• Mar 17th 2006, 02:19 PM
Sebastianasdf
“A turtle, sunbathing in a square patch of grass, is 13 feet from one of the corners of the patch of grass, 17 feet from the corner diagonally opposite that one, and 20 feet from a third corner. Find the area and perimeter of the grass patch. Assume the land is flat.”

• Mar 17th 2006, 11:30 PM
CaptainBlack
Quote:

Originally Posted by Sebastianasdf
“A turtle, sunbathing in a square patch of grass, is 13 feet from one of the corners of the patch of grass, 17 feet from the corner diagonally opposite that one, and 20 feet from a third corner. Find the area and perimeter of the grass patch. Assume the land is flat.”

First draw a labelled diagram, like the one attached. This has altitudes added
to the two triangles, one with sides 13ft and 20ft and the other with sides
20ft and 17ft, and the square has side $\displaystyle s$

(note in the diagram $\displaystyle h_1=y$ and $\displaystyle h_2=s-x$)

Now apply Pythagoras's theorem to find $\displaystyle h_1$ from triangles
$\displaystyle ATU$ and $\displaystyle DTU$, these give:

$\displaystyle h_1^2=13^2-x^2$,

and

$\displaystyle h_1^2=20^2-(s-x)^2$,

so:

$\displaystyle 13^2-x^2=20^2-(s-x)^2$

Now doing the same thing for $\displaystyle h_2$ and triangles $\displaystyle DVT$
and $\displaystyle CVT$ gives:

$\displaystyle 20^2-y^2=17^2-(s-y)^2$.

Also (as 20 is the diagonal of a rectangle with sides $\displaystyle y$ and $\displaystyle s-x$):

$\displaystyle y^2+(s-x)^2=20^2$

Now these three equations should be sufficient to solve tor $\displaystyle s$
and from this the perimeter and area are easily found.

RonL
• Mar 18th 2006, 08:14 AM
earboth
Quote:

Originally Posted by Sebastianasdf
“A turtle, sunbathing in a square patch of grass, is 13 feet from one of the corners of the patch of grass, 17 feet from the corner diagonally opposite that one, and 20 feet from a third corner. Find the area and perimeter of the grass patch. Assume the land is flat.”

Hello,

Capt.Black wrote:

"Now these three equations should be sufficient to solve tor s
and from this the perimeter and area are easily found."

Of course he's right, but don't try to find easily the solution for s. The system of equations, which gave you Capt.Black is a really nasty one. When I tried to solve this system, I became pretty soon the impression that I've lost the last parts of my brain.
Here is, what I did:

You've got 3 equations (originally generated by Capt.Black):
E(1): $\displaystyle 13^2-x^2=20^2-(s-x)^2$
E(2): $\displaystyle 20^2-y^2=17^2-(s-y)^2$
E(3): $\displaystyle y^2+(s-x)^2=20^2$

from E(1) follows $\displaystyle (s-x)^2=20^2-13^2+x^2$
put this result into E(3), which will give E(4): $\displaystyle x^2+y^2=13^2$

Calculate x from E(1):$\displaystyle x= -\frac{231}{2 \cdot s}+ \frac{1}{2}s$
Calculate y from E(2):$\displaystyle y= \frac{111}{2 \cdot s}+ \frac{1}{2}s$

Put these results into E(4): $\displaystyle \left( -\frac{231}{2 \cdot s}+ \frac{1}{2}s \right)^2+\left(\frac{111}{2 \cdot s}+ \frac{1}{2}s \right)^2=169$

Expand this equation.
$\displaystyle \frac{53361}{4s^2}-\frac{231}{2}+\frac{1}{4} s^2+\frac{12321}{4s^2}+\frac{111}{2}+\frac{1}{4}s^ 2=169$

Simplify to
$\displaystyle \frac{1}{2}s^2-229+\frac{65682}{4s^2}=0$
$\displaystyle 2s^4-916s^2+65682=0$

This equation has 4 real solutions, but only $\displaystyle s=3 \cdot \sqrt{41}$ fits to your problem.

Greetings

EB
• Mar 18th 2006, 09:20 AM
CaptainBlack
Quote:

Originally Posted by earboth
Hello,

Capt.Black wrote:

"Now these three equations should be sufficient to solve tor s
and from this the perimeter and area are easily found."

Of course he's right, but don't try to find easily the solution for s. The system of equations, which gave you Capt.Black is a really nasty one. When I tried to solve this system, I became pretty soon the impression that I've lost the last parts of my brain.
Here is, what I did:

There are a number of ways of deriving sets of equations from the
figure, and I suspect there are some which will give an easier route
to a solution.

So the challenge for the helpers is to find the simplest set of equations
representing this problem (I have a set which is not simpler but requires
less fuss to derive).

I think that the main point that needs to be learnt with geometry problems
like this that the first thing you do is draw a diagram.

RonL
• Mar 18th 2006, 09:26 AM
CaptainBlack
Quote:

Originally Posted by earboth

Simplify to
$\displaystyle \frac{1}{2}s^2-229+\frac{65682}{4s^2}=0$
$\displaystyle 2s^4-916s^2+65682=0$

This equation has 4 real solutions, but only $\displaystyle s=3 \cdot \sqrt{41}$ fits to your problem.

You could tell the poster that you solved this last equation using the
substitution: $\displaystyle u=s^2$ and then used the quadratic formula
on the resulting quadratic, which gives 2 solutions for $\displaystyle u$ and
so four solutions for $\displaystyle s$ of which all but this one you give are
unphysical (or false) solutions which can be rejected by looking at the detail
of the question ($\displaystyle s$ must be positive, and the tortoise must be
inside the square).

RonL
• Mar 18th 2006, 09:48 AM
earboth
Quote:

Originally Posted by CaptainBlack
...

I think that the main point that needs to be learnt with geometry problems
like this that the first thing you do is draw a diagram.

RonL

Hello,

I agree with you 100% at this point. I always recommend a drawing even at other mathematical or physicist(?) problems. And I didn't intend to critcize you when I quoted you, but I wanted only to demonstrate to sebastianasdf where he could continue.

I actually didn't realize that I should have mentioned substitution to get the solutions because I thaught that this is a standard method with biquadratic equation. Next time I'll not forget.

Greetings

EB