Question

LM is tangent to circle C at point M. If CN=3x-2 NL= 4x-6 amd CL= 34 what is the length of LM? I need to know the formula or someone to explain how to work this problem. I'm lost

Answer 1

these look like line measurements for a triangle but uts hard to tell with a picture to actually see.

Answer 2

yes its a circle with a triangle from it. it has one side c to l and another m to l does that help

Answer 3

if you can draw it in "paint" and post it in here, itd be more helpful. Paint is a basic program on many computers.

Answer 4

Can you add any detail to this?

Answer 5

It is a triangle with one point in the middle which is C and M is on the radius and the common point of this triangle is L.

Answer 6

C is in the middle of the line? like this?

Answer 7

No, the picture is of a circle hang on a minute

Answer 8

its the top one

Answer 9

a^2 + b^2 = c^2

(3x-2)^2 + LM^2 = [(3x-2)+(46-6)]^2

does this make sense to you?

Answer 10

46-6 sposed ta be 4x-6 lol

Answer 11

LM^2 = (7x-8)^2 - (3x-2)^2

Answer 12

LM^2 = 49x^2 -112x +36 - 3x^2 -12x +4

LM^2 = 46x^2 -124x +40

Answer 13

8*8 = 64 not 36.... you better do the math :)

Answer 14

I messed up a few points there when I was expanding them.... rechack the math cause I know its wrong :)

Answer 15

hang on let me liik
yes but the formula is correct?

Answer 16

LM^2 = 49x^2 -112x +64 -3x^2 +12x -4

the formula is correct, that I know.

Answer 17

thank you sooo much I will be needing more help later, Im having problems finding this in my math book

Answer 18

the pic you sent, the radius is (3x-2) so just use it as the base, then add the other 2 lengths together to get the hypotenuse, and find the missing leg...

Answer 19

thanks

Answer 20

im going to work this out and get back with u

Answer 21

ok..

Answer 22

how did u get 112x

Answer 23

when you add the lengths together for the hypotenuse you get:

(3x-2) + (4x-6)

3x -2 +4x -6 = 7x -8

(7x-8)^2 = 49x^2 - 56(2)x +64

49^2 -112x +64

Answer 24

let me check

Answer 25

oh I see where now

Answer 26

I think Im still lost... I believe I am doing the x^2 wrong

Answer 27

doing the x^2 wrong.... now your sounding as crazy as me...

Answer 28

Im afraid to ask

Answer 29

so what you have there is a simple triangle that has a length of 2 sides and you want to find the third side....

one side has been given to you in 2 seperate equation, and the shortest leg is simply the length of the radius which is one of the equations they gave you for the other....

plug it into the pythagoreum thrm and get your answer :)

Answer 30

ok let me do it

Answer 31

I believe the final answer is 16. but I will have to recheck the process again. I have to leave for now. thanks for all the help I'll be back again.