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the less semicircle, draw FH parallel to BC, meeting the other semicircle in H, let fall HE perp. to AD, and AE will be the breadth of the walk.

For AD.DI=DF-EH-AE. EG. To each side of this equation add A E-AD. DG, then AD. DI +AE-AD.DG=AE. EG+A E-AD DG; or AD. DG-AD.DI-AE-AD. DG-AE. EG -A E, and AD. DIZAD.DG-AG.AE+AE', therefore (AD-A E). (DC or DG—A E)=AD.DI half the area of the rectangle ABCD.

Calculation. Because AD.DI=100×40 DF2=

EH 4000, and HO2-HE' E O2=4100, therefore EO 64.031, and A O-EO AE=25.969 feet.

Otherwise by Messrs. Baines, Hine, Maffett, and Winward.

Make AB the sum of the length

and breadth of the garden, on

which as a diameter

semicircle AD B.

describe the

From B draw the tangent BE

equal to the side of a square whose area is equal to half the area of the garden, draw DE parallel to A B, and from D let fall the perp. DC, so shall CB be the breadth of the walk.

For CBX (A B-C B)=CD', by a known property of the circle.

Calculation. In the right angled triangle COD we have OD and CD given to find OC 64.03124237, and OB-OC=CB=25.9687, &c. yards is the breadth of the walk. This question is in effect the same as the following lineal section, viz. given the sum and rectangle of two right lines, to determine those lines.

This question was answered also by Messrs. W. Allen, W. Bean, B. Brooke, J. Cattrall, W. Dunn, E. S. Eyres, A. Nesbit, W. Putsey, S. Shaw, W. Thornton, and J. Tomlinson.

6. Qu. (48) Answered by Mr. Gawthrop.

The distance which B runs on the circumference is easily found 7260 yards, and A runs 14520 on a right line. But B has to run the sum of the chords of 1°30′,

3°, 4° 50', &c. to 180° as he returns with his stones; therefore, put s, s', s", &c. the nat. sines of 45', 1° 30′, 2° 15', &c. then 2s, 2s', 2s", &c. will be the chords of double those arcs, and putting the radius of the semicircle=r, we shall have the sum of all the chords= 2rs + 2rs' + 2rs", &c. =2rx (8 + s'′+5'' ....)=2rx 76.89328 5874 yards, and 7260+587413134, the distance B must run, then 8:7:: 14520: 12705, the distance B runs while A runs 14520; therefore, 13134 -12705 — 429 is the distance that B has to run when A has finished, consequently B will lose.

"

The same by Messrs. Dunn, Hine, Maffett, and Winward. First (240+2) x 60=14520 yards, the distance run by A; and in order to find the distance run by B, we must find the greatest and least chords; these, by trig. are easily found=76.3805 and .9998, the diam. being 76,394. Then, p. 8, Introduction to Hutton's Tables, as the least chord in a semicircle is to the diameter, so is the sum of the least chord diameter and greatest chord, to double the sum of all the chords." From whence we get 5874.272, the distance travelled by B on the chord lines, and (120+1)× 60=7260 is the distance on the semicircle, therefore 7260+5874.272 =13134.272 B's whole distance, but 8:7:: 14520 : 12705=the distance run by B while A performs his whole journey, of course A is the winner.

Again by Messrs. Cattrall and Hirst.

By adding together all the nat. sines of half the arcs on which the stones lay, that is the nat. sines of 45', 1° 30′, 2° 15′, &c. to 180°, we get 76.89328, and 76.89328 × 2 .153.78656 sum of all the chords to radius 1. But 120 3.141638.19278—rad. of the semicircle, which multiplied by 153.78656 we have 5873.538, to which add half the distance run by A=121 × 120÷2=7260, and we have 13133.538 yards for the distance run by B, that of A being 14520, which x gives 12705, so that when A had completed his task, B had still 13133.538-12705428.538 yards to run, consequently B lost the wager.

Other similar answers were sent by Messrs. Baines,

Butterworth, Nesbit, Putsey, Rylando, Shaw, Tomlinson, and De Weight.

7. Qu. (49) answered by Mr. J. Gawthrop, and
Rylando.

Let A be the port sail-
ed from, and P the place
to which the ship is
bound, also let C be the
place where the ship fell
in with the current, then
while the wind would
carry the ship from C to
B, the current would car- P

HB

G

ry her to D, of course D is the place of the ship when she has sailed 60 miles in the current. Now in the right angled triangle DHB, we have DB 20, and the B 67° 30′, to find DH=18.4775, and H B= = 7.6536; also, in the right angled triangle DGP, we have DG=H G+DH=138.4775, and the ▲ D= 22° 30', to find PG 57.3592, and DP 149.8869; the direct distance to P. But PF-AC+CB+HB+ PG 161.0128 the departure, therefore, (PF+ AF) AP=200.8, is the distance from P to A, hence the angle FPA is found 36° 41' 47", and the course 53° 18′ 13′′ N. E. In this solution, the distance sailed in the current is supposed to be measured by the log.

