PROBLEM XIX. To find the fuperficies and folidity of the five regular or Platonic bodies. RULE. Multiply the square of the given side into the corresponding tabular area for the superficies. And Multiply the cube of the given fide by the proper tabular folidity, for the folidity of the given body. This table exhibits the area and folidity of any of the above bodies, the fide being unity. The areas of the above figures are so related to those of regular polygons, and their solidities to problems already treated of, that we shall leave the construction of the table for the exercife of the learner. २०० EXAMPLE I. Fig. 97. Required the area and folidity of a tetraedron, whose fide is Ex. 2. Required the superficial and folid content of a hexae dron, whose fide is 6. Fig. 98. Anf. Superficies 216 Solidity 216 Ex. 3. Required the area and folidity of an octraedron, whose fide is 3. Fig. 99. Superficies 31.176918 Ans. Solidity 12.7279215 Ex. 4. Required the superficies and folidity of the icosaedron, whose side is 2. Fig. 100. Ans. Superficies 34.641 Solidity 17-4535 Ex. 5. Required the superficies and folidity of a dodecaedron, the side being 4. Fig. 101. Surface Ans. Sortie 133.62848 PROBLEM XX. Fig. 102. To find the furface and folidity of a cylindric ring. RULE Multiply the circumference of the ring by its length for the fuperficies. Multiply the area of a section of the ring by the curve, for the solidity. EXAMPLE EXAMPLE I. Required the furface and folidity of a cylindric ring, whose curve is 12, and the diameter of the ring 3 inches. A CONE may be cut various ways; and, according to the different pofitions of the cutting plane, the five plane figures following will arifc, viz. the circle, the ellipse, the parabola, the kyperbola, and the triangle. DEFINITIONS. 1. The fection is a circle, when the cone is cut parallel to the bafe. 2. If the section is obliquely to the base, it will form an ellipfe. Fig. 102. 3. If the plane cut parallel to one of the fides, the section will be a parabola. Fig. 103. 4. The sectioni s an hyperbola, when the cutting plane meets the oppofite cone, and makes another section fimilar to the for mer. 5. The section forms a triangle, when the plane passes through the vertex and meets the bafe. 6. The vertex of any section is the point in which the plane meets the opposite side of the cone. 7. The transverse axis is a line drawn between two vertices. 8. The centre of an ellipfe is the middle point of the tranf verse. 9. The conjugate axis is drawn through the centre perpendicular to the transverse. 10. The ordinate is a line perpendicular to the axis. 11. The abscissa is that part of the axis intercepted between. the ordinate and the vertex. 12. The axis of a parabola is a right line drawn from the vertex, so as to divide the figure into two equal parts. 13. The transverse diameter of an hyperbola is that part of the axis, intercepted between the vertices of the opposite sections. : PROBLEM I. To defcribe an ellipfe. It is a known property of the ellipfe, that any two lines. drawn from the foci, meeting in any point of the curve, are together equal to the transverse diameter. Hence the following method of defcribing an ellipse. Find the points x y in the transverse, which you are to consider as your foci; there fix two pins, and take a string equal to the transverse, and fasten its ends each to a pin, then stretchthe string with a pencil, and move it round within the thread, so shall its path defcribe an ellipfe. Ee When When the transverse and conjugate diameters are given, the foci may be found thus. Draw the transverse AB, and conjugate CD so as they may bisect each other at right angles in the point E, and with the distance AE or EB, and centre C or D, describe arches, cutting the transverse in the points x y, so shall x and y be the foci. PROBLEM II. To find the length of the elliptic curve. RULE. Multiply the sum of the transverse and conjugate diameters by 3.1416, and half the product will be the circumference nearly. EXAMPLE I. Required the length of an elliptic curve, whose conjugate is 40 and transverse 60 feet. Ans. 109.956 feet. Ex. 2. What is the length of the circumference, when the diameters are 30, 40 feet? Ex. 3. Required the circumference of an ellipfe, whose tranfverse diameter is 20, and conjugate 10 yards. Ans. 282.744 feet. Ex. 4. What is the periphery of an ellipfe, whose axis are 36 feet and 24 feet? Anf. 94.248. |