solids

[wised]

Find the volume of a frustum of a right pyramid whose lower base is a square with a side 5 in, whose upper base is a square with a side 3 in, and whose altitude is 12 in. Round to the nearest whole number.

 

[soroban]

Hello, wised!

Find the volume of a frustum of a right pyramid whose lower base is a square with a side 5 in,
whose upper base is a square with a side 3 in, and whose altitude is 12 in.
Round to the nearest whole number.

There is a formula for the volume of a frustum of a pyramid:

\(\displaystyle \L\;\;\;V\;=\;\frac{h}{3}\left(B_1\,+\,\sqrt{B_1B_2}\,+\,B_2\right)\)

\(\displaystyle \;\;\)where \(\displaystyle h\) is the height and \(\displaystyle B_1,\,B_2\) are the areas of the two bases.

In your problem: \(\displaystyle \,h\,=\,12,\:B_1\,=\,9,\:B_2\,=\,25\)

Therefore: \(\displaystyle \:V\;=\;\frac{12}{3}\left({9\,+\,\sqrt{9\cdot25}\,+\,25\right)\:=\:4(25\,+\,15\,+\,9)\;=\;196\) in\(\displaystyle ^3\).

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

You can invent your own formula for this problem.

The volume of a pyramid is: \(\displaystyle \:V\;=\;\frac{1}{3}BH\)
\(\displaystyle \;\;\)where \(\displaystyle B\) is the area of the base and \(\displaystyle H\) is the height.


Consider the entire pyramid . . . the side view looks like this:

              E
              *
             /|\
            / | \
           /  |x \
          /   |   \
        B*----+----*C
        /     |F    \
       /    12|      \
      /       |       \
    *---------+---------*
    A         G         D

\(\displaystyle \text{We have: }\,BC\,=\,3,\:AD\,=\,5,\:FG\,=\,12.\;\;\text{Let }x\,=\,EF\)

From similar triangles, we have: \(\displaystyle \L\,\frac{x\,+\,12}{5}\:=\:\frac{x}{3}\;\;\)\(\displaystyle \Rightarrow\;\;x\,=\,18\)


Hence, the entire pyramid has: \(\displaystyle \,B\,=\,25,\,H\,=\,30\)
\(\displaystyle \;\;\)and its volume is: \(\displaystyle \:V_1\;=\;\frac{1}{3}(25)(30)\,=\,250\) in\(\displaystyle ^3\)

The upper pyramid has: \(\displaystyle \,B\,=\,9,\,H\,=\,18\)
\(\displaystyle \;\;\)and its volume is: \(\displaystyle \:V_2\;=\;\frac{1}{3}(9)(18)\;=\;54\) in\(\displaystyle ^3\)


Therefore, the volume of the frustum is: \(\displaystyle \:V\;=\;250\,-\,54\;=\;196\) in\(\displaystyle ^3\).

 

[Denis]

Find the volume of a frustum of a right pyramid whose lower base is a square with a side 5 in, whose upper base is a square with a side 3 in, and whose altitude is 12 in. Round to the nearest whole number.

Well, you spelled it right; many think the word is "frustrum"

Do you at least know the formula for volume? If not, go here (scroll down a bit):
http://www.mathsisfun.com/geometry/pyramids.html

Calculate volume of the full pyramid, then similarly volume of the smaller pyramid
(with base 3 in square); difference will be the frustrum, whoops frustum's volume.