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Conical and pyramidal frustum
If we intersect a pyramidal solid or a cone with a plane parallel to its base, we get a pyramidal frustum (Figure 56) and a conical frustum (Figure 55), respectively.
An easy observation based on the figures is that the plane figure of the intersection is similar to the base face. This is because the planar intersection can be dilated into the base using the apex as centre. This property has an important role in determining the volume of the pyramidal frustum.
This also means that the small cut off pyramid is similar to the original pyramid. The same dilation that can be used to map the planar intersection to the base maps the small pyramid into the original one.
\latex{\frac{m_{2} }{m_{1} }=\lambda}
\latex{m}
\latex{m_{1}}
\latex{m_{2}}
Figure 55
Example 1
Where should we cut the right square pyramid parallel to its base to halve its volume?
Solution
Let the distance of the cutting plane from the apex of the pyramid be \latex{ x } and denote the altitude of the pyramid by \latex{ m }. Our goal is to find the value of the ratio \latex{\frac x m}. This ratio is nothing else but the ratio of the similarity which transforms the original pyramid into the small cut off pyramid. (Figure 56)
To determine the volume, first we need to compute the altitude of the solid. This can be done by applying the Pythagorean theorem for the right angled triangle seen in Figure 51.
\latex{\frac{V}{V_{1} }= \lambda^{3}}
\latex{\frac{m}{x}= \lambda}
\latex{ V }
\latex{ V_1 }
\latex{ x }
\latex{ m }
Figure 56
Let us denote the volume of the original pyramid by \latex{ V } and the volume of the cut off pyramid by \latex{ V_1 }. The ratio of these, according to the criteria, is
\latex{\frac{V_1}{V}=\frac 1 2}.
We already know that the ratio of the volume of similar solids is equal to the cube of the ratio of the similarity, that is,
\latex{\frac{V_1}{V}=\left(\frac{x}{m}\right)^3}.
From this, the ratio in question can be determined:
\latex{\frac x m=\sqrt[3]{\frac 1 2}=\frac 1{\sqrt[3]{2}}}.
This means that to halve the volume, we have to cut the pyramid at the \latex{\frac 1{\sqrt[3]{2}}} part of altitude with a plane parallel to the base.
The volume and the pyramidal frustum
When determining the volume of the pyramidal frustum, we use the fact that it can be expressed as the difference of the volumes of the original and the cut off pyramid. Let us use the notation of Figure 57.
The volume of the original solid is
\latex{V_1=\frac{T\times m_1}{3}}.
The volume of the cut off solid is
\latex{V_2=\frac{t\times m_2}{3}}.
Therefore the volume of the pyramidal frustum is \latex{V = V_1 – V_2}, its altitude is \latex{m = m_1 – m_2}.
Let the ratio of similarity of the two pyramids be denoted by \latex{\lambda}.
\latex{\frac{m_{1} }{m_{2} }=\lambda}
\latex{\frac{T}{t}=\lambda^{2}}
\latex{\frac{V}{V_{1} }=\lambda^{3}}
\latex{ T }
\latex{ m_2 }
\latex{ m_1 }
\latex{ t }
Figure 57
Using the correspondences valid for similar solids,
\latex{\lambda=\frac{m_1}{m_2}}, \latex{\lambda^2=\frac{T}{t}}, \latex{\lambda^3=\frac{V_1}{V_2}}.
The difference of the volumes is
\latex{V=V_1-V_2=\lambda^3\times V_2-V_2=V_2\times(\lambda^3-1)=\\=\frac{t\times m_2}{3}\times(\lambda-1)\times(\lambda^2+\lambda+1).}
Multiply the first expression in brackets by \latex{ m_2 }, the second one by \latex{ t } to get
\latex{V=\frac 1 3\times(\lambda\times m_2-m_2)\times(\lambda^2\times t+\lambda\times t+t).}
\latex{V=\frac 1 3\times(\lambda\times m_2-m_2)\times(\lambda^2\times t+\lambda\times t+t).}
Using the equalities based on the similarity, we have
\latex{V=\frac 1 3\times(m_1-m_2)\times\left(T+\sqrt{\frac{T}{t}\times t+t}\right).}
Thus the following equality holds for the volume of the pyramidal frustum:
\latex{V=\frac{m}{3}\times(T+\sqrt{T\times t}+t)}.
To determine the surface area of the pyramid, one has to add the areas of the bounding faces. In case of the pyramidal frustum, this can be divided into three groups, the areas of the base and top faces and the area of the lateral surface:
\latex{A=A_{\text{base}}+A_{\text{top}}+A_{\text{lateral surface}}}.
