A frustum of a pyramid is 16 cm square at the bottom ,6 cm square at the top ,and 12 cm high.Find the volume of the frustum
Consider the whole pyramid, before the top (with height h) was cut off to make the frustrum. Using similar triangles, the missing height is 36/5. So, the volume of the frustrum is
1/3 * 16^2 * (12 + 36/5) - 1/3 * 6^2 * 36/5 = 1552 cm^3
A little bit helpful
Hmmmmmmmmm.....
Why did the frustum go to the baseball game?
Because it heard there would be lots of "py-raised" cheers!
Now, let's calculate the volume of the frustum:
First, we need to find the areas of the top and bottom bases. The area of the bottom base (A₁) is 16 cm², and the area of the top base (A₂) is 6 cm².
The formula for finding the volume of a frustum is V = (1/3) * h * (A₁ + A₂ + √(A₁ * A₂)).
Plugging in the values we know:
V = (1/3) * 12 cm * (16 cm² + 6 cm² + √(16 cm² * 6 cm²))
V = (1/3) * 12 cm * (22 cm² + √(96 cm⁴))
V = (1/3) * 12 cm * (22 cm² + 9.8 cm²)
V = (1/3) * 12 cm * 31.8 cm²
V = 4 cm * 31.8 cm²
V = 127.2 cm³
So, the volume of the frustum is 127.2 cm³.
To find the volume of a frustum of a pyramid, you can use the formula:
\[V = \frac{1}{3}h\cdot(A_b + A_t + \sqrt{A_b \cdot A_t})\]
Where:
- \(V\) is the volume of the frustum
- \(h\) is the height of the frustum
- \(A_b\) is the area of the bottom base
- \(A_t\) is the area of the top base
In this case, we are given that the area of the bottom base (\(A_b\)) is 16 cm², the area of the top base (\(A_t\)) is 6 cm², and the height (\(h\)) is 12 cm.
Substituting these values into the formula, we have:
\[V = \frac{1}{3} \cdot 12 \cdot (16 + 6 + \sqrt{16 \cdot 6})\]
We can simplify this equation further:
\[V = \frac{1}{3} \cdot 12 \cdot (22 + \sqrt{96})\]
Now, let's calculate the value of \(V\):
\[V \approx \frac{1}{3} \cdot 12 \cdot (22 + \sqrt{96}) \approx \frac{1}{3} \cdot 12 \cdot (22 + 9.8) \approx \frac{1}{3} \cdot 12 \cdot 31.8 \approx 127.2 \, \text{cm}^3\]
Therefore, the volume of the frustum is approximately 127.2 cm³.
make a sketch to show the complete pyramid
edge of bottom = 4
edge of top of frustum = √6
height of imagined missing pyramid --- h
by ratios:
h/(h+4) = √6/4
4h = √6h + 4√6
h = 4√6/(4 + √6)
Volume of complete pyramid = (1/3)(16)(4√6/(4 + √6)) = ...
Volume of missing pyramid = (1/3)(6)(4√6/(4 + √6)) = ....
volume of frustum = the difference of these two volumes
copy error:
h/(h+4) = √6/4 should have been h/(h+12) = √6/4
followed by the necessary changes.
btw, oobleck interpreted it as a square base with sides 16 cm, whereas I interpreted "16 cm square" to have a base of 16 cm^2
mmmhhh?