Volume of a Pyramid
The volume of a pyramid can be expressed as \(\frac{1}{3}Ah,\) where \(A\) is the base area of the pyramid and \(h\) is the height of the pyramid. Refer to the image below.
hi
What is the volume of a pyramid with a height of 10 and a square base with sides of length 12?
Since the area of the base is \( 12 \times 12 = 144 \), the volume of the pyramid is
\[ \frac{1}{3} A \times h = \frac{1}{3} \times 144 \times 10 = 480. \ _\square\]
In a playground, some children used \(200 \text{ cm}^3\) of sand to build a pyramid. If the side length of the square base is \(10\text{ cm},\) what is the height of the pyramid?
Since the area of the base is \(A= 10 \times 10 = 100 \), the volume of the pyramid in \(\text{cm}^3\) is
\[ \frac{1}{3} A \times h = \frac{1}{3} \times 100 \times h = 200. \]
Thus, the height of the pyramid is \(h=6 \text{ cm}.\) \(_\square \)
hi
In the above diagram, if the base of the pyramid is a square, what is the volume of the pyramid?
In order to get the volume of the pyramid, we need to find the side length of the base by cutting the pyramid into half.
hi
The above diagram is the cross section of the pyramid cut through \(A, B, D\) and \(C.\) Since the side length of the base is equal to \(\lvert\overline{BC}\rvert,\) which is twice the length \(\lvert \overline{CD}\rvert ,\) we use the Pythagorean theorem as follows to calculate \(\lvert\overline{BC}\rvert:\)
\[\begin{align} \lvert \overline{CD} \rvert^2 &= \lvert \overline{AC} \rvert^2 - \lvert \overline{AD} \rvert^2 \\ &= 25 - 16 = 9. \\ \Rightarrow \lvert \overline{CD} \rvert &= 3 \\ \Rightarrow \lvert \overline{BC} \rvert &= 2 \times \lvert \overline{CD} \rvert = 6. \end{align}\]
Thus, the side length of the base is \(6\text{ cm}.\) Then the volume of the pyramid is
\[\frac{1}{3} A \times h = \frac{1}{3} \times 6 \times 6 \times 4 = 48~ (\text{cm}^3).\ _\square\]
hi
Find the volume of the blue part in the above pyramid.
The volume of the blue part is
\[\text{(volume of whole pyramid)} - \text{(volume of small pyramid on top)}.\]
Since the height ratio between the blue part and the small pyramid on top is \(1 : 1,\) the side length of the base is \(20 \text{ cm}.\) Let \(A\) be the volume of the blue part of the whole pyramid, then
\[\begin{align} A &= \frac{1}{3} (20 \times 20) \times (15 + 15) - \frac{1}{3} (10 \times 10) \times (15) \\ &= 4000 - 500 \\ &= 3500 \ (\text{cm}^3). \ _\square \end{align}\]
A regular pyramid of square base \(7\times 7\) is cut such that another regular pyramid of square base of \(5\times 5\) is removed, leaving a shape with the height of 3 between the parallel square planes as shown above.
What is the volume of this shape?
A square \(ABCD\) has points \(E\) and \(F\) as its midpoints on \(AD\) and \(AB\), respectively. The square is then folded such that the vertices \(A\), \(B\), and \(D\) joined together become a new vertex of the pyramid with triangular base \(EFC\).
If the square has a side length of \(6\text{ cm}\), what is the volume of the pyramid (in \(\text{cm}^3\))?
Inspiration.
Proof by Integration
Derive the formula for the volume of a pyramid using calculus.
First, we want to find \( A(x) \), where \(A(x)\) is the function of the areas of the cross section of the pyramid.
Consider this:
Let \(h\) be the height of the solid, and \(z\) a constant such that \({ z }^{ 2 }\propto A\). Also, let \(x\) and \(y\) be variables which would be used later to define \(A(x)\).
Since \({ z }^{ 2 }\propto A\), we can set \(A=k{z}^{2}\), where \((k)\) is another constant.
From the image above, it can be seen that both triangles are similar. So, finding the equation of \(y\) with respect to variable \(x\), we have \[\frac { z }{ h } =\frac { y }{ x } \Rightarrow y =\frac { xz }{ h }.\] Therefore, \[A(x)=k{y}^{2}=\frac{k{z}^{2}{x}^{2}}{{h}^{2}}.\]
Now, here comes the integrating part.
The volume of the object is \[\begin{align} \int _{ b }^{ a }{ A(x)dx }&=\int _{ 0 }^{ h }{ \frac { k{ z }^{ 2 }{ x }^{ 2 }dx }{ { h }^{ 2 } } } \\ &=\frac { 1 }{ 3 } k{ z }^{ 2 }h. \end{align}\] Recall that \(A=k{ z }^{ 2 }\), which gives \[\frac { 1 }{ 3 } k{ z }^{ 2 }h=\boxed{\frac { 1 }{ 3 } Ah}.\]
Since the volume is based on the area of the cross section, the point at the top of the pyramid can be anywhere and this formula would still work. \( _\square \)