Finding the area using Pythagoras

[Tonia96TT]

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 of the field that is 50m from one corner of the field, 30m from another corner, and 40m from a third corner (the attached diagram shows the configuration). The goal is to find the size of the field (either the area of the length of one side – it is a square field).

Note that the 30m and 50m lines do not make the diagonal line of the square field.

Homework Equations

Pythagoras c^2=a^2 + b^2

The Attempt at a Solution

I have constructed several right angled triangles and looked for common sides to try and solve simultaneous equations but I end up with 3 equations and 4 unknowns and am thus unable to solve this.

Any help (especially in the form of a detailed explanation) would be greatly appreciated. By the way, I am not a student with a homework assignment – this is a problem I came across and am embarrassed that at my age I cannot figure it out!

Kind regards,
Tonia

 

[vela]

What are the equations you come up with?

 

[tiny-tim]

Welcome to PF!

Hi Tonia Welcome to PF!

(try using the X² icon just above the Reply box )

… I end up with 3 equations …

What about a + b = c + d ?

 

[Tonia96TT]

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.

Thanks in advance to all who take the time to reply.
Kind regards,
Tonia

 

[vela]

Let's call the length of the side of the square L. Then you have 5 equations:

1. A + B = L
2. C + D = L
3. A² + C² = 50²
4. B² + C² = 40²
5. B² + D² = 30²

Subtracting equation (4) from (3) gives

A² - B² = 50² - 40²

Similarly, subtracting equation (5) from (4) gives

C² - D² = 40² - 30²

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 solve for L.

 

[Tonia96TT]

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 solve for L" that throw me for a loop, because I end up with quadratic equations for which I cannot isolate any single variable. I would appreciate if you could walk me through your final steps and show how to solve the simultaneous equations listed.

Thanks very much to you and anyone else who takes the time to help.
Kind regards,
Tonia

 

[hgfalling]

So you have, after vela's last comment:

A² - B² = 50² - 40²

Using Eq1, we get:

A² - (L-A)² = 50² - 40²
A² - (L² - 2AL + A²) = 50² - 40²

Can you go from here?

 

[Tonia96TT]

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 equation (A and L). If I try to solve for A in terms of L (or vice versa) and then try to plug in that value in one of the other equations I end up with a very messy equation that I cannot solve. If someone could please walk me through the next steps (ie either showing me the equations how to isolate a single variable or working through the math in detail) I would appreciate it.

Thanks as always to those who take the time to respond.
Kind regards,
Tonia

 

[vela]

What do you get for A and for C? Then what do you get when you plug those results into equation (3)? It may look messy, but don't be intimidated. It works out quite easily.

 

[Tonia96TT]

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 same thing, and I still cannot isolate a single variable. I am not sure I agree with your statement that "It works out quite easily", and would urge you to show me the complete solution in support of this.

(I appreciate that folks are trying to walk me through this step by step with the goal of having me find the answer but I have been racking my brains over this for the better part of a week. Gentle nudges are lovely but it is time to call in the heavy artillery. I CANNOT find the solution, and would ask someone to spell it out for me as I no longer believe the solution is a simple one, and I challenge anyone to show me that it is.

Thanks again to all those you take the time to respond.
Kind regards,
Tonia

 

[eumyang]

Hi Vela;
Thanks again for trying to help. For A I get A=(L^2-900)/L.

That's not what I have. I got
$$A = \frac{L^2 + 900}{2L}$$.
(I found it easier to leave in this form instead of simplifying further.)

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 same thing, and I still cannot isolate a single variable.

But you WILL end up an equation with a single variable, L. Solve for C in the same matter that you solved for A. You'll get that C = (some.expression.involving.L). Plugging in for BOTH A and C gets you
$$A^2 + C^2 = 50^2$$
$$\left( \frac{L^2 + 900}{2L} \right)^2 + (some.expression.involving.L)^2 = 50^2$$
(Of course, you'll have to find for yourself what "some.expression.involving.L" is.)

You mention quartics, and while it is true that your equation will be a quartic, the equation is still quadratic in form. Suppose you have a quartic that looks like this:
$$x^4 - 10x^2 + 29 = 0$$

If you let y = x^2, you'll get this:
$$y^2 - 10y + 29 = 0$$
... which is a quadratic. Here, you would solve for y, and then substitute y=x^2 back in, and solve for x. You'll have to do something similar eventually in this problem.

I am not sure I agree with your statement that "It works out quite easily", and would urge you to show me the complete solution in support of this.

