SOLUTION: four grapefruit (considered spheres) 6 in. in diameter are placed in a square box whose inside base dimensions are 12 in. in the space between the first 4 grapefruit a fifth of the

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Question 1181643: four grapefruit (considered spheres) 6 in. in diameter are placed in a square box whose inside base dimensions are 12 in. in the space between the first 4 grapefruit a fifth of the same diameter is placed. how deep must the box be so that the top will just touch the fifth grapefruit.
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
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four grapefruit (considered spheres) 6 in. in diameter are placed in a square box
whose inside base dimensions are 12 in. in the space between the first 4 grapefruit
a fifth of the same diameter is placed. how deep must the box be so that the top
will just touch the fifth grapefruit.
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(a)  The centers of the first 4 spheres are vertices of a square with the side length of 2*3 = 6 inches.


(b)  The diagonal of this square is  6%2Asqrt%282%29 inches long.

     The half of this diagonal is  3%2Asqrt%282%29  inches long .



(c)  Now consider the right angled triangle formed by the centers of the 1st and 5th spheres as the hypotenuse,

     vertical line from the center of the 5th spere as the leg, 

     and the horizontal line from the center of the 1st sphere coinciding with the diagonal of the above mentioned square.



(d)  The hypotenuse of this triangle is 2 times the radius of 3 cm, i.e. 6 inches.

     Its leg, parallel to the diagonal of the square, is  3%2Asqrt%282%29  inches long.

     Having it, you just can calculate the length of the vertical leg. It is


          sqrt%286%5E2+-+%283%2Asqrt%282%29%29%5E2%29 = sqrt%2836+-+9%2A2%29 = sqrt%2818%29 = 3%2Asqrt%282%29.


          ( It is interesting to note, that it MEANS that the line connecting centers of the 1st and the 5th square
            is inlined 45 degrees to horizon (!) )


(e)  Thus we found that the center of the 5th sphere is elevated  3%2Asqrt%282%29  inches over the plane of the centers 
     of the first four spheres.


(f)  Combining the gained information, we see that the depth of the box must be equal to 

               3+%2B+3%2Asqrt%282%29+%2B+3 = 6+%2B+3%2Asqrt%282%29 inches = 10.243 inches (rounded).


ANSWER.  The depth of the box should be equal to  6+%2B+3%2Asqrt%282%29 inches = 10.243 inches  (rounded).

Solved, answered and carefully explained.