Question

Help. Question will be in a attachment below. Medal will be rewarded.

Answer 1

Which way do you want to solve this? Trignometry or geometry?

Answer 2

geometry

Answer 3

Ok, let's assume that \( RT = x\), then \(RX=\frac{1}{2}x\) right?

Answer 4

Apply Pythagorean theorem, what will \(TX\) be based on x?

Answer 5

well...wouldn't it stay 6 or would it become a different number?

Answer 6

No. RX will be different

Answer 7

okay, I'm kinda confused, what would it make TX?

Answer 8

Ok. In the right triangle RTX, \(TX^2+RX^2=TR^2\Rightarrow TX^2=TR^2-RX^2=x^2-\large(\frac{1}{2}x) ^2 \normalsize =\large\frac{3}{4}\normalsize x^2\)

Answer 9

TX = 6 -> TX^2= 6^2 = 3/4 x^2

Answer 10

just wondering how do you get 3/4 x^2 from 6^2?

Answer 11

Do you get that \(TX^2= \large \frac{3}{4} \normalsize x^2\)?

Answer 12

yes

Answer 13

Ok. You are given that \(TX = 6\) right? Then what \(TX^2\)?

Answer 14

that would equal 36

Answer 15

Isn't it 36?

Answer 16

yeah

Answer 17

So \(TX^2= \large \frac {3}{4} \normalsize x^2\) and \(TX^2=36\). Can you solve for x?

Answer 18

ok

Answer 19

So \(x\) is the side TR. And \(RX= \large \frac{1}{2} \normalsize x\), the \(RX\) is ...

Answer 20

18

Answer 21

No. \(\large \frac{3}{4} \normalsize x^2 =36\Longrightarrow x^2=48 \Longrightarrow x =\sqrt{48}=4\sqrt3 \)

Answer 22

oh, I'm still really confused as to how you got this answer. But it's fne, besides the time for the quiz already finishe :/

Answer 23

I am sorry. Hope you will get all of those

Answer 24

I gotta go. Have a good one

Answer 25

you too, bye