pyramids coincide, the pyramids themselves will evidently coincide, and likewise the solid angles at their vertices. From this, some consequences are deduced. First. Two triangular spherical pyramids are to each other as their bases : and since a polygonal pyramid may always be divided into a certain number of triangular ones, it follows that any two spherical pyramids are to each other, as the polygons which form their bases. Second. The solid angles at the vertices of these pyranids, are also as their basés ; hence, for comparing any two sclid angles, we have merely to place their vertices at the centres of two equal spheres, and the solid angles will be to each other as the spherical polygons intercepted between their planes or faces. The vertical angle of the tri-rectangular pyramid is formed by three planes at right angles to each other : this angle, which may be called a right solid angle, will serve as a very natural unit of measure for all other solid angles. If, for example, the the area of the triangle is of the tri-rectangular triangle, then the corresponding solid angle will also be of the right solid angle. PROPOSITION XXI. THEOREM The surface of a spherical polygon is measured by the sum of all its angles, minus two right angles multiplied by the number of sides in the polygon less two, into the tri-rectangular triangle. А From one of the vertices A, let diago D nals AC, AD be drawn to all the other vertices; the polygon ABCDE will be divided into as many triangles minus two as E it has sides. But the surface of each triangle is measured by the sum of all its an 'B gles minus two right angles, into the trirectangular triangle; and the sum of the angles in all the triangles is evidently the same as that of all the angles of the polygon; hence, the surface of the polygon is equal to the sum of all its angles, diminished by twice as many right angles as it has sides less two, into the tri-rectangular triangle. Scholium. Let s be the sum of all the angles in a spherical polygon, n the number of its sides, and T the tri-rectangular triangle; the right angle being taken for unity, the surface of the polygon will be measured by -2 (n—2,)) T, or (8—2 n+4) T APPENDIX. THE REGULAR POLYEDRONS. A regular polyedron is one whose faces are all equal regular polygons, and whose solid angles are all equal to each other. There are five such polyedrons. First. If the faces are equilateral triangles, polyedrons may be formed of them, having solid angles contained by three of those triangles, by four, or by five: hence arise three regular bodies, the tetraedron, the octaedron, the icosaedron. No other can be formed with equilateral triangles ; for six angles of such a triangle are equal to four right angles, and cannot form a solid angle (Book VI. Prop. XX.). Secondly. If the faces are squares, their angles may be arranged by threes: hence results the hexaedron or cube. Four angles of a square are equal to four right angles, and cannot form a solid angle. Thirdly. In fine, if the faces are regular pentagons, their angles likewise may be arranged by threes: the regular dodecaedron will result. We can proceed no farther : three angles of a regular hexagon are equal to four right angles; three of a heptagon are: greater. Hence there can only be five regular polyedrons; three formed with equilateral triangles, one with squares, and one with pentagons. 2 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, erectos perpendicular to the A plane ABC ; terminate this perpendicular in S, so that AS=AB; draw SB, SC: the pyramid S-ABC will be the tetraedron required. B For, by reason of the equal distances OA, OB, OC, the oblique lines SA, SB, SC, are equally re moved from the perpendicular So, and consequently equal (Book VI. Prop. V.). One of them SA=AB; hence the four faces of the pyramid S-ABC, are trian- A gles, equal to the given triangle ABC. And the solid 'angles of this pyramid are all equal, because each of them is formed by three equal plane angles : hence this pyramid is a regular tetraedron. Construction of the Hexaedron. Let ABCD be a given square. On the I H G base ABCD, construct a right prism whose altitude AE shall be equal to the side AB. The faces of this prism will evidently be E E equal squares; and its solid angles all equal, each being formed with three right angles : hence this prism is a regular hexaedron or cube. B The following propositions can be easily proved. 1. Any regular polyedron may be divided into as many regular 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 the inscribed and of the 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 sides of the polyedrons. 4. If a regular polyedron is 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. a a APPLICATION OF ALGEBRA. TO THE SOLUTION OF GEOMETRICAL PROBLEMS. A problem is a question which requires a solution. A geo metrical 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, and the relations which subsist between the known and unknown 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 particular directions. 1st. Draw a figure which shall represent all the given parts, and all the required parts. Then draw such other lines as will establish the most simple 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 the 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 following problems.* * The following problems are selected from Hutton's Application of Algebra o Geometry, and the examples in Mensuration from his treatise on that subject. PROBLEM I. 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, x+y=s=9. с and xo=y2 +c? (Bk. IV. Prop. XI.) From 1st equ: x=S-Y and x'=s?_2sy+ya B 'A By subtracting, 0=s_2sy-ca 2sy=s SP_C hence, y= =4=AC 2s Therefore x+4=9 or x=5=BC. or PROBLEM II. In a right angled triangle, having given the hypothenuse, and the sum of the base and perpendicular, to find these two sides. or Put BC=a=5, BA=, AC=y and the sum of the base and perpendicular=s=7 Then x+y=s=7 and æ?+ya=a? From first equation X=S-Y æ’=s?—2sy + ya Hence, y?=a?—52 +2sy—y? 2y2—2sy=a?—52 a2_52 ye_sy= 2 By completing the square yé-sy+is=ja-132 or y=is+Va?—so=4 or 3 Hence x= SF Va+s?=3 or 4 or or |