Just a quick note to thank all those who took the time to respond and who offered their help with the calculations. I was looking for a 'simple" solution (since this problem appears in a grade 8 math book) that did not involve solving ugly quartics and could not find one myself, hence my post...
Hi Vela;
Thanks again for trying to help. For A I get A=(L^2-900)/L. If I plug this value into any of the other equations I get a nasty quadratic or even quartic (eg substitute the above into A^2 + C^2 = 50^2 and I get ((L^2-900)/L)^2 + C^2 = 50^2. Trying to find substitutions for C yield the...
Hi hgfalling;
Thanks for your response. I am able to generate the equations you listed by myself, but it is the next steps that baffle me. For example, A^2 - (L^2 - 2AL + A^2) = 50^2 - 40^2 reduces to L^2 -2AL = 900. I still don't see how that helps because I still have 2 unknowns in the...
Thank you for your reply, Vela. I am able to reach the steps you pointed out on my own. It is the next steps "Now use equations (1) and (2) to eliminate, respectively, B and D from these two equations. You can then solve for A and C in terms of L. Then plug those results into equation (3) and...
Hi Everyone;
Thanks for your responses. Below are the equations I came up with (see new diagram to accompany this).
Length of one side of field = A+B = C+D
A^2 + C^2 = 50^2
B^2 + C^2 = 40^2
B^2 + D^2 = 30^2
I am still unable to solve for A+B or for C+D. Any help would be appreciated...
Homework Statement
Hello;
I am trying to solve what must be a ridiculously simple problem involving Pythagoras’ theorem and some simple geometry but I seem to be missing something. The problem is this:
A farmer has a square field and has set up a drinking trough at a point in the middle...