# Math Help - Couldn't solve this question about right triangles proportions.

1. ## Couldn't solve this question about right triangles proportions.

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 .

2. ## Re: Couldn't solve this question about right triangles proportions.

Originally Posted by KnowSeeker

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 .

$S_{\text{KJL}}=\frac{\text{KH}\cdot\text{JL} }{2}$ (1)

Also we can write the same area as:

$S_{\text{KJL}}=\frac{\text{JK}\cdot\text{KL} }{2}$ (2)

$(\text{KL})^2=(\text{JL})^2-(\text{JK})^2$ (By Pythagorean theorem) (3)

From (1) and (2) and (3), you get:

$\text{KH}\cdot\text{JL}=\text{JK}\cdot \sqrt{(\text{JL})^2-(\text{JK})^2}$

Try to solve that...

3. ## Re: Couldn't solve this question about right triangles proportions.

brilliant and thanks for the help
btw since kl isn't squared aren't we suppose to square root the (JL)^2-(JK)^2
thanks again and solving it.

4. ## Re: Couldn't solve this question about right triangles proportions.

Originally Posted by KnowSeeker
brilliant and thanks for the help
btw since kl isn't squared aren't we suppose to square root the (JL)^2-(JK)^2
thanks again and solving it.
You are absolutely right! I have already fixed that...

5. ## Re: Couldn't solve this question about right triangles proportions.

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 .

6. ## Re: Couldn't solve this question about right triangles proportions.

Hi KnowSeeker,
KH is the geometric mean of JH and HL.12^2 =x(40-x).Solve for x and then using Pythagoras findJK

bjh

7. ## Re: Couldn't solve this question about right triangles proportions.

Originally Posted by KnowSeeker
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 .

Use the fact that $JK = \sqrt{(JK)^2}$ to rewrite the RHS as $\sqrt{(JK)^2[(JL)^2-(JK)^2]} = \sqrt{(JK)^2(JL)^2 - (JK)^4}$

You can then square both sides to give $(KH)^2(JL)^2 = (JK)^2(JL)^2 - (JK)^4$ which is the same as $(JK)^4 - (JK)^2(JL)^2 - (KH)^2(JL)^2 = 0$. Strange as it sounds you can solve that using the quadratic formula although it may help to sub known values first.

8. ## Re: Couldn't solve this question about right triangles proportions.

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:

$p+\frac{12}{p}=40$

Multiply by p to get: $p^2+12=40p$

$p^2-40p+12=0$

from here get p and q.then the distance JK=sqrt(JH^2+KH^2) where JH is the smaller of p and q.

9. ## Re: Couldn't solve this question about right triangles proportions.

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
$x=20-\sqrt {278}$
${\it JK}=4\,\sqrt {10}$

Update:
i just noticed the reply above

10. ## Re: Couldn't solve this question about right triangles proportions.

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

11. ## Re: Couldn't solve this question about right triangles proportions.

Here's a case where all lengths are integers:
JK = 15, LK = 20, JL = 25, HK = 12, JH = 9, LH = 16.

Practice with that.

i'll do that
thanks