Finding the area using Pythagoras

**Tonia99** · December 24th 2010, 12:18 AM

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 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.

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

**earboth** · December 24th 2010, 12:51 AM

Originally Posted by Tonia99

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 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.

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

1. Define a coordinate system with the origin at the lower left vertex of the square and the coordinate axes on two sides of the square.
Let a denote the length of the square side. Then the vertices have the coordinates:
A(0, 0), B(a, 0), C(a, a), D(0, a)

2. The circles around A, B and C intersect at the trough. You'll get the system of non-linear equations:

$\left|\begin{array}{rcl}x^2+y^2&=&50^2 \\ (x-a)^2+y^2&=&40^2 \\ (x-a)^2+(y-a)^2&=& 30^2 \end{array}\right.$

3. You'll get two solutions. The most plausible one is:

$(x, y, a) = \left( \frac2{13} \cdot \sqrt{59345 - 1040\cdot \sqrt{14}}\ ,\ \frac4{13}\cdot \sqrt{260\cdot \sqrt{14} + 11570}\ ,\ 10\cdot \sqrt{{4\sqrt{14} + 17} \right)$
approximately
(x, y, a)=( 36.22862195, 34.45993254, 56.53903920)

4. My results look very strange so please check my calculations.

**Tonia99** · December 25th 2010, 02:51 AM

Hello Earboth;

Thank you for taking the time to reply. Your answer for length “a” is correct but unfortunately I am still somewhat confused as to how you arrived at it. I understand your point 1, but am not sure what you mean (or why it is necessary) to invoke circles as mentioned in point 2. Also, if the length of one side of the field is “a”, then I would have thought that equation 2 in your list should be (a-x)^2 + y^2 = 40^2 and not (x-a)^2.

I was able to generate 4 simultaneous equations (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 or any additional details on Earboth’s solution would be greatly appreciated.

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

**Archie Meade** · December 25th 2010, 05:15 AM

Originally Posted by Tonia99

Hello Earboth;

Thank you for taking the time to reply. Your answer for length “a” is correct but unfortunately I am still somewhat confused as to how you arrived at it. I understand your point 1, but am not sure what you mean (or why it is necessary) to invoke circles as mentioned in point 2.

The trough-point is where 3 circles with centres at 3 of the field corners
with radii 30, 40, 50 intersect

Pythagoras' theorem gives the equations of circles,
the set of all points at a fixed distance from the centre.

Also, if the length of one side of the field is “a”, then I would have thought that equation 2 in your list should be (a-x)^2 + y^2 = 40^2 and not (x-a)^2.

I was able to generate 4 simultaneous equations (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 or any additional details on Earboth’s solution would be greatly appreciated.

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

In using A, B, C, D you have 4 unknowns.

To solve a system of 4 unknowns, you need 4 clues.

However $A+B=C+D\ \Rightarrow\ D=A+B-C$

Hence you only need solve for 3 unknowns as the 4th depends on them
(you can write any of the 4 in terms of the other 3).

This is the approach Earboth took using x, y and "a".

In your sketch, you could write

$A^2+C^2=50^2$

$(S-A)^2+C^2=40^2$

$(S-C)^2+(S-A)^2=30^2$

where $S$ is the side of a square and $A+B=S\Rightarrow\ B=S-A$

and $C+D=S\Rightarrow\ D=S-C.$

That way, you can more easily work with 3 unknowns.

**Tonia99** · December 25th 2010, 07:01 AM

Hello Archie Meade;

Thank you for your help, but I am still confused as to how to generate the correct answer. First off, I think there might be an error with your first equation (according to my sketch, I think it should be A^2 + C^2 = 50^2, not A^2 + 40^2 = 50^2).

Secondly, I am able to generate the set of equations you listed but I still cannot isolate any one of the variables. For example, C^2 = 50^2 – A^2 (rearranged first equation). If I substitute this into the 2nd equation, I get (S-A)^2 + (50^2 – A^2) = 40^2. Multiplying out the first term I get (S^2 – 2AS + A^2) + (50^2 – A^2) = 40^2 which reduces to S^2 – 2AS = 40^2 – 50^2. At this point I get stuck because I don’t know how to isolate any one of the variables.

Thanks again for taking the time to respond, and I would appreciate any more help with how to get the final answer.
Kind regards,
Tonia

**Archie Meade** · December 25th 2010, 01:33 PM

Yes, that was a typo.

The equations are non-linear and require some patience to solve.

Earboth's idea to find the point of intersection of 3 circles is good.

