SOLUTION: martin wants to ship a souvenir baseball bat that is 16 inches long in this box. um i don't have the pic of the box but i will describe it to you. its 12 inches at the base. 9 inch

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Question 964177: martin wants to ship a souvenir baseball bat that is 16 inches long in this box. um i don't have the pic of the box but i will describe it to you. its 12 inches at the base. 9 inches on the other base and 8 inches on the height if that makes sense. the question asks: will martin be able to ship the bat in the box? show your work and explain your answer please... i am just so confused. the second question asks: martin finds another box he could use to ship the same bat in. the box is a cube, and has the least whole number edge lengths that will allow the bat to fit. what is the edge length of that box?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
This is the base of the box:
The length of the diagonal AC (in inches) is
green%28d%29=sqrt%2812%5E2%2B9%5E2%29=sqrt%28144%2B81%29=sqrt%28225%29=15 (so says Pythagoras).
It is clear that you cannot lay down the bat on tha base of the box,
along that diagonal, but there may be a way.
However, you could rest one end of the bat on one corner of the base of the box,
and see if the other end can fit above the opposite corner of the base,
but still within the box.
Let's see what that box would look in almost 3-D:
It does not look realistic, because I did not draw true perspective.
For true perspective I should draw shorter the lines are supposed to be further back,
for example making CD shorter than AE.
Instead, I made them the same length.
Maybe you could rest one end of the bat on one corner of the base of the box (A),
and see if the other end can fit above the opposite corner of the base (C),
but still within the box (maybe at D, or a little below D along line CD).
We just need to measure the distance AD to see if the bat will fit.
Let me draw the other edges and diagonals, as if the box were see-through
(I will just make them different colors, instead of black):
Since AC is horizontal and CD is vertical, angle ACD is a right angle, and triangle ACD is a right triangle.

So, according to the Pythagorean theorem
AD=sqrt%2815%5E2%2B8%5E2%29=sqrt%28225%2B64%29=sqrt%28289%29=17 .
So the box is big enough, and the bat fits.

Now, what about a square box, with side length x inches?
The length (in inches) of a diagonal of the base (like AC) is
sqrt%28x%5E2%2Bx%5E2%29=sqrt%282x%5E2%29 .
That diagonal and a vertical edge of length x inches form a right angle,
and with a line connecting opposite corners of the box (like AD) they form a right triangle.
The distance between opposite corners of the box is
.
Since sqrt%283%29=about+1.73 ,
for x=9 , 9sqrt%283%29=about15.6 ,
so a cube-shaped box of edge length 9 inches is too small,
but for x=10 , 10sqrt%283%29=about17.3 ,
so a cube-shaped box of edge length highlight%2810%29 inches is large enough to fit the bat.