It's against forum policy to show complete solutions. How do we know if you're telling the truth when you stated that this was not a homework question?

(I appreciate that folks are trying to walk me through this step by step with the goal of having me find the answer but I have been racking my brains over this for the better part of a week. Gentle nudges are lovely but it is time to call in the heavy artillery. I CANNOT find the solution, and would ask someone to spell it out for me as I no longer believe the solution is a simple one, and I challenge anyone to show me that it is.

Well, the solution is not "simple" in the sense that the answers are not integers. According to WolframAlpha you'll get multiple solutions, all of which are irrational. Of course, the negative ones will have to be eliminated.

But don't give up! I left you off with this:
$$A^2 + C^2 = 50^2$$
$$\left( \frac{L^2 + 900}{2L} \right)^2 + (some.expression.involving.L)^2 = 50^2$$

Find out what C is, and plug it in at "some.expression.involving.L" and simplify into a quartic. Show us what you got.

 

[Mentallic]

Tonia, I've tried it myself and the algebra is just horrifyingly long and complicated.

If we treat this problem in coordinate geometry, then we are trying to solve these 3 equations:

$$x^2+y^2=50^2$$

$$(x-L)^2+y^2=40^2$$

$$(x-L)^2+(y-L)^2=30^2$$

Where L is the length of a side of the square. These are just 3 circles with centres at (0,0) (L,0) and (L,L) respectively and radii of the lengths mentioned. To find an L such that they all intersect at a point, we solve it accordingly:

from the first equation we have $$y^2=50^2-x^2$$

and substituting into the second gives $$(x-L)^2+50^2-x^2=40^2$$

Solving for x gives $$x=\frac{L^2+30^2}{2L}$$

We will keep a hold of this equation, substitute the first equation into the 3rd now and solve for x there as well. Then when we have x=f(L) from the 3rd equation, we can finally let both functions of L be equal and solve it. The algebra gets really tricky though when trying to solve for x in the 3rd equation...

$$(x-L)^2+(\sqrt{50^2-x^2}-L)^2=30^2$$

expanding and simplifying gives us $$L\sqrt{50^2-x^2}=L^2-Lx+800$$

and now squaring just leads us down a long-winding path. If you have the motivation to do so, you can solve for x since it's in quadratic form, but I'm noticing that once we use the quadratic formula, we end up with a polynomial of degree 6 in L. I could have made a mistake somewhere but either way, a numerical solution will probably be best.

 

[vela]

Well, the solution is not "simple" in the sense that the answers are not integers.

Right. I meant that at this point in the problem, there's no particularly difficult or tricky algebraic technique needed to get to the final answer. Anyone whose had a year of algebra should be able to work out the rest of the problem. It's straightforward algebra now.

It's against forum policy to show complete solutions. How do we know if you're telling the truth when you stated that this was not a homework question?

It's not even an issue of whether we think you're telling the truth or not. The philosophy of the forum is that the student needs to do the work and we're just here to provide guidance when you get stuck. It's supposed to be a dialogue. I'll admit, Tonia, I'm a bit frustrated with this thread since you seem very reluctant to show any work.

 

[zgozvrm]

Note that if the diagram was not given as part of the problem (that is, it is left up to the student to figure it out), then it could be drawn such that the 40m and 30m distances are swapped. This would result in a 2nd answer.

 

[vela]

Good catch!

 

[Tonia96TT]

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. In the end I have to agree with vela that "It's straightforward algebra now", and also with mentallic's comment that the "algebra is just horrifyingly long and complicated". Indeed. I was able to generate most of the equations that folks have offered as starting points, but struck a barrier when it came time to solve those ugly quadratics and quartics. I was wondering if I was missing something elegant and simple, but I guess the solution requires brute and straightforward calculations. So be it.

I do want to address the comment "How do we know if you're telling the truth when you stated that this was not a homework question?". I had to laugh when I saw this, as I am 50 years old and it has been a while since I've had any homework (or done much algebra). As I mentioned in a previous comment, I came across this problem in a grade 8 math book and because I like to keep my brain cells active I thought I would give it a go. As for vela's comment that "I'll admit, Tonia, I'm a bit frustrated with this thread since you seem very reluctant to show any work", it is not any reluctance on my part to show my work, it is more a case that I am new to these forums and find it very difficult to write endless equations by typing them out.

In the end I do want to extend my gratitude and appreciation for the help that was offered.

Kind regards,
Tonia