Fun with triangles (847 hits)
Category: Science & EnvironmentalRating: 0.01 on 38 reviews (Rate this item) (V)
Submitted by Kamargö (View user info) at 2005-02-10 19:39:26 EST
Ohhhhh über, I have a big problem... I have to solve this triangles for h and I think it cannot be done because h is negative, what do you think?
I used both Pythagoras and Tales theorems, but I am sad to say that I am stumped, can you help me out?
User Reviews
Submitted by Axolotl (user info) at 2005-07-07 22:27:19 EDT (#)
Ranking: 0
Are you kidding me? Math?
The hell if I'm ever going to use math ever again.
Submitted by zakalwe (user info) at 2005-02-11 19:16:25 EST (#)
Ranking: 0
Shouldn't it be 28/20 = (2h+4)/h?
Submitted by williamson (user info) at 2005-02-11 19:13:01 EST (#)
Ranking: 0
Submitted by DrPresident (user info) at 2005-02-10 22:18:26 (#)
Ranking: -2
This problem is simple.
One triangle nested in another as such the bottom of each triangle is parallel to the other means the triangles are similar.
Therefore, the ratio of the first triangle to the second is 20:28, or 5:7.
Thus, a simple proportion can be made:
5/7 = h/(h+4)
Cross multiply, and you get:
7h = 5h + 20
Simplify:
2h = 20
h = 10
Elementary.
----------------------
Wrong. It is not the 7 that equals h+4 but the remaining 2. (Hard to explain without pointing where I mean). Just try and draw the damn triangle on paper. You can not do it.)
Submitted by JonnyX (user info) at 2005-02-11 19:08:57 EST (#)
Ranking: 0
um , let me get my ruler - uh, it's about a third of an inch, is that good?
Submitted by UberPirate (user info) at 2005-02-11 10:06:52 EST (#)
Ranking: -2
fartface
Submitted by Kamargo (user info) at 2005-02-11 10:03:05 EST (#)
Ranking: 0
Yes, h=10 if and only if BC=h+4 if not, it's an impossible triangle...
Thanks über
Submitted by lordofthedance (user info) at 2005-02-10 23:28:45 EST (#)
Ranking: 0
Yep. I suck.
Submitted by JohnGalt (user info) at 2005-02-10 23:08:18 EST (#)
Ranking: 2
Submitted by lordofthedance (user info) at 2005-02-10 22:58:20 (#)
Ranking: 2
That's not right.
If h was 10 that would mean that for the smaller triangle 20 = 100 + x^2. Therefore, x would be the square root of -80. Since I doubt one of the sides of the triangle is an imaginary number .....it's a big call saying its elementary when you're wrong.
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I'd say forgetting that you square all three numbers, including the 20, when making such a smart ass reply is a pretty big call as well.
Submitted by JohnGalt (user info) at 2005-02-10 23:04:37 EST (#)
Ranking: 2
Only two possible answers here:
All of side AC = h+4 and h=10, or this isn't a valid triangle.
I mean all of side BC = h+4, not AC
Submitted by JohnGalt (user info) at 2005-02-10 23:01:14 EST (#)
Ranking: 2
Going with the answer of h=10 from before...
AD = 2*(sqrt 147) - 10*(sqrt 3) ~= 6.93
DB = 10*(sqrt 3) ~= 17.32
AB = 2*(sqrt 147) ~= 24.25
AE = 10
EC = 4
AC = 14
AC = 28 (given)
If that's all correct then
h^2 + (10*(sqrt 3))^2 = 20^2 and (h+4)^2 + (2*(sqrt 147))^2 = 28^2
Check:
h^2 + (10*(sqrt 3))^2 = 20^2
100 + 300 = 400...looks good to me
(h+4)^2 + (2*(sqrt 147))^2 = 28^2
196 + 588 = 784...also looks good
Check the ratios:
h/h+4 = 10/14 = 5/7
20/28 = 5/7
10*(sqrt 3) / 2*(sqrt 147) = (sqrt 300/588) = (sqrt 100/196) = 10/14 = 5/7
Only two possible answers here:
All of side AC = h+4 and h=10, or this isn't a valid triangle.
