PROBLEM XIX. To find the superficies and folidity of the five regular or Platonic bodies. RULE. Multiply the square of the given fide into the corresponding tabular area for the superficies. And Multiply the cube of the given side by the proper tabular solidity, for the solidity of the given body. Names. Containing fides. Area. Solidity. 1. Tetraedron 4 equilateral 1.732051 0.117851 3.464102 | 0.471405 1 2 equal pentagons 20.6457297.663119 20 equal equilat. tria. | 8.660254 | 2.181695 This table exhibits the area and folidity of any of the above bodies, the side being unity. The areas of the above figures are so related to those of regular polygons, and their folidities to problems already treated of, that we shall leave the construction of the table for the exercise of the learner. EXAMPLE I. Fig. 97. Required the area and solidity of a tetraedron, whose fide is 30. Anf. Solidity Ex. 2. Required the superficial and solid content of a hexacdron, whose fide is 6. Fig. 98. S Superficies 216 216 Ex. 3. Required the area and solidity of an octraedron, whose fide is 3. Fig. 99. Anf. (Superficies 31.176918 Solidity 12.7279215 Ex. 4. Required the superficies and solidity of the icofaedron, whose fide is 2. Fig. 100. . Ans. S Superficies 34.641 Solidity 17.4535 Ex. 5. Required the superficies and folidity of a dodecaedron, the fide being 4. Fig. 101. Surface 33.03312 Anf Solidity 139.62848 PROBLEM XX. Fig. 102. To find the surface 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 surface and folidity of a cylindric ring, whose curve is 12, and the diameter of the ring 3 inches. A Coxe may be cut various ways; and, according to the different pofitions of the cutting plane, the five plane figures following will arisc, 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 ek lipse. 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 opposite cane, 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 base. 6. The vertex of any fection 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 ellipse is the middle point of the tranfverse. 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 abfciffa 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 fections, PROBLEM I. To describe an ellipfis It is a known property of the ellipse, 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 describing 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 streich the string with a pencil, and move it round within the thread, so thall its path describe an ellipse. Еe 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 fhall x and y be the foci. 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 curyć, whose conjugate is 40 and transverse 60 feet. Ex. 2. What is the length of the circumference, when the diameters are 30, 40 feet? Anf. 109.956 feet. Ex. 3. Requited the circumference of an ellipse, whose transverse diameter is 20, and conjugate 10 yards. Anf. 282.744 feet. Ex. 4. What is the periphery of an ellipse, whose axis are 36 feet and 24 feet? Anf. 94.248. PROBLEM |