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Inscribed solids (extra-curricular topic)
The parallelepiped and the tetrahedron
The prism with a parallelogram base is called parallelepiped. This solid is centrally symmetric, its parallel edges are equal in length and all of its faces are parallelograms. The cube and the cuboid are in this category.
Take a parallelepiped with edges \latex{ a }, \latex{ b }, \latex{ c } and let the angle between edges \latex{ a } and \latex{ b } be \latex{\alpha} and the angle between edge \latex{ c } and base \latex{ ABCD } be \latex{\beta}. (Figure 65)
The volume of the solid is, according to the formula for the volume of prisms,
Take a parallelepiped with edges \latex{ a }, \latex{ b }, \latex{ c } and let the angle between edges \latex{ a } and \latex{ b } be \latex{\alpha} and the angle between edge \latex{ c } and base \latex{ ABCD } be \latex{\beta}. (Figure 65)
The volume of the solid is, according to the formula for the volume of prisms,
\latex{V=A_{base}\times m.}
The area of the base of the solid is equal to the area of a parallelogram:
\latex{A_{base}=a\times b\times \sin \alpha .}
The altitude can be determined using the following equality from triangle\latex{ AEP } in Figure 65:
\latex{m=c\times \sin\beta.}
Thus the volume of the given parallelepiped is
\latex{V=a\times b\times c\times \sin \alpha \times \sin \beta.}
The surface area of the parallelepiped can be obtained by adding the double of the area of the three different types of parallelograms.
There is a very useful relation between the parallelepiped and the tetrahedron.
Take the endpoints of two non-parallel face diagonals of two parallel faces of the parallelepiped. Let this four point be the vertices of a tetrahedron. It can easily be seen that the edges of this tetrahedron coincide with the diagonals of the parallelepiped. (Figure 66)
We can assign a tetrahedron to every parallelepiped and we can assign such a parallelepiped to every tetrahedron. This solid is called the tetrahedron's enclosing parallelepiped.
There is a very useful relation between the parallelepiped and the tetrahedron.
Take the endpoints of two non-parallel face diagonals of two parallel faces of the parallelepiped. Let this four point be the vertices of a tetrahedron. It can easily be seen that the edges of this tetrahedron coincide with the diagonals of the parallelepiped. (Figure 66)
We can assign a tetrahedron to every parallelepiped and we can assign such a parallelepiped to every tetrahedron. This solid is called the tetrahedron's enclosing parallelepiped.
\latex{ D }
\latex{ P }
\latex{ B }
\latex{ A }
\latex{ b }
\latex{ m }
\latex{ E }
\latex{ a }
\latex{ H }
\latex{ G }
\latex{ C }
\latex{ F }
Figure 65
\latex{ E }
\latex{ m }
\latex{ D }
\latex{ A }
\latex{ B }
\latex{ C }
\latex{ G }
\latex{ H }
\latex{ F }
Figure 66
Example 1
Prove that the volume of any tetrahedron is one third of the volume of its enclosing parallelepiped.
Solution
Let the area of the parallelogram \latex{ ABCD } be denoted by \latex{ T }, and the altitude of the parallelepiped be denoted by \latex{ m }. Thus its volume is
\latex{V_{p}=T\times m.}
The volume of the tetrahedron can be obtained by subtracting the volume of the cut off tetrahedrons from the volume of the parallelepiped. There are four tetrahedrons cut off, and they all have the same volume as the area of their bases (half of the area of the parallelogram \latex{ ABCD }) and their altitudes (the height \latex{ m } of the parallelepiped) are equal.
The total volume of the four cut off tetrahedrons:
\latex{4\times \frac{1}{3}\times \frac{T}{2}\times m=\frac{2\times T\times m}{3}.}
Therefore the volume of the tetrahedron is
\latex{V_{t}=T\times m-\frac{2\times T\times m}{3}=\frac{T\times m}{3}.}
Example 2
The lengths of the skew pairs of edges in a tetrahedron are \latex{5\, cm}, \latex{\sqrt{34}\, cm} and \latex{\sqrt{41}\, cm}. Determine the volume of the solid.
