediate algebra so i don't think i am up to the task i have a long way to go .
you can solve this by simple algebra.first find JH and LH.then you can use Pythagorean theorem.
let p=JH and q=LH
then pq=12 and p+q=40
from the frist equation we get:
q=12/p
substitute into the second equation and we get:
Multiply by p to get:
from here get p and q.then the distance JK=sqrt(JH^2+KH^2) where JH is the smaller of p and q.
thank you both for you responses
i think bjhopper solved it the way the book intended it to be solved
i missed the fact that both segments can be interpreted as x and 40-x
continuing this will end up giving x(40-x)=122
-x^2+40x-122=0
using maple
and the book asks for
Update:
i just noticed the reply above
Hello again Know Seeker,
The quadratic equation is x^2-40x +144 =0
this factors (x-36)*(x-4) = 0
x=36 or 4 and 4 is the short segment of JL
JK^2 = 12^2 +4^2 = 160 JK = rad 160 or 4rad10