Submitted by lordofthedance (user info) at 2005-02-10 22:58:20 EST (#)
Ranking: 2
That's not right.
If h was 10 that would mean that for the smaller triangle 20 = 100 + x^2. Therefore, x would be the square root of -80. Since I doubt one of the sides of the triangle is an imaginary number .....it's a big call saying its elementary when you're wrong.
Submitted by d_prime (user info) at 2005-02-10 22:36:19 EST (#)
Ranking: 0
*so* elementary
Submitted by DrPresident (user info) at 2005-02-10 22:18:26 EST (#)
Ranking: -2
This problem is simple.
One triangle nested in another as such the bottom of each triangle is parallel to the other means the triangles are similar.
Therefore, the ratio of the first triangle to the second is 20:28, or 5:7.
Thus, a simple proportion can be made:
5/7 = h/(h+4)
Cross multiply, and you get:
7h = 5h + 20
Simplify:
2h = 20
h = 10
Elementary.
Submitted by MyNameIsTim (user info) at 2005-02-10 22:07:56 EST (#)
Ranking: 0
my highschool math teacher used to fuck with us and give us problems like this...ones that do not have an answer, but the answer is "this does not have an answer"
that could be the case...
Submitted by drfeggphd (user info) at 2005-02-10 21:38:55 EST (#)
Ranking: 0
Submitted by drfeggphd (user info) at 2005-02-10 21:19:56 (#)
Ranking: 0
i get about 10.8964, using 5h^2 + 16h + 16 = 784
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the reason it's so difficult is DA is about -5.875
but at least h is positive
Submitted by williamson (user info) at 2005-02-10 21:20:49 EST (#)
Ranking: 0
Damn ubersite and it's incapability to allow the text drawing of Triangles!!
Submitted by williamson (user info) at 2005-02-10 21:20:10 EST (#)
Ranking: 0
Submitted by williamson (user info) at 2005-02-10 21:18:33 (#)
Ranking: 0
it's impossible. Just accept it.
Another impossible Triangle: Triangles sides = 3cm, 4cm and 8 cm.
Just try to draw the triangle and you can see without maths it aint possible.
...../\
..../..\ <--- This is "h" which is greater than "h+4": Impossible!
.../....\
../__20__\
./........\
/____28____\
Submitted by drfeggphd (user info) at 2005-02-10 21:19:56 EST (#)
Ranking: 0
i get about 10.8964, using 5h^2 + 16h + 16 = 784
Submitted by williamson (user info) at 2005-02-10 21:18:33 EST (#)
Ranking: 0
it's impossible. Just accept it.
Another impossible Triangle: Triangles sides = 3cm, 4cm and 8 cm.
Just try to draw the triangle and you can see without maths it aint possible.
/\
/ \ <--- This is "h" which is greater than "h+4": Impossible!
/ \
/__20__\
/ \
/____28____\
Submitted by Rawrg (user info) at 2005-02-10 21:00:00 EST (#)
Ranking: 0
I've now come to the conclusion that if f(x) is true then h = ghey.
Submitted by Freight_Train (user info) at 2005-02-10 20:46:34 EST (#)
Ranking: 0
The problem must be worded wrong. Ok..
Two triangles:
20 is to 28 as
h is to 2h+4
then, 28h=20(2h+4)
-12h=80
h=-6.67
This triangle cannot exist
Submitted by Kamargo (user info) at 2005-02-10 20:24:36 EST (#)
Ranking: 0
Yup, that is the diagram that I have... and no, I dont have a graph. calc. sucks to be me...
Submitted by zakalwe (user info) at 2005-02-10 20:21:52 EST (#)
Ranking: 0
Don't think there is enough information to solve. Is the problem complete?
