Gecko problem

G

Gamest

New member
Joined
Jan 31, 2011
Messages
9
"A gecko is in a room that is 12 feet long, 10 feet wide, and 8 feet tall. The gecko is currently on a side wall (10' by 8'), one foot from the ceiling and one foot from the back wall (12' by 8') The gecko spots a fly on the opposite side wall, one foot from the floor and one foot from the front wall. What is the length of the shortest path the gecko can take to reach the fly assuming that it does not jump and can only walk across the ceiling and the walls? Express your answer in simplest form."

My idea was to make a net of the solid, and use the pythagorean theorem to find the length. The answer I got was (22^2+6^2)^(1/2)=2*130^1/2, but the correct answer was 2*113^1/2. I don't have any more ideas. Thanks for helping.
 
A gecko is in a room that is 12 feet long, 10 feet wide, and 8 feet tall. The gecko is currently on a side wall (10' by 8'), one foot from the ceiling and one foot from the back wall (12' by 8') The gecko spots a fly on the oppositeside wall, one foot from the floor and one foot from the front wall. What is the length of the shortest path the gecko can take to reach the fly assuming that it does not jump and can only walk across the ceiling and the walls? Express your answer in simplest form."

My idea was to make a net of the solid, and use the pythagorean theorem to find the length. The answer I got was 22^2+6^2=2*130^1/2, but the correct answer was 2*113^1/2. I don't have any more ideas. Thanks for helping.

Here is a similar problem which will guide you in the solution of your problem.

>In a rectangular room, 30' x 12' x 12', a spider is on the middle of an end
>wall, one foot from the ceiling. A fly is on the end of the wall of the far
>side, in the middle and one foot above the floor, too frightened to move.
>What is the shortest distance that the spider must travel to get to the fly?
>( Hint: the distance is less than 42 )


Unfold the room into 2 dimensions. You will find that there are four ways of unfolding, shown below, that provide a continuous surface over which you can draw a straight line from the spider to the fly. In each of the layouts, draw a right triangle, the hypotenuse of which connects the spider and fly, and the two legs are parallel to the edges of the room. I'll leave the math to you. The diagram isn't exactly correct but I think you get the idea.
The four possible distances compute to be 42 ft., 40.718 ft., 43.17 ft., and 40 ft. Can you find the unfolding that produces the minimum of 40 ft.?
_____________
I I
I I
I I
I Spider I 12' high
I S I
I_____________I______________
I I I
I I I S to F distance = 40.718 feet
I I I
I I I
I I I 30' long
I I I
I I I
I I I
I____________I_______________I
I I
I I
I F I Fly
I I
I I
I______________I
12' wide

____________
I I
I I
I Spider I 12'
I S I
I___________________________________________
I I I I
I I I I
I I I I
I I I I
I I I 30' long I
I I I I
I I I I
I I I I
I___________ _I____ __________I______________ I
I I I
I I F I
I I Fly I
I I I
I______________I_______________I
12' wide


Good luck.
 
Hello, Gamest!

This is a classic problem, usually involving a spider and a fly.

A gecko is in a room that is 12 feet long, 10 feet wide, and 8 feet tall.
The gecko is currently on a side wall (10 x 8), one foot from the ceiling and one foot from the back wall (12 x 8).
The gecko spots a fly on the opposite side wall, one foot from the floor and one foot from the front wall.
What is the length of the shortest path the gecko can take to reach the fly
assuming that it does not jump and can only walk across the ceiling and the walls?
Express your answer in simplest form.

My idea was to make a net of the solid, and use the Pythagorean theorem to find the length.
The answer I got was 222+62 = 2130\displaystyle 22^2+6^2\,=\,2\sqrt{130}
But the correct answer was 2113\displaystyle 2\sqrt{113}
I don't have any more ideas. Thanks for helping.
Click to expand...

Consider the room to be a rectangular "box".
There are a number of ways to "flatten" the box and trace the gecko's path.

Evidently you had the gecko walk from the Left wall, across the Back wall, then the Right wall to the fly.
Code: 
      :  9  : 1 : - - - - 12  - - - :  9  : 1 :
      *-----*---*-------------------*---------* -
      |     |   |                   |     E   | 1
      |     o - | - - - - - - - - - | - - o   * -
      |     G   | *                 |     |   | :
      |         |       *           |     |   | :
    8 |         |             *     |     |   | 6
      |         |                   *     |   | :
      |         |                   |     o - | -
      |         |                   |    F|   | 1
      *---------*-------------------*-----*---* -
      : - 10  - : - - - - 12  - - - :  9  : 1 :

You have:  GE=22,  EF=6\displaystyle \text{You have: }\:GE = 22,\;EF = 6
. . Hence:  FG = 222+62 = 520 = 2130\displaystyle \text{Hence: }\:FG \:=\:\sqrt{22^2 + 6^2} \:=\:\sqrt{520} \:=\:2\sqrt{130}


Of course, there are other ways to "unfold" the room.

The shortest path is: down the Left Wall, down the Back wall, across the Floor, and up the Right wall.

Code: 
      :  9  : 1 : - - - 12- - - - :
      *-----*---*-----------------*
    1 |     |   |                 |
    - |     o - |                 |
    7 |     G   |   *             | 1 :  7  : - -
      *---------*---------*-------*---*-----* -
          10    |               * |         | 9
                |                 | - o     | -
                |                 |   |     | 1
                *-----------------*---*-----* -
                : - - - 12  - - - : -  8  - :

From G to F ⁣:  {horizontal displacement:14vertical displacement: 16}\displaystyle \text{From }G\text{ to }F\!:\;\begin{Bmatrix} \text{horizontal displacement:} & 14 \\ \text{vertical displacement: } & 16 \end{Bmatrix}

Therefore:  GF = 142+162 = 452 = 2113\displaystyle \text{Therefore: }\:GF \:=\:\sqrt{14^2+16^2} \:=\:\sqrt{452} \:=\:2\sqrt{113}

 
You must log in or register to reply here.
Top