Solution
It is obvious from the data that the faces of the tetrahedron are congruent triangles. Take a look at the enclosing parallelepiped of the solid (Figure 67). Since the skew edges are equal, it follows that the diagonals of any bounding parallelogram of the enclosing parallelepiped are equal. This means that these parallelograms are rectangles, that is, the enclosing parallelepiped is a cuboid. This property holds for any tetrahedron with faces being congruent triangles.
Let \latex{ a }, \latex{ b } and \latex{ c } denote the lengths of the edges of the cuboid. According to the Pythagorean theorem,
Let \latex{ a }, \latex{ b } and \latex{ c } denote the lengths of the edges of the cuboid. According to the Pythagorean theorem,
\latex{a^{2}+b^{2}=25, a^{2}+c_{2}=34, b^{2}+c^{2}=41.}
The sum of the three equation gives
\latex{2\times (a^{2}+b^{2}+c^{2} )=100,}
\latex{a^{2}+b^{2}+c^{2}=50.}
\latex{a^{2}+b^{2}+c^{2}=50.}
Since \latex{b^{2}+c^{2}=41,} it follows that \latex{a^{2}=9,} from which \latex{a=3\, cm.} Substituting into the above equations gives the lengths of the other edges:
\latex{b=4\, cm} and \latex{c= 5\,cm}.
We already know that the volume of the tetrahedron is one third of the volume of the cuboid, therefore
\latex{V_{t}=\frac{1}{3}\times a\times b\times c=20\, cm^{3}.}
\latex{ a }
\latex{ b }
\latex{ c }
Figure 67
The inscribed sphere
There exist solids in which a sphere can be placed which touches every face of the solid. This solid is called the inscribed sphere of the solid.
Note: Of course, not every solid has an inscribed sphere. It is enough to consider a cuboid with two edges of different lengths.
Take a polyhedron which has an inscribed sphere. Connect the centre of the sphere with the vertices of the polyhedron. (Figure 68)
Note: Of course, not every solid has an inscribed sphere. It is enough to consider a cuboid with two edges of different lengths.
Take a polyhedron which has an inscribed sphere. Connect the centre of the sphere with the vertices of the polyhedron. (Figure 68)
This way the polyhedron was divided into pyramids whose bases are the faces of the polyhedron and whose altitudes are all equal to the radius of the sphere.
Therefore the volume of the solid is the sum of the resulting pyramids:
\latex{V=V_{1}+V_{2}+...+V_{n}=}
\latex{=\frac{1}{3}\times A_{1}\times r+\frac{1}{3}\times A_{2}\times r+...+\frac{1}{3}\times (A_{1}+A_{2}+...+A_{n} )\times r.}
Figure 68
volume of the inscribed
circle
volume of the inscribed
circle
Since the sum of the area of the faces is nothing else than the surface area of the solid, the volume of the solid is
\latex{V=\frac{1}{3}\times A\times r.}
It can be shown that the obtained equality is also true for solids whose bounding faces are not polygons but plane figures bounded by curves.
To prove the formula, we have divided the solid and the volume was determined as a sum. Farkas Bolyai asked and studied a lot of problems concerning the decomposition of polyhedra. One of his problems was to decide whether an arbitrary tetrahedron can be cut into pieces and rearranged into a prism or not. The answer found in the first half of the \latex{ 20 }th century proved that usually this is not possible.
To prove the formula, we have divided the solid and the volume was determined as a sum. Farkas Bolyai asked and studied a lot of problems concerning the decomposition of polyhedra. One of his problems was to decide whether an arbitrary tetrahedron can be cut into pieces and rearranged into a prism or not. The answer found in the first half of the \latex{ 20 }th century proved that usually this is not possible.
