solid content of this segment is x-r x x + 2r X &c and the solid content of ABC the given hemisphere

2

is cr3. Therefore by the question x-rxx+2rxzc cr3, which reduced gives x2 = 3r and hence

=

x= r√3

=

:

=

10/3 = 17.3205 inches x — r = 10 × (√3 — 1) = 7.3205 inches the dimensions required.

EI and
GE,

The same answered by William Thompson, Charleston,

South-Carolina.

Put OA = OB = r = 10 the radius of the hemispherical loaf ABC, c = 3.1416 and x OG = BE the thickness of the crust; then will r — x = OE the radius of the sphere DEFI of which DEF the interior soft part of the loaf is evidently a segment; 2x GE the height of this segment and 2rx DF the diameter of its base. Fur

2

ther, by mensuration r— 2x X 2r -X

= solidity of the same segment and cr3 solid content of the hemispherical loaf ABC: and therefore by

the question r

+ 3rxa - x·3 =

each side of this

2

2x X 2r x = r3, or fr3 — £r2 x r3. Now r3 •2x added to equation gives r3

r2x, or r—x

2

xr-x=r-x x2. Hence r—x ⇒r √3 = 5√3 = 8.660254 inches = OE, x = rx (1 − √3) 1.339746 the thickness of the crust, r 2x 7.320508 = GE the height of the interior soft segment and 2√21 -2rx 17·11197 = DE the diameter of the base of the same segment.

V. QUEST. answered by James Temple, New-York. It is well known that a heavy body falling from a state of rest, descends in the first second of time, through a space s = 16 feet; and that sound moves

with the uniform velocity v = 1142 feet per second.

x

Put the height required, then = time of the

sounds ascent, and therefore by the question

square of the time of the ball's descent. the spaces described by falling bodies are as

of the times of falling, we have 8 : x :: 1" :

972

Now since the squares

8

and hence x = = 729791 feet = 138.218 Eng

lish miles the height required.

VI. QUEST. answered by John Smithis, Philadelphia. The land will be had at the cheapest rate possible when the sum paid for any given quantity is a miniPut f = 50 rods = half the given perimiter, the breadth of the rectangle; then will length, px the area, 1+ 2x = val

mum.

and

ue of the land, and

rod, a minimum, the fluxion of which equated with O and reduced gives x2 + px = 2. Whence x = 1⁄2ƒ × ( √/3 - · 1) = 18.30127, the breadth and -x=31·69873 the length required.

The same answered by the Proposer.

When the quantity purchased for a given sum of money is a maximum the land will be had at the cheapest rate possible. (Here this ingenious gentleman makes the same substitution as in the preceding sox2 lution, and proceeds.). Therefore

tity purchased for one dollar = a maximum; from the fluxion of which reduced, we have x2 + px = 2. Hence x = 18.30127 and the length and breadth required.

31.69873

VII. or PRIZE QUEST. answered by Robert Adrain, York-Town, Pennsylvania.

METHOD 1. Let ABC represent the triangular

F B

A E P

D

field, AB and BC the given sides, E and D the places of the trees, so that ABD and EEC may be right angles, and ED the given distance of the two trees. Instead of the numbers given in the question it will here be more convenient to take the least integers in the same ratio, viz. 50, 78 and 35, or 5, 7.8 and 3-5; and after obtaining a solution in terms of these numbers, the lengths of the unknown lines will be had by proportion, and consequently the area required will be easily This being premised, let AB 5a, BC

found. =7.8

= b, ED = 3.5e and BP =x; then

202

PE=

BP2

PC

and PD =

BP2

AP

=

202

3, and the abso

3.5. Here it is easy to perceive that x therefore by proportion 5: 24 :: 3 : 144 lute length of the perpendicular BP; consequently the base or third side AC = 53.76, BE and BD the distances of the two trees from the obtuse angle B are 15.6 and 18 respectively, and the area of the field 387.072 square chains.

COROLLARY.

The sides AB, AP and BP are as 5, 4 and 3 and the sides BC, PC and BP are as 13, 12 and 5.

This discovers the curious artifice used in the compo sition of this elegant problem.

METHOD 2. Let AB, BC and ED
b and c respectively, also let b2 + a2
n and the base AC = x; then by

theorems of geometry we have CP =

be denoted by

=

■m, b2 — a2 the common x2 + n

2x

and

2x X x2

2x x x2 + 22 and this equation by an easy reduction becomes x2 + ex4 2mx3 + n2x en2 = 0: which by assuming a = 50, b = 78 and e= 35 gives x5 + 35x4 17168x3+12845056 x .449576960 = 0. Hence by any of the common methods, we find x =112, and therefore by proportion 50: 24:: 112: 53-76 the absolute length of the base AC, from which the other parts of the question are easily found.

The same, ansivered by the proposer, G. Baron, N.Y.

Let ABC represent the field and D and E the two hemlock trees. Produce AB and CB to G and F, draw CG perpendicular to AG and AF perpendicular to CF. Put AB = α = 24, BC 37.44, DE =d=16·8, CG = x and BG= y. Then by similar

= b

triangles BC: CG :: AB : AF =

ay

::AB: BF = ay Again by similar triangles

AG: GC:: AB : BD =

CB: BE =

abx

b2 + ay

And since by the question

GBD and CBE are right angles, the angle EBD = supplement of ABC; of consequence the natural sine of EBD = natural sine of ABC, and therefore by mensuration, the area of the triangle EBD area of ABC :: BD × BE: BC × BÅ :: ED : AC a + y × b2 + ay axz

AB X CG

2

ax

2

=

X d. Again by mensuration

area of ABC. Now since

the triangle AGC is right-angled at G we have AC = ✔AG2 + GC' = ✔a2 + 2ay + y2 + x2, which equated with the former value of AC gives ab2 + a2 + b2 ×y+ay2

axa

xd=√a2+2ay+y2+x2 ̧ Further by right angled triangles x2 = 62 - y2. From these two equations we have

ab2 +a2+b2 × y+ay2

ab2 ay

Hence by reduction, and a restoration of the values of the known quantities we obtain an equation from which y is easily found 19.008 exactly; therefore

=

√b ty x b. ・y=32·256,

b2 + ay

ax

a+y

= 15.6 = BE, ✅a+y + x2

53.76 AC the base, and

ax

387·072 square

2