Please help... so confused
HOLD ON! im going to draw a picture!

- anonymous

Please help... so confused
HOLD ON! im going to draw a picture!

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- anonymous

|dw:1328575316576:dw|
solve for x

- precal

use \[a^2+b^2=c^2\]

- anonymous

well i think that would work but we're supposed to make it a proportion

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## More answers

- precal

usually it involves the small triangle vs the large triangle

- anonymous

ya but i cant do that... at least i dont think i can

- precal

do you know about the angle on top? Is is split in the middle? Is it 45 degrees since the larger part is 90 degrees? I can't tell by your drawing. I know it is hard to draw

- anonymous

well the whole triangle is 90 degrees but it doesnt tell you if its split through the middle

- precal

|dw:1328575806467:dw| if so we can assume the side that is in between them is x...

- anonymous

but how can you do that since x is the length of the side on the bottom?

- precal

Is the drawing from a book or a worksheet? the problem with geometry is that you can not assume anything

- anonymous

its in the book and ya thats why i dont like it!

- precal

|dw:1328575902416:dw|
We can assume that 90 degrees, the other one we can assume is the 45 degrees because that line that splits the triangle has a special name but I can't recall it at the moment.

- precal

You can scan the picture and attach the file. but I think I might have it figured out

- anonymous

well its online so give me a sec

- precal

\[x^2+x^2=(x+3)^2\]
\[2x^2=x^2+6x+9\]
\[x^2-6x-9=0\]
not that does not work !!!!!

- precal

I meant that does not work!!! Sorry can't seem to type as well

- anonymous

there is a picture of the problem!

##### 1 Attachment

- precal

ok upload it, I will be back in about 5 mins

- precal

ok you got my attention.

- precal

ok x+3 is the leg of the big triangle
the hypotenuse is 12-x for the big triangle

- anonymous

well the problem never tells you anything about the angles or anything and as you can tell the line doesnt go right through the right angle symbol

- precal

ok x is the leg of the small triangle
and x+3 is the hypotenuse of the small triangle
there is our proportion

- precal

|dw:1328576462186:dw|

- anonymous

but how did you get 12-x since 12 is the whole LARGE triangle's hypotenuse?

- anonymous

im sorry im just super confused!

- precal

|dw:1328576638458:dw|

- anonymous

so (x+3)(x+3)=12 x

- precal

No, I think we are ok
cross multiply
\[12x=(x+3)^2\]

- precal

yes now solve it

- anonymous

x^2+3x+3x+9=12x
x^2+6x+9=12x
x^2+9=6x
right? but i dont know what do to next because we havent done anything like this in a while!

- precal

x=3 I used the quadratic formula

- precal

\[x^2-6x+9=0\]

- anonymous

ok but how did you get x=3?

- precal

|dw:1328576877634:dw|

- precal

use the quadratic formula

- anonymous

see all of that confuses me and i dont think we've done that yet!

- anonymous

well thanks for your help but i dont think im going to get it.... but thank you for your trying... :)

- precal

Is this a challenge question?

- anonymous

i dont think so.... but online they sometimes leave off something like a picture or a number so that might be what happened... thanks for you patients.

- precal

You will understand this eventually, your teacher has to teach quadratics. One day it will make sense.

- anonymous

haha ok thanks :)

- precal

Oh well another reason why computers will not replace teachers. I am sure that it is an oversight of the program. Hang in there :)

- anonymous

thank you and again thanks for your help1

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