Magic Plumbing is needing to ship out a new water pipe to replace a broken one in the Smith’s house. The only box they could find has dimensions of 20 in x 16 in x 12in. The pipe they need to...
Magic Plumbing is needing to ship out a new water pipe to replace a broken one in the Smith’s house. The only box they could find has dimensions of 20 in x 16 in x 12in. The pipe they need to ship is 24 inches long. Will it fit in the box? Explain your answer.
This can be solved by finding the length of a few right triangles.
The diagonal distance from one corner of the box to the opposite corner, as shown in the diagram, is the hypotenuse of a right triangle. If we labelled the corners of the box in the diagram like so:
...then the triangle is formed by GBF. I would recommend that you label the corners on your diagram just like this, so you can see where these shapes are located.
For the triangle GBF, we want to find out if the hypotenuse, GB, is long enough to fit the 24 inch pipe. To find the hypotenuse, we can use the pythagorean theorem; the sum of the squared sides equal the hypotenuse squared. However, we first need to know the length of the sides, which are the lengths BF and GF.
BF is easy: this is just the height of the box, 12 inches.
GF is a little more complicated because this is the hypotenuse of a triangle as well; the triangle GFH. To find THIS hypotenuse, we need to square the lengths FH and GH.
So, we're adding the squares of FH and GH, then finding the root. This will be GF. Then we're adding the squares of GF and 12, and then finding the root. That will be our final answer, the length of GB.
FH is the width of the box, 16 inches.
GH is the length of the box, 20 inches.
So our full equation is: √((20²+16²) + 12²)
This comes out to 28 inches, so there's plenty of room for the pipe.
The length of a diagonal from opposite corners of a rectangular prism (parallelpiped) is:
`d=sqrt(a^2+b^c+c^2) ` where a,b, and c are the dimensions of the "box".
The length of the diagonal is `d=sqrt(20^2+16^2+12^2)=sqrt(800)=20sqrt(2)~~28.28 ` in.
As long as the pipe has a small enough diameter will fit in the box.
(The approximate maximum diameter is 6.25in)