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Math Help - Sunbathing Turtle Problem Please Help

  1. #1
    Sebastianasdf
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    Sunbathing Turtle Problem Please Help

    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.

    How can I solve the problem? please help I don't understand
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  2. #2
    Grand Panjandrum
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    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.

    How can I solve the problem? please help I don't understand
    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 s

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

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

    <br />
h_1^2=13^2-x^2<br />
,

    and

    <br />
h_1^2=20^2-(s-x)^2<br />
,

    so:

    <br />
13^2-x^2=20^2-(s-x)^2<br />

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

    <br />
20^2-y^2=17^2-(s-y)^2<br />
.

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

    <br />
y^2+(s-x)^2=20^2<br />

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

    RonL
    Attached Thumbnails Attached Thumbnails Sunbathing Turtle Problem Please Help-tortoise.png  
    Last edited by CaptainBlack; March 17th 2006 at 11:47 PM.
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  3. #3
    Super Member
    earboth's Avatar
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    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.

    How can I solve the problem? please help I don't understand
    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): 13^2-x^2=20^2-(s-x)^2
    E(2): 20^2-y^2=17^2-(s-y)^2
    E(3): y^2+(s-x)^2=20^2

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

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

    Put these results into E(4): \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.
    \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
    \frac{1}{2}s^2-229+\frac{65682}{4s^2}=0
    2s^4-916s^2+65682=0

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

    Greetings

    EB
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  4. #4
    Grand Panjandrum
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    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
    Last edited by CaptainBlack; March 18th 2006 at 09:27 AM.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by earboth

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

    This equation has 4 real solutions, but only s=3 \cdot \sqrt{41} fits to your problem.
    You could tell the poster that you solved this last equation using the
    substitution: u=s^2 and then used the quadratic formula
    on the resulting quadratic, which gives 2 solutions for u and
    so four solutions for 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 ( s must be positive, and the tortoise must be
    inside the square).

    RonL
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  6. #6
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    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
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