I need help with d). Thank you very much.

[shell sleet]

you can do 2 pythagorean theorems in KNT and MTK

[edgy tartan]

Does the semi colon means it is a scale factor

[sudden bison]

TNxMT = 11
KN/KM=6/5
using euclids theorem its possible to see that (TN+MT)TN=KN and (TN+MT)MT=KM. Being provided by the propotion KN/KM= 6/5 you can see that its possible to write this as (TN+MT)TN/(TN+MT)MT=KN/KM=6/5 which if simplified would yield the result TN/MT=6/5.
if TN+MT =11 than TN must be 6 and MT must be 5.
KT^2= TNxMT so KT= sqrt30 (by the euclids theorem)
and KM^2=30+25 and KN^2= 30+36 so KM=sqrt55 and KN=sqrt66

(Terribly sorry for bad English and not using texit idk how it works so i hope this is atleast understandable.)

[untold fog]

Thank you for this answer and your time, but I need to use the triangle similarity theorem and solve it that way. I haven't even learned the Euclid theorem yet.

sorry for not saying this before

[sudden bison]

euclids theorem is just a fancy name for similarities
the big triangle KMN and the small triangles TMK and TKM are all similar
all of the similarities derived from this are called euclids theorem (atleast thats what i understood)
you're welcome

[untold fog]

Oh okay! I am not English so we don't call it that way in my class

Hello, can you explain this more? " (TN+MT)TN=KN and (TN+MT)MT=KM."

I dont understand why we add TN and MT, multiplicate it by TN and its equal to KN

[shell sleet]

To use the similarities you want you can see that KTN is similar to KMN as K angle=NTK angle=90o and the N angle is common. So you would get that KN/MN=TN/KN=KT/KM=6/5. Doing a pythagorean theorem on TNK you get TN^2+KT^2=KN^2 and by doing a pythagorean in KMT you get KT^2+MT^2=KM^2. By subtracting these two you get TN^2-MT^2=KN^2-KM^2 so (TN-MT)(TN+MT)=KN^2-KM^2 so 11MN= KN^2-KM^2. By doing a pythagorean on KMN you get KN^2+MN^2=MN^2 so by adding this and the previous one you get 11MN+MN^2=2KN^2. (KN/MN=6/5 so KN=6/5*MN) so 11MN+MN^2=72/25MN^2 so 11MN=47MN^2 so 11MN=47 so MN=47/11. Then you can easily find the rest

[untold fog]

Can you write out the subtraction part? I lost after this part

[shell sleet]

Well you have KT^2+MT^2=KM^2 (1)

and TN^2+KT^2=KN^2 (2)

and then you substract (1) from (2)

and you get KT^2+MT^2-TN^2-KT^2=KM^2-KN^2

so MT^2-TN^2=KM^2-KN^2

so (TN-MT)(TN+MT)=KN^2-KM^2

TN-MT is 11 as the exercise has given us

TN+MT=MN

so 11MN=KN^2-KM^2 (3)

By doing a pythagorean on KMN you get MN^2= KN^2+KM^2 (4)

By adding (3) and (4) you get: 11MN + MN^2= 2KN^2 (5)

The exersice gave us KN/MN=6/5 so KN=MN*6/5

by subsituting that in (5) you get:

11MN + MN^2=2MN^2 * 36/25

and you solve this equation...

[untold fog]

Thank you so much for your time, this explanation is perfect, you made it easy to understand.

[shell sleet]

no problem

[untold fog]

Wait, I have a question: is it correct to MT^2-TN^2 make it into (TN-MT)(TN+MT), instead of (MT-TN)(MT+TN). Can you also show how you would solve the equation?

[shell sleet]

well no it would be the opposite thing. you could do -(TN-MT)(TN+MT)

yes the equation

lets say MN=a so i dont write MN

[shell sleet]

11a+a^2=72/25a^2

so 11a=72/25a^2-a^2

11a=72/25a^2-25/25a^2

11a=47/25a^2

since a is not 0 we can divide by a

so 11=47/25a

(multiply by 25/47)

so a is 11*25/47...whatever that is

[untold fog]

Well, now I understand everything. Thank you!!!

[shell sleet]

np

[karmic falcon]

.solved