Very good answers to this question were sent by Messrs. Allen, Baines, Bean, Brooke, Butterworth, Cattrall, Dunn, Eyres, Hine, Hirst, Maffett, Nesbit, Shaw, Thornton, De Weight, and Winward.

8. Qu. (50) Answered by Messrs. Butterworth, Gawthrop, and Rylando.

The pressure of the atmosphere being taken off at the bottom, or against the orifice, it must be added to the pressure of the water in the vessel; therefore, reckoning the pressure of the air=$4 feet of water, the velocity of the effluent water will be the same as from a vessel full of water, the altitude of which is 38 feet, and the time of evacuation will be the same as that in which the surface of the water in this vessel

would be depressed 4 feet. This being premised, put the area of the base of the vessel A, that of the orifice a, 38 h, g=16, and x=4 feet; then, by art. 705, Marrat's Mechanics, we have the time of disA charging the vessel to the depth of 4 feet=

h-x)=12m. 45s.

a√g

Other answers were given by Messrs. Brooke, Dunn, Harvey, Hine, Hirst, Maffett, Nesbit, Patsey, Tomlinson, and Winward.

9. Qu. (51) Answered by Mr. A. Nesbit.

Put a=x-2, b=2x-2, and c(x−2)+(2x−2) =3x-4; then (a + b) × c = 9x2-24x+16 is a sqr. and a2+26=x2 is a sqr. it therefore only remains necessary to make b2+2a, or 4x2-6x a sqr. suppose it= (2x+2p)2=4x2—8 px+4p2; then, by reduction, x=

4 p2 8p-6

64

26

76

12 Let p 4, then = ; hence, a= b= and 26' 20' c= which are three numbers answering the condi26'

88

tions of the question.

Otherwise by Mr. A. Hirst.

Put abc then the first formula (a+b) x c is a sqr. again, put a2+2b=(a+n)3, and by reduction we 2b-n2 which being subsituted in the last for

have a
2n
mula it becomes b'+

2n

26-n2

a sqr. which equate to

m2n + n2

(m-b) and we get b =

Take n = 1, and

2mn+2

m=2, and we have a, b, and c=2, which numbers will answer the conditions of the question.

Otherwise by Messrs. Hine, Maffett, and Winward. When ca+b, then (a + b) x c is evidently a square; we have then only to make the other two for mulæ squares. Put b=2a+2, then a'+26a2+4a+4 is a square, and b2+2a4a2 + 10a + 4 must be a square, equate it to (2a-n)2=4a2-4an+n2, and we

have a=

n24

4a+10'

where n may be taken at pleasure

provided it be greater than 2.

And thus nearly it was answered by Messrs. Baines, Dunn, Eyres, Gawthrop, Harvey, Putsey, Rylando, Shaw, and Tomlinson.

10. Qu. (52) Answered by Mr. J. Gawthrop, and
Mr. S. Shaw.

Let ABC represent the cone, and put .7854-p, FG=4DE=x, CF 10, and AF FB=1; then, GB=1−x, AG=1+r, and F B : CF :: BG: EG

D

n

F G

E

B

10-10x, also (AG+EG)=AE =√(101-198x + 101x2) the transverse diameter of the section; and by A conics (AB× DE)=2x is the conjugate; therefore, A Ex2/xxp the area of the elliptic section is to be a maximum, or A Ex/x, that is √(101x198x2+101x3), or 101x-198x+101r3, must be a max.; which put into fluxions, and reduced,*

132

we have x- x; whence x=

66

101

-

2867

30603

101 =34739; 2x=1.1787; A E-6.6637, and the area of the elliptic section 6,1689 square feet, which at 8s. per foot will come to £2. 9s. 4d. Now the solidity of the cone is easily found = 10,472 feet, which at 4s. per foot comes to £2. 1s. 10d. and their difference, which is 7s. 5d. is the loss the carpenter sustained by accepting the second proposal.

Again, by Mr. A. Nesbit, of Farnley.

Put Cnx, then Fn=10-r, and by sim. triangles CF:AB:: CnDE=. Also A B-DE=1—

[ocr errors][merged small]

x =GB, whence AG = 1+ and A E'=AG’+

GE2x2

DE .4r

10'

19.60396x+100. But, by conics, AB X sqr. of the conjugate, and the area of the elliptic section being as the rectangle of these squares, we have - 19.60396 + 100 ×.4, or 4 x3

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