A kosarad üres
Volume of the pyramid and the cone
It is worth starting the inspection of the volume of pyramidal solids with the simplest of those, which is the volume of the triangular pyramid. In the following, this solid will be called tetrahedron.
The volume of the tetrahedron
It is easy to see that every tetrahedron can be transformed into a prism with the help of appropriately chosen vertices and planes. Such a possible extension is shown in Figure 49.
The volume of the resulting solid can be determined since it is a prism whose base is the triangle \latex{ ABC }. Thus its volume equals to the product of the area of the base and the altitude.
The original tetrahedron is part of this solid. The extra solids created during the extension are also tetrahedrons. These are determined by points \latex{ DEFC } and \latex{ BCDE }.
While we won't take a closer look at the details of it, these tetrahedrons satisfy the conditions of Cavalieri's principle. This means that these three tetrahedrons have the same volume, and since the sum of these quantities equals to the volume of the prism, we can conclude that the volume of the tetrahedron \latex{ ABCD } can be determined using the following formula:
\latex{ E }
\latex{ D }
\latex{ A }
\latex{ C }
\latex{ B }
\latex{ F }
Figure 49
\latex{V_{\text{tetrahedron}}=\frac{A\times m}{3}}.
Here \latex{ A } denotes the area of the base of the tetrahedron and m denotes its altitude.
The volume of the pyramid
If the base is an arbitrary polygon, then the solid is called a pyramid. Since every polygon can be divided into triangles, pyramids can be divided into pyramids with triangles as their bases, that is, into tetrahedrons.
Let us denote the area of the resulting triangles by \latex{t_1, t_2, …, t_n}, then the sum of these equals the area \latex{ A } of the base (Figure 50). Therefore the volume of an arbitrary pyramid is
\latex{V=\frac{t_1\times m}{3}+\frac{t_2\times m}{3}+\dots+\frac{t_n\times m}{3}=(t_1+t_2+\dots+t_n)\times\frac{m}{3}}, from which
\latex{ t_2 }
\latex{ t_1 }
\latex{ t_3 }
Figure 50
\latex{V=\frac{A\times m}{3}}.
Both for tetrahedrons and other pyramids determining the surface means only the problem of computing and adding the area of triangles and a polygon. The triangles acting as sides of the pyramid together determine the lateral surface of the pyramid.
Example 1
Determine the volume and surface of a regular tetrahedron.
Solution
Let a denote the length of the edges. Since every face of the regular tetrahedron is an equilateral triangle, and the surface consists of four of them, the size of the whole surface is
\latex{A=4\times T=4\times\frac{a\times\frac{a\times\sqrt3}{2}}{2}=4\times\frac{a^2\times\sqrt3}{4}=a^2\times\sqrt3}.
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{ m }
\latex{ a }
\latex{ a }
\latex{ a }
\latex{ a }
\latex{ C }
\latex{ B }
\latex{ A }
\latex{ D }
Figure 51
The foot of the solid's altitude coincides with the centroid \latex{ E } of triangle \latex{ ABC }, thus
\latex{AE=\frac 2 3\times\frac{\sqrt3\times a}{2}=\frac{\sqrt3\times a}{3}}.
The altitude can be determined the following way:
\latex{m^2=a^2-\left(\frac{\sqrt3\times a}{3}\right)^2=\frac{2}{3}\times a^2}, from which \latex{m=\sqrt{\frac{2}{3}}\times a=\frac{\sqrt6}{3}\times a}.
Therefore the volume of the regular tetrahedron is
\latex{V=\frac{T\times m}{3}=\frac{\frac{a^2\times\sqrt3}{4}\times\frac{\sqrt6\times a}{3}}{3}=\frac{\sqrt{18}\times a^3}{36}=\frac{\sqrt2\times a^3}{12}}.
The volume and the surface of the cone
Just as the volume of the pyramid can be used to determine the volume of the cylinder, the volume of the cone can be expressed with the help of the volume of the pyramids.
It is trivial that as we increase the number of sides of the polygon base of a pyramid, it will resemble more and more to a cone (Figure 52). It can be conjectured that the formula for the volume of pyramids will be valid for cones as well.
