3 20% By the similar cones we have* 3:1:: 203 : 3 = the 203 cube of the height of the top sections ; wherefore ♡ 13.867 the upper part. 2x 203 2 x 203 Also 3 : 2 :: 203 : and y 13.867 = 3 3 3.604 the middle part; wherefore the lower part will be 2.528. 3. Three men bought a tapering piece of timber, which was the frustrum of a square pyramid; one side of the greater end was 3 feet, one side of the less end 1 foot, and the length 18 feet ; what is the length of each man's piece, supposing they paid equally, and are to have equal shares ! * This proportion as well as all others of the kind, may be ex: 20 pressed thus : ¥3: 91::20: =the height of the top section 93 and, in some instances, this is the more convenient method Let ABCDE be a section of the pyramid (when completed) passing through the vertex, and bisecting the oppo site sides of the base, and let IL and MN represent the required sections. Draw EF to the middle of AB, and draw CG parallel to it. Then by similar triangles BG (1 foot) : GC (18) :: BF (1.5) : FE(27) and FE_FH=EH=9, the altitude of the pyramid EDC. Hence Prob. VIII. of solids the solidities of the two pyramids EAB and EDČ will be found 81 and 3 cubic feet respectively, and 81—3=78=the solidity of the frustrum 78 ABCD. Also =26, the solidity of each person's share 3 which added to the solidity of EDC, will give the solidity of EIL=29, and the double of it added to EDC will give the solidity of EMN=55. Now in the similar pyramids, EDC(3): EIL(29) :: EH : EOS, the cube root of which will give EO and EO—EH =HO the length of the part adjacent to the less end= 10.172. Again EIL(29) : EMN(55) :: EOS: EP the cube root of which will give EP and EP-EO=OP the length of the middle part=4.559. Lastly, EF-EP=PF the length of the part adjacent to the larger end=3.269. 4. If a round pillar, 7 inches over, have 4 feet of stone in it; of what diameter is the column, of equal length, that contains 10 times as much? The solidities of cylinders, prisms, parallelopipedons, foc. which have their altitudes equal, are to each other as the squares of their diameters or like sides. The same remark is applicable to frustrums of a cone or pyramid when the altitude is the same, and the ends proportional. Hence, As 4 : 40, or As 1:10 :: 72 : 490=the square the required diameter, and ✓ 490=22.1359 the diameter equired. 5. There is a mill-hopper, in the form of a square pyramid, whose solid content is 134 feet; but one foot is gut. off its perpendicular altitude to make a passage for the grain, from the frustrum or hopper to the mill-stone: the sides of its greater and less end are in proportion of 4 to 1. Required the content in dry or corn measure. Ans. 10.7292 bushels. 6. The ditch of a fortification is 1000 feet long, 9 feet deep, 20 feet broad at bottom, and 22 at top; how much water will fill the ditch, allowing 282 cubic inches to make gallon? Ans. 115812734 gallons. 7. A person having a frustrum of a cone 12 inches in height, and the diameters of the greater and smaller ends 5 and 3 inches respectively, wishes to know the diameter of a frustrum of the same altitude, that would contain 3666 cubic inches, and have its diameters in the same proportion as the smaller one. Ans. The greater diameter 24.4002, and less 14.6401. OF THE REGULAR BODIES. A REGULAR BODY is a solid contained under a certain number of similar and equal plane figures. The whole number of regular bodies which can possibly be formed is five. 1. The Tetraedron, or regular pyramid, which has four triangular faces. 2. The Hexaedron, or cube, which has six square faces. 3. The Octaedron, which has eight triangular faces. 4. The Dodecaedron, which has twelve pentagonal faces. 5 The Icosaedron, which has twenty triangular faces. If the following figures are made of pasteboard, and the lines be cut half through, so that the parts may be turned up and glued together, they will represent the five regular bodies here mentioned. PROBLEM I.. To find the solidity of a tetraedron. RULE.* Multiply t of the cube of the linear side by the square ront of 2, and the product will be the solidity. * Demon. From one angle c of the tetraedron ABCN, let fall the perpendicular ce, upon the opposite side, and draw Ae. Then ad-ae=ce*; and since the point e is equally distant from the three angles A, B, and n, $ Ace= ({ AB) Aeas is shown in the demonstration of the rule for regu. lar polygons, page 61. Consequently Ac — Ac = {AC ce, or ce = AC ✓ . But the area of the triangle ANB=\ABV3=$ AC 3; and therefore } AC V $ (t ce) x fac 3(AANB)= AC/2. Q. E. D. If be put = length of the linear edge, then will l'13 =whole surface of the tetraedron. The rule for the hexaedron, or cube, has been given before. |