# conundrum

• Aug 16th 2006, 07:54 PM
wardreese
conundrum
I originally was presented with this conundrum as a challenge on a website, but the time limit has passed to present an answer. Still, I am not sure how to solve the conundrum.
The Farmer has 3 fields. One field is an equilateral triangle. One field is a circle. One field is a square. The square field is 75% larger in area than the triangular field. The square field is 50% larger in area than the circular field. In order to completely fence all three fields exactly 4000 meters of fencing is required. What is the total area of all three fields?
This is presented to me as a conundrum. I have worked the problem thus far:
a= circle area
x= triangle area
n= square area
n=x+(.75x)
n=a+(.5a)
or n=1.75x=1.5a
2 times Pi times the radius is a circle's circumference
4 times the length of a side is a square's perimeter
3 times the length of a side is an equilateral triangle's perimeter
Pi times the radius of the circle squared is the circle's area
The length of the side of the square squared is the square's area
Base times Height of a triangle is the triangle's area
I am sort of fuzzy about how to find the height of the equilateral triangle, but I suspect I need to know the Pythagorean theorem, which I don't remember from college.
My equation to find all the perimeter's is:
s is measure of side of square field
t is measure of side of triangular field
4000=(2*Pi*r)+(4s)+(3t)

PS: I already posted this on an algebra tutor website, but I got an answer that seems wrong, since the person who answered it tried to solve the equation for the perimeter using the area equation variables.
"a= circle area
x= triangle area
n= square area
so n=1.75*x=1.75x
n=1.5*a=1.5a
and x+a+n=4000
x+a+1.75x=4000
a+2.75x=4000
a=4000-2.75x
x+a+1.5a=4000
2.5a+x=4000
2.5*(4000-2.75x)+x=4000
10000-6.875x+x=4000
6.875x-x=10000-4000
5.875x=6000
x=1021.28
and thus
a=4000-2.75x
a=4000-(2.75*1021.28)
a=4000-2808.51
a=1191.49
and check this out
n=1.5a=1.5*1191.49=1787.23 (let pick this one b'coz it makes 4000 even)
n=1.75x=1.75*1021.28=1787.24
x+a+n=4000
1021.28+1191.49+1787.23=4000 "
• Aug 16th 2006, 08:59 PM
CaptainBlack
Quote:

Originally Posted by wardreese
I originally was presented with this conundrum as a challenge on a website, but the time limit has passed to present an answer. Still, I am not sure how to solve the conundrum.
The Farmer has 3 fields. One field is an equilateral triangle. One field is a circle. One field is a square. The square field is 75% larger in area than the triangular field. The square field is 50% larger in area than the circular field. In order to completely fence all three fields exactly 4000 meters of fencing is required. What is the total area of all three fields?
This is presented to me as a conundrum. I have worked the problem thus far:
a= circle area
x= triangle area
n= square area
n=x+(.75x)
n=a+(.5a)
or n=1.75x=1.5a
2 times Pi times the radius is a circle's circumference
4 times the length of a side is a square's perimeter
3 times the length of a side is an equilateral triangle's perimeter
Pi times the radius of the circle squared is the circle's area
The length of the side of the square squared is the square's area
Base times Height of a triangle is the triangle's area
I am sort of fuzzy about how to find the height of the equilateral triangle, but I suspect I need to know the Pythagorean theorem, which I don't remember from college.
My equation to find all the perimeter's is:
s is measure of side of square field
t is measure of side of triangular field
4000=(2*Pi*r)+(4s)+(3t)

Let the side of the triangle be $t$, the side of the square be $s$, and the radius of the circle be $r$.

Then the area of the equilateral triangle (by Herons formula - google for
it if you don't know it) is:

$A_{t}=\sqrt{(3/2)t\ (t/2)^3}=\frac{t^2}{2}\sqrt{3}$

The area of the circle is:

$A_c=\pi\ r^2$,

and the area of the square is:

$A_s=s^2$.

We are told that:

$A_s=1.75 A_t=1.5 A_c$

which allows us to write $t$ and $r$ in terms of $s$.

We are also told that:

$4000=2 \pi r + 3 t +4 s$

which when we replace $t$ and $r$ with what they are in terms of $s$ we are left with
a single linear equation for $s$ which is trivial to solve.