Little rusty on stuff like this.
Submitted by Rawrg (user info) at 2005-02-10 20:20:43 EST (#)
Ranking: 0
REVISION, I'm fucking stupid.
---
The way this needs to be solved is... hold on, gotta bust out the character map...
Nevermind, assume that S is the square root symbol...
Where x is BD
Where y is BA
S( h^2 + x^2 ) = 20
S( (2h+4)^2 + y^2 ) = 28
I assume the easiest way to find the true value would be to combine the functions and solve for the three, although I don't see any method of solving them other than trial and error... Got a graphing calculator? The solver can typically point plot until it hits...
Submitted by Rawrg (user info) at 2005-02-10 20:19:47 EST (#)
Ranking: 0
The way this needs to be solved is... hold on, gotta bust out the character map...
Nevermind, assume that S is the square root symbol...
Where x is BD
Where y is BA
S( h^2 + x^2 ) = 20^2 = 400
S( (2h+4)^2 + y^2 ) = 28^2 = 784
I assume the easiest way to find the true value would be to combine the functions and solve for the three, although I don't see any method of solving them other than trial and error... Got a graphing calculator? The solver can typically point plot until it hits...
Submitted by darko (user info) at 2005-02-10 20:18:07 EST (#)
Ranking: 2
nevermind, im a dumbass
Submitted by darko (user info) at 2005-02-10 20:17:42 EST (#)
Ranking: 2
i dont see you accounting for e-c
Submitted by Kamargo (user info) at 2005-02-10 20:13:45 EST (#)
Ranking: 0
I used Thales's theorem where BA/BD = BC/BE = AC/DE (proportionality) so...
BC=2h+4
BE=h
AC=28
DE=20
Solving that I get h=-6.667 and that cannot be possible...
Do you see any flaws there?
Submitted by darko (user info) at 2005-02-10 20:12:08 EST (#)
Ranking: 1
i think you are misinterpreting it. There is no way the hypotonuse is 20 on the small figure, then one side more than doubles and the hypotenuse only goes to 28. I could be wrong though, I interpreted the same way as Rawrg, that the entire thing was H+4 not just that fragment.
Submitted by Gollman (user info) at 2005-02-10 20:08:51 EST (#)
Ranking: -2
It can't be done in any other method other than trial and error, in my opinion. there are not enough values to work out the value of h.
Submitted by Rawrg (user info) at 2005-02-10 20:06:49 EST (#)
Ranking: 0
Whoops, I thought the whole distance was H+4. Dang.
16 or 12 works for the first triangle, but the second one makes it quite a bit harder...
Submitted by Kamargo (user info) at 2005-02-10 19:58:47 EST (#)
Ranking: 0
I dont think so, if h=16 then BC=36 and that's bigger than the hypothenuse (AC)
Submitted by darko (user info) at 2005-02-10 19:52:42 EST (#)
Ranking: 0
yeah, 16 is what i got for h too
Submitted by Rawrg (user info) at 2005-02-10 19:46:34 EST (#)
Ranking: 2
H = 16
Submitted by Sassmasterr (user info) at 2005-02-10 19:45:25 EST (#)
Ranking: -2
actually, i could've tackled this back in the day, but my math skills have faded to simple multiplication
i know you don't care, but -2 for your first post
Submitted by Kamargo (user info) at 2005-02-10 19:43:07 EST (#)
Ranking: 0
Well, thats what I get when I try to solve it, and I know that h CANNOT be negative... I get that h= -6.6666 and that isn't possible
ohh, and it's Thales not Tales
Submitted by Sassmasterr (user info) at 2005-02-10 19:42:37 EST (#)
Ranking: 1
no fun can be had with triangles...it's a scientific fact
Submitted by zakalwe (user info) at 2005-02-10 19:40:24 EST (#)
Ranking: 0
h is negative? What are you talking about?