Example 2
The base edge of a regular quadrilateral pyramidal frustum has length of \latex{ 32\, cm }, that of the top face has length of \latex{ 20\, cm } and the length of the side edges is \latex{ 14\, cm }. Determine its volume and surface area.
Solution
Let the base of the pyramidal frustum be the square \latex{ ABCD } while its top face be the square \latex{ A_1B_1C_1D_1 }. Let \latex{A’_1B’_1C’_1D’_1} denote the orthogonal projection of the top face onto the base face. Since the pyramidal cone is regular, the projected image is symmetric to the centre of the base face. Its sides are parallel to the edges of the base square. (Figure 58)
The line \latex{A’_1B’_1} intersects the edge \latex{ BC } orthogonally in point \latex{ T }, therefore
\latex{ D }
\latex{ D_1 }'
\latex{ B_1 }'
\latex{ A_1 }'
\latex{ A }
\latex{ B }
\latex{ T }
\latex{ C }
\latex{ C_1 }'
\latex{ B_1 }
\latex{ C_1 }
\latex{ D_1 }
\latex{ A_1 }
Figure 58
\latex{BT=\frac{1}{2}\times (32-20)=6\, cm.}
It follows from the right angled triangle \latex{BTB_{1}} that
\latex{B_{1}T=\sqrt{14^{2}-6^{2} }=\sqrt{160}=4\times\sqrt{10}\, cm},
which equals to the altitude of the side trapezoid face, thus the area of this face is
\latex{\frac{32+20}{2}\times 4\sqrt{10}=104\times \sqrt{10}\, cm.}
Therefore the surface of the pyramidal frustum is
\latex{A=32^{2}+20^{2}+4\times 104\times \sqrt{10}=416\times \sqrt{10}\approx 2739.5\, cm^{2}.}
The length of the altitude can be determined using the triangle \latex{B_{1},TB'_{1}:}
\latex{m=B_{1}B'_{1}=\sqrt{B_{1}T^{2}-B'_{1}T^{2} }=\sqrt{160-6^{2} }=\sqrt{124}=2\times \sqrt{31}\, cm.}
Apply the formula for the volume of the pyramidal frustum to find that the volume of the solid is
\latex{V=\frac{2\times \sqrt{31} }{3}\times (32^{2}+\sqrt{32^{2}\times 20^{2} }+20^{2} )=1376\times \sqrt{31}\approx 7661.2\, cm^{3}.}
The volume of the conical frustum
The formula for the volume of pyramidal frustum can be applied for conical frustums as well. If the segment connecting the base and top discs of the frustum is orthogonal to the plane of the base face, then it is called a right conical frustum, otherwise it is said to be oblique.
If the radius of the base disc is \latex{ R } and the radius of the top disc is \latex{ r }, and the altitude of the frustum is denoted by \latex{ m } (Figure 59), then it follows from the already proven formula that
\latex{V=\frac{m}{3}\times (R^{2}\times \Pi +\sqrt{R^{2}\times \Pi \times r^{2}\times \Pi }+r^{2}\times \Pi ),} from which
\latex{ R }
\latex{ r }
\latex{ m }
\latex{ a }
Figure 59
\latex{V=\frac{m\times \Pi }{3}\times \left(R^{2}+R\times r+r^{2} \right).}
When determining the surface area of the right conical frustum, one has to keep in mind that the lateral surface can be laid into the plane along the top and base faces, and the lateral surface determines an annulus segment. (Figure 60)
The middle arc of the resulting annulus segment is the arithmetic mean of the circumferences of the two discs:
The middle arc of the resulting annulus segment is the arithmetic mean of the circumferences of the two discs:
\latex{\frac{2\times R\times \Pi +2\times r\times\Pi }{2}=(R+r)\times \Pi ,}
and its width is the generatrix \latex{ a } of the conical frustum. Therefore the area of the lateral segment is
\latex{A_{lateral segment}=(R+r)\times \Pi \times a.}
Using this result, the surface area of the conical frustum is
\latex{A=R^{2}\times \Pi +r^{2}\times \Pi +(R+r)\times \Pi \times a,} from which
\latex{ r }
\latex{ R }
\latex{ a }
Figure 60
\latex{A=\Pi \times (R^{2}+r^{2}+(R+r)\times a ).}
Example 3
A right conical frustum with altitude \latex{m =10\, cm}, generatrix \latex{a = 15\,cm} and volume \latex{V=1,000\times \Pi\, cm^{3}} is given. Find the radii of the base and top faces.
Solution
Let \latex{ r } and \latex{ R } denote the radii of the top and base faces, respectively. Take the axial section of the solid, and let us select a right angled triangle \latex{ ABC } as seen in Figure 61.