RonL
• Aug 17th 2006, 12:59 AM
wardreese
Okaaaayyyyy............?
I know that Heron's formula is something I should have been able to find out for myself, and I really appreciate the freebie there. :) What I don't understand is how the radius, the length of side of the square, and the length of side of the triangle can be determined from knowing the relationship between them, since you don't have the area of any of them, or their total area, or even what proportion of the line of 4000 meters each perimeter equals. How do you reduce three variables to a single solvable variable? :confused:
Oh, and for that matter, I am not a user of higher mathematics on any regular basis. I can make jewelry, throw a pot, overhaul an aircooled engine, repair a jet turbine, repaint a car, discuss Campbell's theories of mythology or write a story that is worth reading, but I haven't ever encountered a linear equation outside a classroom. I do this kind of thing for fun, and to expand my experience, but my level of experience is that I have forgotten most of what is in high school or college algebra due to lack of application. It may seem really basic to you, but could you lead me through the rest of the setup for this problem? I really don't understand where you left me. I feel like I am standing in the middle of a road intersection in the middle of Nebraska with nothing but fields around and no road signs. You really would be showing the method to me for the first time.
Thank you for taking the time with this.
wardreese
• Aug 17th 2006, 02:11 AM
CaptainBlack
Quote:

Originally Posted by wardreese
I know that Heron's formula is something I should have been able to find out for myself, and I really appreciate the freebie there. :) What I don't understand is how the radius, the length of side of the square, and the length of side of the triangle can be determined from knowing the relationship between them, since you don't have the area of any of them, or their total area, or even what proportion of the line of 4000 meters each perimeter equals. How do you reduce three variables to a single solvable variable? :confused:

$A_s=1.75\ A_t$,

so:

$s^2=1.75\ \frac{t^2}{4}\sqrt{3}$,

so:

$s \approx 0.615 t$.

Similarly:

$A_s=1.5\ A_c$,

so:

$
s^2=1.5\ \pi\ r^2
$

so:

$
s \approx 2.171 r
$
.

Now:

$
4000=2 \pi r + 3 t +4 s
$
,

so

$
4000=2 \pi \frac{s}{2.171} + 3 \frac{s}{0.615} +4 s
= 11.77 s$

so:

$
s=4000/11.77=339.8\ \mbox{m}
$
.

Now given that we know $s$ we can find $r$ and $t$, and from them the areas.

RonL
• Aug 17th 2006, 07:45 AM
Soroban
Hello, wardreese!

I'll try to give you a walk-through . . .

Quote:

The Farmer has 3 fields: one is an equilateral triangle, one is a circle, one is a square.
The square field is 75% larger in area than the triangular field.
The square field is 50% larger in area than the circular field.

In order to completely fence all three fields exactly 4000 meters of fencing is required.
What is the total area of all three fields?

Let $s$ = side of the equilateral triangle.
. . Then its area is: . $A_t \:=\:\frac{\sqrt{3}}{4}s^2$ . . . a standard formula **
. . and its perimeter is: . $P_t \:=\:3s$

Let $r$ = radius of the circle.
. . Then its area is: . $A_c\:=\:\pi r^2$
. . and its perimeter is: . $P_c\:=\:2\pi r$

Let $x$ = side of the square.
. . Then its area is: . $A_s\:=\:x^2$
. . and its perimeter is: . $P_s\:=\:4x$

We are told that: . $A_s\:=\:175\% \times A_t \:=\:\frac{7}{4}\,A_t\quad\Rightarrow\quad A_t\:=\:\frac{4}{7}\,A_s
$

. . This means: . $\frac{\sqrt{3}}{4}\,s^2\:=\:\frac{4}{7}\,x^2\quad \Rightarrow\quad s\;=\;\frac{4}{\sqrt{7\sqrt{3}}}\,x
$

We are told that: . $A_s\:=\:150\% \times A_c\:=\:\frac{3}{2}\,A_c\quad\Rightarrow\quad A_c\:=\:\frac{2}{3}\,A_s$
. . This means: . $\pi r^2\:=\:\frac{2}{3}\,x^2\quad\Rightarrow\quad r \:= \:\sqrt{\frac{2}{3\pi}}\,x$

The perimeter is: . $P_t + R_c + R_s \:=\:4000$

So we have: . $3\left(\frac{4}{\sqrt{7\sqrt{3}}}\,x\right) + 2\pi\left(\sqrt{\frac{2}{3\pi}}\,x\right) + 4x \;= \;4000$

Now solve for $x$ . . . then determine the total area.

• Aug 17th 2006, 07:59 AM
galactus
Here was my take on the problem. Soroban and I agree :)

• Aug 17th 2006, 08:49 AM
CaptainBlack
Opps, slight typo which has propagated through:

Quote:

Originally Posted by CaptainBlack
Let the side of the triangle be $t$, the side of the square be $s$, and the radius of the circle be $r$.

Then the area of the equilateral triangle (by Herons formula - google for
it if you don't know it) is:

$A_{t}=\sqrt{(3/2)t\ (t/2)^3}=\frac{t^2}{2^2}\sqrt{3}$

The area of the circle is:

$A_c=\pi\ r^2$,

and the area of the square is:

$A_s=s^2$.

We are told that:

$A_s=1.75 A_t=1.5 A_c$

which allows us to write $t$ and $r$ in terms of $s$.

We are also told that:

$4000=2 \pi r + 3 t +4 s$

which when we replace $t$ and $r$ with what they are in terms of $s$ we are left with
a single linear equation for $s$ which is trivial to solve.

RonL