Take a cone with area of base being \latex{ T } and altitude being \latex{ m }, then it can be shown that its volume can be expressed as follows:
\latex{V=\frac{T\times m}{3}}, from which
\latex{ r }
\latex{ m }
Figure 52
\latex{V=\frac{r^2\times\pi\times m}{3}}.
This equality can be proven for arbitrary conic solid as well, but it requires the use of advanced mathematical tools.
Let us note here that the previous formulas for the volume of the prism and the cone can also be used if we are to determine the volume of an oblique prism or oblique cone.
There are more complications when we are to determine the surface. The most important thing that we have to accept for now, is that the lateral surface of a cone can laid in the plane, just as the base disc, and thus it determines a sector. This is due to the fact that every generatrix of the rotational cone are of equal length. (Figure 53)
This generatrix determines the radius of the sector, and its arc is equal to the base disc's circumference \latex{2 \times r\times \pi}. Therefore its area is
Ageneratrix\latex{=\frac{2\times r\times\pi\times a}{2}=r\times \pi\times a}.
Thus the surface of a rotational cone is
\latex{A=r^2\times\pi+r\times\pi\times a}, from which we obtain
\latex{2\times r\times\pi}
\latex{ r }
\latex{ a }
\latex{ a }
Figure 53
\latex{A=r\times\pi\times(r+a)}.
Example 2
What is the surface and volume of the straight circular cone whose laid lateral surface is a third of a full disc with radius of \latex{ 9\, cm }?
Solution
Let the generatrix of the cone be denoted by \latex{ a }, the radius of its base by \latex{ r } and its altitude by \latex{ m }. (Figure 54)
The area of the lateral surface can be determined using the formula for the area of a sector:
Alateral surface\latex{=\frac{a^2\times\pi}{3}\approx84.82 \text{ cm}^2}.
Since the circumference of the base disc is equal to the arc of the sector determined by the lateral surface, the radius of the cone can be expressed:
\latex{2\times r\times\pi=\frac{2\times a\times\pi}{3}}, from which \latex{r=\frac a 3=3 \text{ cm}}.
The surface of the cone is, therefore,
\latex{A=r^2\times\pi+}Alateral surface\latex{\approx}
\latex{\approx28.27 \text{ cm}^2+84.82\text{ cm}^2=113.09 \text{cm}^2}.
\latex{\approx28.27 \text{ cm}^2+84.82\text{ cm}^2=113.09 \text{cm}^2}.
To find the volume, first we have to determine the altitude. It can be done using the Pythagorean theorem:
\latex{m=\sqrt{a^2-r^2}\approx8.49\text{cm}.}
Thus the volume of the solid is
\latex{V=\frac{r^2\times \pi\times m}{3}\approx80 \text{ cm}^3}.
\latex{120^{\circ}}
\latex{ m }
\latex{ a }
\latex{ r }
\latex{ r }
\latex{ a }
\latex{ a }
Figure 54
Example 3
Which one of the straight circular cones with a given surface area has the largest volume?
Solution
Let us denote the length of the generatrix of the cone by \latex{ a }, the radius of its base disc by \latex{ r } and the altitude by \latex{ m }. The surface area of the cone is \latex{A = r\times \pi\times(r + a)}. The length of the generatrix can be expressed from this:
\latex{a=\frac{A}{r\times \pi}-r}.
The altitude satisfies the Pythagorean theorem:
\latex{m^2=a^2-r^2}.
Now take a look at the volume of the solid:
\latex{V=\frac{r^2\times \pi\times m}{3}}.
The maximum value of this expression depends on the maximal value of the product \latex{r^2 \times m}. Because of the identities we have, it is better to study the square of this product instead. After the proper substitutions,
\latex{f(r)=r^4\times m^2=r^4\times(a^2-r^2)=r^4\times\left[\left(\frac{A}{r\times \pi}-r\right)^2-r^2\right]=\\=r^4\left(\frac{A^2}{r^2\times \pi^2}-\frac{2\times A}{\pi}\right)=-\frac{2\times A}{\pi}\times r^4+\frac{A^2}{\pi^2}\times r^2}.