Example 3
The radius of the inscribed sphere of a tetrahedron is \latex{ r } and the altitudes of the tetrahedron are \latex{m_{1},m_{2},m_{3}} and \latex{m_{4}.} Prove that
\latex{\frac{1}{r}= \frac{1}{m_{1} }+\frac{1}{m_{2} }+\frac{1}{m_{3} }+\frac{1}{m_{4} }.}
Solution
The surface area of the tetrahedron is nothing else than the sum of the area of the faces:
\latex{A=A_{1}+A_{2}+A_{3}+A_{4}.}
Express the volume of the solid using the radius of the inscribed sphere and the altitudes:
\latex{V=\frac{}{3}\times A\times r, V=\frac{1}{3}\times A_{1}\times m_{1}, V=\frac{1}{3}\times A_{2}\times m_{2},}
\latex{V=\frac{1}{3}\times A_{3}\times m_{3}, V=\frac{1}{3}\times A_{4}\times m_{4}.}
Now express the surface area of the solid and the areas of the faces:
\latex{A=\frac{3\times V}{r}, A_{1}=\frac{3\times V}{m_{1} }, A_{2}=\frac{3\times V}{m_{2} },}
\latex{A_{3}=\frac{3\times V}{m_{3} }, A_{4}=\frac{3\times V}{m_{4} }.}
\latex{A_{3}=\frac{3\times V}{m_{3} }, A_{4}=\frac{3\times V}{m_{4} }.}
Substitute these equations into the equation linking the surface area and areas of faces, and simplify by \latex{3\times V} to get
\latex{\frac{3\times V}{r}=\frac{3\times V}{m_{1} }+\frac{3\times V}{m_{2} }+\frac{3\times V}{m_{3} }+\frac{3\times V}{m_{4} },}
\latex{\frac{1}{r}=\frac{1}{m_{1} }+\frac{1}{m_{2} }+\frac{1}{m_{3} }+\frac{1}{m_{4} }.}
The circumscribed sphere
There exist polyhedra, for which it is possible to place a sphere – the circumscribed sphere – such that it is incident to all vertices of the polyhedra. For example, every tetrahedron and every cuboid have circumscribed sphere.
Note: Of course not every solid has a circumscribed sphere. For example usually there is no circumscribed sphere for a parallelepiped.
Note: Of course not every solid has a circumscribed sphere. For example usually there is no circumscribed sphere for a parallelepiped.
Example 4
Two congruent pyramidal frustums, with base edges being \latex{ 10\, cm } and top edges being \latex{ 5\, cm }, are being carved out of a ball of radius \latex{r=15\, cm}, which encloses them. How many percent is the waste in the process?
Solution
It is clear that we can carve out two congruent pyramidal frustums only if none of them contain the centre of the sphere as an interior point. First determine the volume of one frustum.
Take the pyramidal frustum \latex{ ABCDEFGH } and send a line through the centre \latex{ O } of the sphere perpendicular to the planes of the base and top faces. Let \latex{ K } and \latex{ L } be the intersection points. These points coincide with the centres of the squares, thus they are halving the diagonals of the squares. (Figure 69)
Take the pyramidal frustum \latex{ ABCDEFGH } and send a line through the centre \latex{ O } of the sphere perpendicular to the planes of the base and top faces. Let \latex{ K } and \latex{ L } be the intersection points. These points coincide with the centres of the squares, thus they are halving the diagonals of the squares. (Figure 69)
It follows that
\latex{AK=\frac{a\times \sqrt{2} }{2}, EL=\frac{b\times \sqrt{2} }{2}.}
The length of the segment \latex{ KL } determines the altitude of the solid. This can be obtained by determining the lengths of segments \latex{ OK } and \latex{ OL } in the right angled triangles \latex{ AOK } and \latex{ EOL }, respectively, as \latex{ KL = OL – OK }.
