Again,

52.36:11.358::√562-√562:1.0855=the length of the

slant height of the lower part.

68. An elliptic inclosure has diameters of 840 and 612 links within its wall, which is 14 inches thick; required the area it incloses and covers.

69. A block of marble, in the form of a square pyramid, weighs 18 tons, the perpendicular height being twice the diagonal of the base; required its dimensions and volume in cubic feet.

A cubic inch of the marble is assumed to weigh 1.6 ounces avoirdupois.

Ans. {Side,

(Side, 4.35325 feet. 233.334 cubic feet.

70. A pail containing 2.1215 cubic feet is 12 inches in depth; what are its top and bottom diameters, they being in the proportion of 5 to 3?

Ans. 14.64 and 24.4 inches.

71. If a sphere, the diameter of which is 4 inches, is depressed in a conical glass full of water, the diameter of which is 5 inches and altitude 6, it is required to know the quantity of water which will run over? Ans. 26.272 inches.

By Construction, Fig. 132.

Draw a section of the glass, as A B C, and of the sphere, def.

Then will A B= 5; Ci=6; Bi

=

5÷22.5; A C and B C = √ (Ci2+B i2)=6.5; and oƒ=4÷2 (half diameter of sphere)=2.

The triangles B i C and of C are similar, for they have the common angle C, and the right angles i and ƒ; hence, their remaining angles are equal.

Therefore, Bi: B C::of:o C; that is, as 2.5: 6.5::2:5.2=the depth of the glass from where the centre of the sphere rests.

Consequently, 5.2-6.8, and 2+.8=2.8 inches of the sphere immersed, and 4-2.8 1.2 inches of it above the glass.

Hence, the volume of the segment immersed (by rule 2, p. 177) deducted from that of the sphere=26.2722 cubic inches the result required.

72. If a sphere is depressed in a conical glass full of water, the diameter of which is 5 inches and altitude 6, so that its upper edge is in a line with the rim of the glass, while its side rests upon the side of the glass; required the quantity of water that will overflow from the glass.

By Construction, Fig. 133.

Draw a section of the glass, as A B C, and of the sphere, as def.

Then will C e=6, B C=2=2.5, and, by preceding example, B C

=6.5.

Hence, o e and of are equal, and the line B o, bisecting Be of, is a hypothenuse common to both triangles, B e o, Bƒo.

Consequently, B c and B f are equal to 2.5, and B C-Bƒ=Cƒ= 6.5-2.5

4.

..6 (Ce): 2.5 (e B)::4 (Cf): 1.667 (fo), and 1.667×2=3.334 inches, the diameter of the sphere, the volume of which (by rule, p. 176)=19.387 inches the quantity of water that will overflow from the glass.

73. If a sphere, the diameter of which is 4 inches, is depressed in a conical glass full of water, the diameter of which is 5 inches and altitude 6, it is required how much of the vertical axis of the sphere is immersed in water.

By Construction, Fig. 134.

Draw a section of the glass, as A B C, and of the sphere, d e f

Let m n be the original level of the water, and n r the level when the sphere is immersed.

Then will the cone n C r cone m Cn+the volume of the segment of the sphere d sƒ=} of cone A B C+the volume of the segment d s fƒ.

AC=✓Ci2(6)+A i2(5÷2)=6.5=length of slant side.

As A i (2.5): A C (6.5):: of, radius of sphere (2): C o (5.2)=distance from the centre of sphere at rest and the bottom of the glass.

Contents of cone (by rule, p. 166)=39.27, † of which=7.854. Put x=s u=the immersed part of the axis of the sphere, and C u=C s +su=3.2+x.

Then, as similar solids are to each other as the cubes of their like dimensions,

(3.2+x)3
216

X 39.27.

63: (3.2+x)3::39.27: cone n Cr; ..cone n Cr= Segment d sf (by rule 2, p. 177)=(4×3—2x) ×x2×.5236. Since Cone n Cr=cone m C n+segment d sf, ::

7.854+(4×3-2x)× x2 ×.5236.

And 25×(3.2+x)3=5×216+4× 6a × (6−x) × x2.

(3.2+x)3
216

× 39.27=

Cube 3.2+x and 25 (3.23+3×3.2x+3×3.2x2+x3)=5×63+4×6a (6−x)×x3.

Actually multiplying the terms in the first member by 25,*

163

5

+3·16x+3·5·16x2+25x3=5·63+4·62 (6-x)x2.†

But 4·62 (6-x)x2=4 · 63x2 −4·63x3.

163

5

+3·16x+3·5·16x2+25x3=5 · 63+4·63x2 —4 · 62x3.

Multiplying by 5 throughout,

163+3 5·162x+3·52 · 16x2+53x3=52 63+4 · 5 · 63x2-4 · 5 · 62x3.

By transposition,

(53 —4 · 5 · 62)x3+(3·52 · 16−4·5 · 63)x2+3 · 5 · 161x=52 · 63 — 163.

..5·132x3-3·5·13·16x2+3·5·16'x=8.163.

..5 (132x3-3 · 13 · 16x2+3·16'x)=8·163.

The first member is now a perfect cube, the root of which is

13x-16=-705.6-8.90265.

13x=16-8.90265=7.09735.

*By Professor G. B. Docharty, New York.

† See Explanation of Characters, page 10, for use of a period be

tween two factors.

from

Then, 63:39.27:: 3.745953: 9.5563=volume of water in cone, n Cr, which is to be deducted the volume of segment d s ƒ, and the remainder should be equal to of 39.27=7.854.

Thus, volume of segment (by rule 2, p. 177)=1.7023, and 9.5563 -1.7023=7.854=result required.

tre.

74. A lady having three daughters had a farm of 450.758 acres, in a circular form, with her dwelling-house in the cenBeing desirous of having her daughters near her, she gave to them three equal parcels of land as large as could be made in three equal circles within the periphery of her farm, one to each, with a dwelling-house in the centre of each; that is, there were to be three equal circles as large as could be drawn within the periphery of the farm; required the diameters of the farm and of the three parcels.

√/450.758 × 43560÷.7854=5000 feet diameter of farm. By Construction, Fig. 135.

A

B

Draw the given circle, with o as its centre, and divide its periphery into three equal parts, as at A B C; connect A o, B o, and C o, and assume d e f as the centres of the required circles.

As the three circles required touch one another and the given circle, the points, as A, B, and C, the centres, def, of the required circles, and o, are necessarily in right lines. Connect d, e, and ƒ.

Then, as def and e of are isosceles triangles, the angle d and the base e ƒ are bisected at right angles in i by the line d i, and e o, in like manner, bisects the angle e.

The triangles, e di, e o i, are equiangular:

Hence, e d: ei=(ed)::eo; oi=(}eo); ..eo=2oi.