The equation of the circle centred at $(0,0)$ is

$(x-0)^2+(y-0)^2=50^2\Rightarrow\ x^2+y^2=2500$

The equation of the circle centred at $(a,0)$ is

$(x-a)^2+y^2=40^2=1600$

The equation of the circle centred at $(a,a)$ is

$(x-a)^2+(y-a)^2=30^2=900$

This leads to

$x^2+y^2=2500$

$x^2-2ax+a^2+y^2=1600$

$x^2-2ax+a^2+y^2-2ay+a^2=900$

which gives

$x^2+y^2=2500$

$2500+a^2-2ax=1600\Rightarrow\ 900+a^2=2ax$

$2500+2a^2-2ax-2ay=900\Rightarrow\ 800+a^2-ax=ay$

leading to

$\displaystyle\ y=\frac{800+a\left(a-\sqrt{2500-y^2}\right)}{a}$

$2a\sqrt{2500-y^2}-a^2=900$

The intersection of these curves gives the values for "a"
(plotted on the x-axis for convenience).

**Tonia99** · December 25th 2010, 04:36 PM

Thank you for your reply, Mr. Meade. Part of the problem I was having is that I was trying to solve the problem using Pythagoras' theorem for triangles, and was unaware of the existence of the equation of a circle in the form you (and Earboth) stated. Having said that, I can follow your logic and math up until the point where you say "The intersection of these curves gives the values for "a"". I can see where these intersect on the graph, but how would one determine "a" mathematically based on the 2 equations you give? Does one have to substitute the value for y given in the first equation into the second equation? This seems to me to result in a very messy equation that I cannot solve.

[By the way, in terms of background I found this problem in a math book aimed at grade 8 students, so I am somewhat surprised at the complexity of the solution.]

I feel I almost understand how to get at the final answer but I am not quite there yet. Any last bits of help would be lovely.

Thanks again ever so much for taking the time to respond in such detail.
Kind regards,
Tonia

**Archie Meade** · December 25th 2010, 05:28 PM

Yes Tonia,

there is another way to arrive at a solution.

It's 2.30AM here, so I will post it in the morning for you.

**Archie Meade** · December 26th 2010, 02:48 AM

1. Add a new triangle by rotating triangle BPD about the vertex D through $90^o$
until vertex B meets vertex C, introducing triangle DCE.

2. Triangle PDE is both right-angled and isosceles

$\Rightarrow|\angle\ DPE|=|\angle\ DEP|=45^o$

$|PE|^2=40^2+40^2=2(40)^2\Rightarrow\ |PE|=40\;\sqrt{2}$

3. If we knew the angle $\theta,$ we could find $|CD|^2$ with the Law of Cosines.

That gives us the field area, or taking the square root gives a side of the square field.

The Law of Cosines also gives us $\theta$ using triangle CPE

$30^2=50^2+\left(40\sqrt{2}\right)^2-2(50)40\sqrt{2}cos\theta$

$\Rightarrow\ 2500+3200-900=4000\sqrt{2}cos\theta$

$\displaystyle\Rightarrow\ cos\theta=\frac{4800}{4000\sqrt{2}}=\frac{6}{5\sqr t{2}}$

$\displaystyle\Rightarrow\theta=cos^{-1}\left[\frac{6}{5\sqrt{2}}\right]$

4. Using triangle CPD

$|CD|^2=50^2+40^2-2(50)40cos\left(\theta+45^o\right)$

we get

$Area=3196.66295471\Rightarrow\ side=\sqrt{3196.66295471}=56.539$

**Tonia99** · December 27th 2010, 01:32 AM

A very elegant solution and one that is satisfying. Thank you very much for your wonderful help and for taking the time to walk me through this, and many thanks also to all the others who took the time to respond. Much appreciated.

Kind regards,
Tonia

**bjhopper** · December 27th 2010, 07:40 AM

Hello Tonia,
I add my thanks to Earboth and Archie for their solutions.
For an 8th grader I think that the diagram should have shown a diagonal wth 50 m and 30m segments or stated it that way. It is then a Pythagoran problem wherein the side length is 30/rad2 + 50/rad2 =80/rad2 =56.569

bjh

**Tonia99** · December 27th 2010, 11:41 AM

Hi bjhopper;

I too wanted to "thank" Earboth and Mr. Meade for their help but I don't know how to do this (I don't see any buttons to select for "thanks"). Any help would be lovely so I can do this for them.

Kind regards,
Tonia

**emakarov** · December 27th 2010, 12:09 PM

The button "Thanks" is located on the left just under a post. Here is a picture of the post #9 above.

If you don't see it, maybe you can post a complaint in the Question, Feedback and Comments forum.

**Tonia99** · December 27th 2010, 01:38 PM

Thanks for the advice, emakarov. I unfortunately do not see any "thanks" button in my browser window (I'm using Firefox). Would this be a problem with my machine and/or configuration? I'm not sure lodging a complaint would do anything (or even be appropriate).

Kind regards,
Tonia

**bjhopper** · December 29th 2010, 10:32 AM

go to replier post and click on thanks lower left

bjh

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