Eric builds a small pyramid for a school project. His pyramid has a height of twelve inches and a square base that measures ten inches on each side. Eric wants to find the smallest cube-shaped box to put his pyramid in so that he can safely bring it to school right side up. What is the volume of this box, in inches cubed?
y r u cheating on alcumus
Sorry but it's wrong :(
hint: put it in on its triangular face
stop cheating on alcumus
juicy nuggets
To find the volume of the cube-shaped box that can fit Eric's pyramid, we need to determine the length of the box's side.
Since Eric's pyramid has a square base that measures ten inches on each side, the length, width, and height of the pyramid's base are all equal to ten inches. Additionally, the height of the pyramid is twelve inches.
To fit the pyramid inside the cube-shaped box, the box's side length should be greater than or equal to the length of the longest diagonal of the pyramid's base. Since the base of the pyramid is a square, the longest diagonal can be found using the Pythagorean theorem.
The length of the longest diagonal of the square base is the hypotenuse of a right triangle formed by two sides of the square base. Let's call the length of the longest diagonal "d".
Using the Pythagorean theorem, we have:
d^2 = 10^2 + 10^2
d^2 = 100 + 100
d^2 = 200
d ≈ 14.14 inches (rounded to two decimal places)
Therefore, the side length of the cube-shaped box should be greater than or equal to 14.14 inches.
Since a cube has all sides equal in length, the volume of the cube-shaped box can be found by cubing the side length.
Let's calculate the volume of the box:
Volume = side length^3
Volume = 14.14^3
Volume ≈ 2862.83 inches cubed (rounded to two decimal places)
Therefore, the volume of the smallest cube-shaped box to safely transport Eric's pyramid is approximately 2862.83 inches cubed.
the smallest box must have an edge length of 12 inches
v = e^3