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Hence, there can only be five regular polyedrons; three formed with equilateral triangles, one with squares, and one with pentagons.
CONSTRUCTION OF THE TETRAEDRON.
Let ABC be the equilateral triangle which is to form one face of the tetraedron. At the point 0, the centre of this triangle, erect OS perpendicular to the plane ABC; terminate this perpendicular in S, so that AS=AB; draw SB, SC; the pyramid S-ABC is the tetraedron required.
For, by reason of the equal distances OA, OB, OC, the oblique lines SA, SB, SC, cut off equal distances estimated from the foot of the perpendic
C ular So, and consequently are equal (B. VI., P. 5). One of them SA=AB; hence, the four faces of the pyramid B S-ABC, are triangles, equal to the given triangle ABC. The triedral angles of this pyramid are all equal, because each of them is bounded by three equal plane angles (B. VI., P. 21, s. 2); hence, this pyramid is a regular tetraedron.
CONSTRUCTION OF THE HEXAEDRON.
Let ABCD be a given square. On the base ABCD, construct a right prism whose altitude AE shall be equal to the side AB. The faces of this prism will evident
D! ly be equal squares; and its triedral angles all equal, each being formed with three equal faces : hence, this prism is a regular hexaedron or cube.
The following propositions can be easily proved.
1. Any regular polyedron may be divided into as many right pyramids as the polyedron has faces; the common vertex of these pyramids will be the centre of the polyedron; and at the same time, that of an inscribed and of a circumscribed sphere.
2. The solidity of a regular polyedron is equal to its surface multiplied by a third part of the radius of the inscribed sphere.
3. Two regular polyedrons of the same name, are two similar solids, and their homologous dimensions are proportional; hence, the radii of the inscribed or the circumscribed spheres are to each other as the edges of the polyedrons.
4. If a regular polyedron be inscribed in a sphere, the planes drawn from the centre, through the different edges, will divide the surface of the sphere into as many spherical polygons, all equal and similar, as the polyedron has faces.
APPLICATION OF ALGEBRA
SOLUTION OF GEOMETRICAL
A PROBLEM is a question which requires a solution. A geometrical problem is one, in which certain parts of a geometrical figure are given or known, from which it is required to determine certain other parts.
When it is proposed to solve a geometrical problem by means of Algebra, the given parts are represented by the first letters of the alphabet, and the required parts by the final letters. The geometrical relations which subsist between the known and required parts furnish the equations of the problem. The solution of these equations, when so formed, gives the solution of the problem.
No general rule can be given for forming the equations. The equations must be independent of each other, and their number equal to that of the unknown quantities introduced (Alg., Art. 103). Experience, and a careful examination of all the conditions, whether explicit or implicit (Alg., Art. 94), will serve as guides in stating the questions; to which may be added the following general directions.
1st. Draw a figure which shall represent all the given parts, and all the required parts. Then draw such other lines as will enable us to establish the necessary relations between them. If an angle is given, it is generally best to let fall a perpendicular that shall lie opposite to it; and this perpendicular, if possible, should be drawn from the extremity of a given side.
2d. When two lines or quantities are connected in the same way with other parts of the figure or problem, it is in general, not best to use either of them separately ; but to use their sum, their difference, their product, their quotient, or perhaps another line of the figure with which they are alike connected.
3d. When the area, or perimeter of a figure, is given, it is sometimes best to assume another figure similar to that proposed, having one of its sides equal to unity, or some other known quantity. A comparison of the two figures will often give a required part. We will add the follow. ing problems.*
In a right-angled triangle BAC, having given the base BA,
and the sum of the hypothenuse and perpendicular, it is required to find the hypothenuse and perpendicular.
Put BA 3, BC= x, AC y, and the sum of the hypothenuse and perpendicular equal to s = 9. Then,
2 + y =s= 9, and (B. IV., P. 11), z? = y + c . From 1st equ:
Y, and % = – 2sy + ya. B
A By subtracting 0 = $. – 2sy -c,
2sy = $c;
4 = AC.
x + 4 = 9, or x = 5 = BC.
* The following problems are selected from Hutton's Application of Algebra to Geometry; and the examples in Mensuration, from his treatise on that subject.
In a right-angled triangle, having given the hypothenuse, and
the sum of the base and perpendicular, to find these two sides.
X = S
Put BC= a = 5, BA = x, AC = y, and the sum of the base and perpendicular = s = 7. Then,
x + y = s = 7, and
2 + y = a. From first equation,
Y, . or,
ac? = ? – 2sy + y; Hence,
ya = a’ – 5% + 2sy -- y,
2 By completing the square yo – sy + s = ja? -45, or,
y = ? ?
js = V1 – 15= 4 or 3. Hence, x = js F VjQ? – 152
= 3 or 4.
In a rectangle, having given the diagonal and perimeter, to find
Let ABCD be the proposed rectangle. Put AC=d=10, the perimeter =2a = 28, or AB +BC= a 14: also put AB = x, and BC= y.
D Then, 2? + y2 = d?, and
2 + y = a. From which equations we obtain,
Α. y= ja + V]d? – ja? = 8 or 6, and
x = jaF VI – ja = 6 or 8.
Having given the base and perpendicular of a triangle, to find the side of an inscribed square.
G AB = b, CD = a, and HE or GH := x: then CI
A HDЕ В We have by similar triangles AB CD
the side of the inscribed square; which, therefore, depends only on the base and altitude of the triangle.
bx = AX,
a + b
In an equilateral triangle, having given the lengths of the three
perpendiculars drawn from a point within, on the three sides : to determine the sides of the triangle.
Let ABC be an equilateral triangle: DG, DE and DF the given per pendiculars let fall from D on the
F sides. Draw DA, DB, DC, to the vertices of the angles, and let fall the perpendicular CH on the base.
H G B DG = a, DE = b, and DF=c: put one of the equal sides AB = 2x; hence, AH = x, and CH VAC AH2 = V 4x? - x? = V3x2 = x 73.
Now, since the area of a triangle is equal to half its base into the altitude, (B. IV., P. 6),
LAB X CH = x x x V3 = x2 V3 = triangle ACB,
triangle BCD, AC X DF = 2 X C