The length of the segment \latex{ AC } is, obviously, \latex{R – r}, therefore an application of the Pythagorean theorem gives
The length of the segment \latex{ AC } is, obviously, \latex{R – r}, therefore an application of the Pythagorean theorem gives
\latex{a^{2}-m^{2}=(R-r)^{2},}
\latex{125=R^{2}-2\times R\times r+r^{2}.}
\latex{125=R^{2}-2\times R\times r+r^{2}.}
On the other hand, it follows from the formula for the volume of the conical frustum that
\latex{V=\frac{m\times \Pi }{3}\times (R^{2}+R\times r+r^{2} ),}
\latex{300=R^{2}+R\times r+r^{2}.}
\latex{300=R^{2}+R\times r+r^{2}.}
Subtracting the two equations above gives \latex{175=3\times R\times r,} from which
\latex{R=\frac{175}{3\times r}.}
Substitute the result into the formula for the volume:
\latex{\left(\frac{175}{3\times r} \right)^{2}+\frac{175}{3\times r}r+r^{2}=300.}
Rearrange to end up with an equation of degree \latex{ 4 }, which can be reduced to a quadratic equation:
\latex{\frac{30,625}{9\times r^{2} }+\frac{175}{3}+r^{2}=300,}
\latex{9\times r^{2}-2175\times r^{2}+30,625=0.}
\latex{9\times r^{2}-2175\times r^{2}+30,625=0.}
This yields two possible solutions first for \latex{r^{2}} and then for \latex{r:}
\latex{r_{1}=3.87\,cm}, \latex{r^{2}=15.05\,cm}.
Finally, \latex{ R } can be determined using this (the two roots determine the same solid):
\latex{R_{1}=15.05\,cm}, \latex{R_{2}=3,87\,cm}.
\latex{ C }
\latex{ R-r }
\latex{ A }
\latex{ m }
\latex{ B }
\latex{ r }
\latex{ a }
Figure 61
Exercises
{{exercise_number}}. The base edge of a regular quadrilateral pyramidal frustum is \latex{ 10\, cm }, the edge of its top face is \latex{ 5\, cm }, the length of the side edges is \latex{ 20\, cm }.
- What is the altitude of the solid?
- What is its volume and surface area?
- What is the angle between one of its side faces and the base face?
{{exercise_number}}. Determine the volume and surface area of the regular quadrilateral pyramidal frustum if we know that
- its base edge is \latex{ 10\, cm }, its side edge is \latex{ 5\, cm } and its altitude is \latex{ 4\, cm } long;
- its base edge is \latex{ 10\, cm }, its side edge is \latex{ 6\, cm } and the edges of the top face are \latex{ 5\, cm } long.
{{exercise_number}}. Determine the volume and surface area of the conical frustum if we know that
- the radius of the base is \latex{ 22\, cm }, that of the top face is \latex{ 14\, cm }, its altitude is \latex{ 3.4\, cm };
- the radius of the base is \latex{ 113\, cm }, that of the top face is \latex{ 87\, cm }, its generatrix is \latex{ 140\, cm } long;
- the radius of the base is \latex{ 45\, cm }, its altitude is \latex{ 28\, cm }, its generatrix is \latex{ 35\, cm } long.
{{exercise_number}}. An isosceles trapezoid is being rotated around its axis of symmetry. What is the volume and surface area of the resulting solid if the bases of the trapezoid are \latex{ 6\, cm } and \latex{ 4\, cm }, and its legs are \latex{ 5\, cm } long?
{{exercise_number}}. The base edge of a square pyramid is \latex{ 10\, cm } long and its altitude is \latex{ 8\, cm }. The solid is being cut by a plane through the midpoint of the altitude, parallel to the base. Determine the volume and surface area of the two resulting solids.
{{exercise_number}}. A regular quadrilateral pyramid is being cut by a plane through the midpoint of the altitude parallel to the base. The volume of the resulting pyramidal frustum is \latex{336\, cm^{3}}, and its altitude is one third of the base edge of the original pyramid. Determine the surface area of the frustum. What is the angle between the plane of the base and one of the side faces of the frustum?
{{exercise_number}}. The altitude of a conical frustum is \latex{ 10\, cm }. It is also known that a sphere can be inscribed into the frustum and that the radius of the base disc is twice the radius of the top disc. What is the volume and surface area of the solid?
{{exercise_number}}. The area of the lateral surface of a right conical frustum is \latex{2,000\, cm^{2}}, the angle between a generatrix and the base is \latex{ 60º }, the radius of the top face is is one third of the radius of the base face. Determine the radius of the top and base faces. What is the altitude and volume of the solid?