Ugh....geometry challenge for you brains out there

[stoner87]

So I go to college for 4 years and get an engineering degree, get a job, etc. Now, I come home to visit my parents, and my dad hands me a simple geometry problem. Thinking I'm such the bigshot college boy, I take a run at it, and can't solve it for all I'm worth!!! I'm sure I could solve this with CAD, but I don't have software here, and would really like to know the mathmatical approach to it. So if any of you can solve it, please LMK your methodology! Or, if you think there are too many unknowns (I kinda doubt it), let me know that, too. And NO, this is not some cheap ploy to have you guys do homework for me.
Thanks!

Linkified

 

[fatbaby]

is it to scale ?

 

[isekii]

i dont get the X part..

is the 62 the base of the original triangle . ? ?

and X the extension ? ?

 

[stoner87]

NOT TO SCALE!!! Just something I drew up really quick. I know you can solve it by scaling it or using CAD pretty easily, I was just disgusted w/myself over not being able to solve it mathematically.

 

[stoner87]

62 is the entire distance across the bottom...X is the horizontal distance between the 2 parallel lines, 4" is the perpendicular distance between them. Basically, once I get the x distance, any angle, etc. I can easily roll through any of the unknowns.

 

[Einz]

Ok, haven't done geo in a while.. but here's my guess:

Slide the 4" line that you drew down left so it forms a right triangle with the 2 hypo's and X as the hypothenuse of the new triangle. Now, you've got 2 similar triangles. So, using proportions:

4/40=x/sqrt((62-x)^2+40^2)

I solved for X and my TI-89 spit out

x= (2(10sqrt(1357)-31))/99 or approx 6.81565

Dunno if I'm right, but that's my 2 cents

 

[gopunk]

Now, you've got 2 similar triangles.

really? i thought the theorem was that all lines had to be parallel to their counterparts?

 

[Pundit]

Einz, I was just about to post that. Good work.

 

[b0mbrman]

Think I got the starting point but you're gonna have to figure out the rest since it's only fair you do at least *part* of your own homework

X is the solution to the pair of equations:

Let Y be the length of the rest of the big triangle side with 40 as a segment
(62-X)/(62) = (40)/(40+Y)
(X^2 - 16)/X = 4/Y

 

[Pundit]

<< Now, you've got 2 similar triangles.

really? i thought the theorem was that all lines had to be parallel to their counterparts? >>


Maybe you're thinking of congruent triangles?

 

[stoner87]

That's the stuff....thanks Einz and b0mbrman...funny how it all starts to slip away when you don't use it for a while.

 

[Einz]

Well... isn't an angle-angle-angle match a condition for similarity? You've got 2 right angles and they have an angle that equal each other since two parallel lines are cut by one line and the two angels should equal.. ok, that might have been confusing.. let me draw a picture... gimme a sec

*EDIT* Good, I wasn't on crack I guess someone else can explain it. My brain hurts too much

 

[gopunk]

<<

<< Now, you've got 2 similar triangles.

really? i thought the theorem was that all lines had to be parallel to their counterparts? >>


Maybe you're thinking of congruent triangles? >>



no, i was thinking of similar triangles. i'll wait and see the SAS einz comes up with though.

yes i hate how this simple stuff gets away from you, it makes you look like an idiot when an eighth grader can beat you at a math problem. 'course then you bring up the fact that you can whoop him in calculus or something but people laugh and say you're a sore loser

 

[Pundit]

<< Think I got the starting point but you're gonna have to figure out the rest since it's only fair you do at least *part* of your own homework

X is the solution to the pair of equations:

Let Y be the length of the rest of the big triangle side with 40 as a segment
(62-X)/(62) = (40)/(40+Y)
(X^2 - 16)/X = 4/Y >>


Another good approach! Damn, ATOTers are pretty smart

 

[Mungla]

For triangles to be similar, you can use AA~ or SAS~. There may be more ways to prove it, but those are what I know.

 

[gopunk]

okay i think they are similar.

my reasoning is as follows:

for the small triangle we created, call the leftmost (and common corner) point B. call the right angle corner A, and call the remaining corner C. A is common to both triangles, as is B. the only thing remaining is to show that C is the same, which will give us AAA, thus proving similar triangles.

C is equal to 180 - 90 - B, since the two lines are parallel. that's just 90 - B, which is the remaining angle.

 

[alee25]

hmmm....
woudl this work?

that whole shape is actually just a piece of a square:

so the area of the square: 62x40 = 2480

so the 2480 should add up to 2 triangles and the quadralaterial in the middle

area of quadralatral: (rearange it into a long rectangle) 4*x (width x lenght)
area of triangle: (62-x)40 * 1/2 (1/2 Bh)

2480= (62-x)40 + 4x

solve for x

edit: hmmmm im getting x = 0 so that cant be right.. anyone know what i did wrong?

 

[911paramedic]

Of course they are similar triangles, he said the hypotaneuse (sp?) was parallel on both, and we know that they will have the same angles by proofs.

parallel lines that intersect a line will have the same angles at their intersecting points. LOL, I used proofs in '82, long time ago.

 

[Capn]

Other solutions have been posted that may be more straightforward, here's mine though:

X/4 = sqrt((62-X)²+40²)/40

Solve for X, excel gives me 6.8156 approx.

 

[Pundit]

<< Other solutions have been posted that may be more straightforward, here's mine though:

X/4 = sqrt((62-X)²+40²)/40

Solve for X, excel gives me 6.8156 approx. >>


That's exactly what Einz did.

Ok, who has more? Don't you guys love this stuff?

 

[Capn]

No his is a little different, I should've read more thoroughly I supposed, dang 1 minute down the drain.

 

[b0mbrman]

<< Ok, who has more? Don't you guys love this stuff? >>


Yeah, I admit I do...Only because it takes me back to a time when it *didn't* take 4 hours to do each problem on my problem sets

 

[stoner87]

Ha ha...I never thought this would drum up this much of a response. Sounds like you guys are starving for math challenges. Similar triangles were all I needed, it's just the algebra got a little hairy with the adjacent of the smaller unit coming out as (x^2 - 16)^.5. Also, I'm a little embarassed to admit I forgot the QUADRATIC EQUATION when it came time to solve the variables (slaps myself).

Anyway, thanks a ton!

 

[dighn]

answer same as Einz's


now how does bomberman's method work... edit n/m

 

[alrocky]

Isn't x just the hypotenuse of a 4" square?