PROBLEM IV. To find the solidity of an icosaedron. RULE.* To 3 times the square root of 5 add 7, and divide the sum by 2; then the square root of this quotient being multiplied by of the cube of the linear side will give the solidity required. * Demon. Let A be a solid angle of the icosaedron, formed by 5 faces, or triangles, whose bases make the pentagon BCDEm. Then, if a perpendicular be demitted from A upon the pentagonal plane BCDEm, it will fall into the centre n, and Bn, by the demonstration of the last problem, will be = AB 5+√5 and the radius of the circle 10 circumscribing one of the faces ABC=}AB√/3. B But the radius of the circumscribing sphere is R-BA BA2 2√AB2-BN2 R= E 2An 2=AB√ And, since R is the hypothenuse of a right angled triangle, one of whose legs is AB √3, the radius of the circle circumscribing the face ABC, and the other r, the radius of the inscribed sphere, we shall But the solid is composed of 20 equal triangular pyramids, each of AB2 whose bases is= √3 by Problem VIII.; therefore 4 20AB3 EXAMPLES. 1. The linear side of the icosaedron ABCDEF is 3; what is the solidity? D solidity of the icosaedron. Q. E. D. If L be put for the linear side, then will 512√3=surface of the icosaedron. 1 Note.-The superficies and solidity of any of the five regular bodies may be found as follows. RULE 1. Multiply the tabular area by the square of the linear edge, and the product will be the superficies. 2. Multiply the tabular solidity by the cube of the linear edge, and the product will be the solidity. Surfaces and Solidities of the Regular Bodies. 22.5=2.61803× 22.5=58.9056=solidity required. 2. Required the solidity of an icosaedron, whose linear side is 1? Ans. 2.1817. P2 OF CYLINDRIC RINGS. PROBLEM I. To find the convex superficies of a cylindric ring. RULE.* To the thickness of the ring add the inner diameter, and this sum being multiplied by the thickness, and the product again by 9.8696, will give the superficies required. EXAMPLES. 1. The thickness of Ac of a cylindric ring is 3 inches, and the inner diameter cd 12 inches; what is the convex superficies? * A solid ring of this kind is only a bent cylinder, and therefore the rules for obtaining its superficies, or solidity, are the same as those already given. For let Ac be any section of the solid perpendicular to its axis on, and then Ac × 3.14159, &c. circumference of that section, and Ac+cd (on) × 3.14159, &c.=length of the axis on. 12+3x3x 9.8696 = 15 × 3 × 9.8696 = 45 × 9.8696 = 444.132 superficies required. 2. The thickness of a cylindric ring is 4 inches, and the inner diameter 18; what is the convex superficies? Ans. 868.52 square inches. 3. The thickness of a cylindric ring is 2 inches, and the inner diameter 18; what is the convex superficies? Ans. 394.784 square inches. PROBLEM II. To find the sclidity of a cylindric ring. RULE.* To the thickness of the ring add the inner diameter, and this sum being multiplied by the square of half the thickness, and the product again by 9.8696, will give the solidity. Whence ac × 3.14159, &c. × Ac + cd × 3.14159, &c. = Ac+cd× Ac× 3.14159, &c. |2=ac+cd× ac× 9.8696, &c. superficies required. * Demon. Ac2× .78539, &c.=AC2> 3.14159 &c. = Ac2 3.14159, &c. area of the section ac; and Ac+cd (on) × 3.14159, &c.=length of the axis on. Therefore Ac+cd × ac2 3.14159, &c. |2 = ac+cd× 4. AC2x 9.8696. Q. E. D. This figure being only a cylinder bent round into a ring, its surface and solidity may also be found as in the cylinder, namely, by multiplying the axis or length of the cylinder by the circumference of the ring, or section, for the surface, and by the area of a section for the solidity. Thus, if c-circumference of the ring, or section, a=area of that section, and t-length of the axis; then will cl=surface of the ring, and alto its solidity. Which rules are the same as for the cylinder, and may be easily converted into those given in the text. These rules are indeed so obvious, as to render any demonstration of them altogether unnecessary. |