\latex{ O }
\latex{ L }
\latex{ F }
\latex{ D }
\latex{ K }
\latex{ B }
Figure 69
By the Pythagorean theorem,
\latex{OK=\sqrt{r^{2}-\left(\frac{a\times \sqrt{2} }{2} \right)^{2} }\approx 13.23\, cm,}
\latex{OL=\sqrt{r^{2}-\left(\frac{b\times \sqrt{2} }{2} \right)^{2} }\approx 14.58\, cm.}
\latex{OL=\sqrt{r^{2}-\left(\frac{b\times \sqrt{2} }{2} \right)^{2} }\approx 14.58\, cm.}
Thus the altitude m of the frustum is \latex{m=14.58\, cm-13.23\, cm=1.35\, cm.}
The volume of the frustum is then
\latex{V=\frac{m}{3}\times (a^{2}+ab+b^{2} )=78.75\, cm^{3}.}
The volume of the sphere is
\latex{V_{sphere}= \frac{4}{3}\times r^{3}\times \Pi \approx 14137.17\, cm^{3}.}
The percentage of the waste is given by the following formula, keeping in mind that two such frustums are being carved:
\latex{\frac{V_{sphere}-2\times V }{V_{sphere} }\approx 0.989=98.9}%
Exercises
{{exercise_number}}. What is the volume of the parallelepiped whose two edges are \latex{ 12\, cm } and \latex{ 24\, cm }, the angle between these edges is \latex{ 30º } and its altitude is \latex{ 10\, cm }?
{{exercise_number}}. The length of the edges of a parallelepiped are \latex{ 3\, cm }, \latex{ 4\, cm } and \latex{ 5\, cm }. The angle between the two shorter edges is \latex{ 60º } and the angle between the longest edge and the plane determined by the other two is \latex{ 45º }. What is the volume of the solid?
{{exercise_number}}. The lengths of the edges of a cuboid are \latex{ 6\, cm }, \latex{ 8\, cm } and \latex{ 10\, cm }.
- Determine the edges of the tetrahedron for which the enclosing parallelepiped is this cuboid.
- What is the volume and surface area of this tetrahedron?
{{exercise_number}}. The surface area of a cuboid is \latex{ 468\, cm^2 }, the ratio of its edges is \latex{2 :3 :4. } Determine the volume of the tetrahedron for which this cuboid is the enclosing parallelepiped
{{exercise_number}}. The faces of a tetrahedron are congruent triangles with edges of lengths \latex{ 3\, cm }, \latex{ 4\, cm } and \latex{ 5\, cm }. What is the volume of the solid? (The result can be surprising!)
{{exercise_number}}. A sphere is inscribed in a regular tetrahedron with edges of length \latex{ 10\, cm }. What percentage of the volume of the tetrahedron is the volume of the sphere?
{{exercise_number}}. The radius of the base disc of a rotational cone is \latex{ r = 5\, cm }, its altitude is \latex{ m = 5\, cm.} What is the radius of its inscribed sphere and what is the surface area of the cone? Does the equality \latex{V=\frac{1}{3}\times A\times r} hold for the volume?
{{exercise_number}}. What is the volume and surface area of the regular quadrilateral pyramid if its base edge is \latex{ a= 32\, cm } and the radius of its inscribed sphere is \latex{r=12\; cm} ?
{{exercise_number}}. Every edge of a regular quadrilateral pyramid are of equal length. How many percent of the volume of the prism is the volume of the solid determined by the touching points of the inscribed sphere?
{{exercise_number}}. Take the inscribed and circumscribed equilateral cones of a sphere with radius \latex{ r }. Determine the ratio of the volumes and the surface areas of the two cones. (A cone is said to be equilateral if the diameter of its base disc is equal to its generatrix: \latex{a=2r} .)
{{exercise_number}}. There is a certain amount of water in a cylinder shaped glass with radius \latex{ R = 5\, cm }. How much will the level of water increase if a ball with radius \latex{r = 4 \;cm} is placed in the water such that it is completely covered?
{{exercise_number}}. At most how high can we fill an equilateral cone shaped bowl, apex pointing downwards, with water such that no water to be spilled even if a ball equal to the cone's inscribed ball is placed in it?