Find the missing lengths of the sides.
where?
Alright. Are you familiar with the notation with the small dashed lines?
yes.
|dw:1448593879046:dw|
Like A = 6 there
yes
ok then nvm lol
Oh okay then. So we know that whatever the answer will be, a and b will be the same. So we can immediately rule out b).
XD, Tanks tho .fedorable.
ok
Now, question, have you ever seen something called the "Pythagorean Theorem?"
yes, a^2 + b^2 = c^2
Alright then. So do we know any of the values in that equation right now?
Yes, I know that c = 12 sqrt(2)
Awesome. Now, we know that a = b right?
right
So whenever we have an equals sign, we can substitute one for the other right? Like if x=y then if x=5 then y=5 right?
the easier way to do is just do 12sqrt(2)/sqrt(2), becuase we know its a 45/45/90, which means the legs are 1/sqrt(2) of the longest side
right @ Hitaro
So we can make this equation a^2 + b^2 = c^2 a^2 + b^2 = (12sqrt( 2 )) ^2 a^2 + a^2 = (12sqrt (2 )) ^2
ok. That makes sense.
So is there any way we can simplify that?
yes, u can add like variables.
Right. So that would be 2a^2 = 12(sqrt2)^2
Think you can take it from there?
\[2a^2 = (12\sqrt{2})^2\]
I would have to get a by itself right... So, 2a^2 = 12(sqrt)2^2 2a^2 / 2 = 12(sqrt)2^2 / 2 a^2 = 6(sqrt)2... No, I don't get it... Im sorry. Could u pls walk me thru it.
I wrote the equation wrong in text form, that's my bad. the 12 is inside the square.
The correct equation is a couple replies up.
Oh! I see it Tanks. Let me try again.
2a^2 = (12sqrt2)^2 here is where i am confused, am I supposed to get rid of the square first or the 2 before the a, to get a^2 by itself?
Right. Right now its a bit difficult to simplify the left side because dividing by two doesn't work. So maybe we can simplify the right side. Can you simplify (12sqrt(2))^2 ?
12sqrt(2) = 16.9 or 17.
Okay hold on. Can you solve (sqrt(2))^2 ?
2
Alright awesome. How would you solve (2sqrt(2))^2?
Sorry I am having a lot of lag on my end.
Yeah. Openstudy can be finicky. So, how might we go about solving (2sqrt(2))^2 ?
Do u mean (12sqrt(2)^2?
Yes.
(my bad)
k, 24.
Can you walk me through your steps to get that?
I used the calculator. :( I am allowed to do that in my school. Unfortunatly I don't know how to get it.
\[2a^2 = (12\sqrt{2})^2\]
Okay. So, let's think this through. Squaring a number is multiplying it by itself. So we're trying to find 12sqrt(2) times 12sqrt(2)
But this is kinda awkward, because we don't have an intution on how to multiply these quite yet. So what I'm going to do, is ask you to subtitute z= sqrt(2). Can you solve 12z times 12z?
144z
Hmm okay. Can you solve 3a times 3a ?
9a^2
Perfect. So what is 12z times 12z again?
144z^2
Awesome. Now, we substituted z for sqrt(2) right? so if z = sqrt(2) what is 144(z^2)?
I think it is 144((sqrt(2)^2)
Right, so that's 144(2) right? So 2a^2 = 144(2)
oh, right
So can you solve for a?
Divide both sides by 2.
?
Right
So, you good? Have you found a?
Ok, so I would end up with a^2 = 144. Then I would end up w/ 12? Right!!
Yep! Awesome job. Hopefully all that muddying around with how to multiply square roots didn't take away from the concept. Could you solve this simpler problem? |dw:1448597486372:dw|
The legs each = 16, right?
well let's remember the steps. What would the pythagorean theorem look like in this situation?
oh, right sorry. 256
Um one sec, a^2 + b^2 = 32^2 or 256
I'm realizing this problem isn't actually easier, I tried to set up one with simple numbers but I think I just made it more complicated.
Im sorry... I thought I was starting to understand.
Either way, you let a =b because the two sides are congruent. so you end up with a^2 +a^2 = 256 2a^2 =256 a^2 = 128 a = sqrt(128)
No no no, what you said was perfectly right
oh! Ok great!!
Like, imagine I was trying to explain division, so I tried to pick and example like 8 divided by 2 But by accident I actually did 1596/1405
Anyways, does each step make sense? feel free to let me know if it doesn't.
Yes! Thank you so much!! You should be a qualified helper!!
Awesome, feel free to let me know if you have any more questions.
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