This expression depends only on the value of \latex{ r }. It can be expressed as a complete square:
\latex{f(r)=-\frac{2\times A}{\pi}\times\left(r^4-\frac{A}{2\times \pi}\times r^2\right)=-\frac{2\times A}{\pi}\times\left[\left(r^2-\frac{A}{4\times\pi}\right)^2-\frac{A^2}{16\times\pi^2}\right]=\\=-\frac{2\times A}{\pi}\times\left(r^2-\frac{A}{4\times\pi}\right)^2+\frac{A^3}{8\times\pi^3}.}
This expression is maximal if the quadratic term is zero, that is,
\latex{r^2=\frac{A}{4\times\pi}}.
Now use the formula for the volume of the cone. The following connection can be established between the radius of the base disc and the length of the generatrix:
\latex{\begin{array}{lcl}r^2=\frac{r\times\pi\times(r+a)}{4\times\pi},\\4r=r+a,\\3r=a.\end{array}}
This means that from the set of rotational cones with a given surface area the volume is the largest for the one whose generatrix is three times the radius of the base disc. The volume of this cone, in terms of the surface area, is
\latex{V=\sqrt{\frac{A^3}{72\times\pi}}.}
DEMOCRITUS (approx. \latex{ 460–370 } BC) was a Greek mathematician and philosopher, whose name is best known for his atomic theory of the universe. His theory says that any material can be decomposed into smaller parts until we reach the level of atoms, which are indivisible.
Exercises
{{exercise_number}}. Determine the volume and surface area of the regular pyramid whose edge of base is \latex{ 10\, cm } long and edge of side is \latex{ 20\, cm } long, and its base is
- a triangle;
- a square;
- a pentagon;
- a hexagon.
{{exercise_number}}. The following are the properties of a right circular cone:
- \latex{r = 5 \text{ cm}, m = 6 \text{ cm}};
- \latex{r = 6\text{ cm}, a= 10\text{ cm}};
- \latex{a= 10\text{ cm}, m = 8\text{ cm}}.
Determine the volume and surface area of the cone.
{{exercise_number}}. The surface area of a regular pyramid is \latex{ 100\, cm^2 }, its base is a square with sides of length \latex{ 5\, cm }. What is its volume?
{{exercise_number}}. The volume of a regular pyramid is \latex{ 1,000\, cm^3 }, its base is a regular hexagon with sides of length \latex{ 6\, cm }. What is the surface area of the pyramid?
{{exercise_number}}. The volume of a rotational cone is \latex{ 1,000\, cm^3 }, the radius of its base is \latex{ 10\, cm }. What is its surface area?
{{exercise_number}}. The surface area of a rotational cone is \latex{ 100\, cm^2 }, its generatrix is \latex{ 10\, cm } long. Determine its volume.
{{exercise_number}}. A pyramid is inscribed into a cube with edges of length \latex{ 10\, cm } such that the centre of one face is connected to the vertices from the opposite face. Find the resulting pyramid's volume and surface area.
{{exercise_number}}. Determine the volume and surface area of the solid determined by the centres of the faces of a cube with edges of length \latex{ 10\, cm }.
{{exercise_number}}. The sides of a triangle are of length \latex{ 3\, cm }, \latex{ 4\, cm } and \latex{ 5\, cm }. What is the surface area of the solid received by rotating the triangle around its longest side?
{{exercise_number}}. The bases of an isosceles trapezoid are \latex{ 14\, cm } and \latex{ 8\, cm } long, its legs are \latex{ 5\, cm } long. Rotate the trapezoid around its
- longer;
- shorter base.
Determine the volume and surface area of the resulting solid.
{{exercise_number}}. Rotate a regular tetrahedron with edges of length a around one of its altitudes by \latex{ 60º }. Determine the volume and surface area of the intersection of the original and the rotated tetrahedrons.
{{exercise_number}}. Which right circular cone has the smallest surface area among those with a fixed volume?
Puzzle
Approximate the weight of stone in tons used in the building of the Great Pyramid of Giza. (The density of limestone is \latex{ 2.7\, tons } per \latex{ m^3 }.) It is believed that its builder, Pharaoh Khufu ruled for \latex{ 23 } years. How many \latex{ m^3 } of stone was moved each day if we suppose that the pyramid was built for \latex{ 23 } years?
