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the less semicircle, draw FH parallel to BC, meeting the other semicircle in H, let fall H E perp. to AD, and A E 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. EGFAE_AD: DG: or AD.DG-AD.DI-AN=AD. DG-AE.EG -A E, and AD.DI=AD.DGAG, AE+AE, therefore (AD-AE). (DC or DG-AE)=AD. DI -half the area of the rectangle A B C D.

Calculation. Because AD.DI=100 x 40 – DFEH’= 4000, and HO-HE-EOP=4100, therefore EO=64.031, and AO-EO=AE=25.969 feet. Otherwise by Messrs. Baines, Hine, Maffett, and Win.,

ward. · Make A B=the sum of the length and breadth of the garden, on which as a diameter describe the semicircle ADB. From B draw the tangent BE A

CB 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-CB)=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. Pulsey, 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, 3o, 4o SO', &c. to 180° as he returns with his stones ; therefore, put 8, s', s', &c.=the nat. sines of 45', 1° 30', 2° 15', &c. then 2s, 28', 2s", &c. will be the chords of double those arcs, and putting the radius of the semi. circle=r, we shall have the sum of all the chords= 2rs + 2rs! + 2rs, &c. =2r x (s+s + s....)=2rx 76.89328=5874 yards, and 7260+5874—13134, the distance B must run, then 8:7 :: 14520 : 12705, the distance B runs while A runs 14520; therefore, 13134

12705 = 429 is the distauce 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 distavce 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) x 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 Hirsi. 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 x 2 5.153.78656=sum of all the chords to radius 1. But 120-3.1416=38.19273=rad. of the semicircle, which multiplied by 153.78656 we have 5873.538, to which add half the distance run by A=121 X 120%2=7260, and we have 13133.538 yards for the distance run by B, that of A being 14520, which x š gives 12705, so that when Ahad completed his task, B had still 13133.538 - 12705 = 428.538 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 sailed 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 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 LB = 67° 30', to find DH=18.4775, and HB= 7.6536; also, in the right angled triangle DGP, we have DG-HGDH = 138.4775, and the LD22° 30', to find PG = 57.3502, and DP=143.8869 ; the direct distance to P. But PF=AC+CB+HB+ PG=161.0128=the departure, therefore, ✓(PF? + AF)=A P=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, Cuttrall, Dunn, Eyres, Hine, Hirst, Maffett, Nesbit, Shaw, Thornton, De Weight, and Winward. 8. Qu. (50) Answered by Messrs. Butterworth, Gaw

throp, 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=94 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=1675, and x=4 feet; then, by art. 705, Marrat's Mechanics, we have the time of dis

A charging the vessel to the depth of 4 feet W he NT-x)=12m. 455.

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

9. Qu. (51) Answered by Mr. A. Nesbit. Put a=x-2,b=2x-2, and c=(x-2)+(28—2) -3..-4; then (a + b) xc=9x2 – 24x+16 is a sqr. and art2b=r? is a sqr. it therefore only remains necessary to make ba + 2a, or 4x2–6x a sqr. suppose it (2x+2p=4x-8 px+4p?; then, by reduction, r=

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12, 76 Let p= 4, then == 0; hence, a= b= and e=%, which are three numbers answering the conditions of the question.

Otherwise by Mr. A. Hirst. Put a +b = c then the first formula (a + b) x cis a sqr. again, puta?+2b=(a+n)', and by reduction we have a= 6", which being subsituted in the last formula it becomes b’ +500

t 2n =a sqr. which equate to (mb): and we get b =

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

Otherwise by Messrs. Hine, Maffett, and Winward. When c li+b, then (u + b) xc is evidently a square; we have then only to make the other two formulæ squares. Put b=2a+2, then a’+2b=a'+42+4 is a square, and b2 + 20 = 40 + 10a +4 must be a square, equate it to (20_)=40*4an+ns, and we

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Take n=1, and

have a=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, Rylanılo, Shaw, and Tomlinson. 10. Qu. (52) Answered by Mr. J. Gawthrop, and

Mr. S. Shar. · Let ABC represent the cone, and put .78543P, FG={DE=x, CF=10, and AF=FB=1; then, GB=-*, AG=1+r, and FB : CF:: BG: EG =10—10x, also v(AG+ EGJ = AE =N(101 — 198x + 101x?) = the trans- l/1M verse diameter of the section; and by ALTUB conics (ABX DE)=2x is the con- " jugate; therefore, A EX2Vxx p=the area of the ela. liptic section is to be a maximum, or A EX x, that is (1012 —-1987° +101x?), or 101x-198x2 +101x), must be a max.; which put into fluxions, and reduced, 132

66 2867 we have zem10*=-*; whence x==

101* ji whence * 101V 30603 =34739; EN x=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. Ys. 4d. Now the solidity of the cone is easily found = 10,472 feet, which at 4s. per foot comes to €2. ls. 10{d. and their difference, which is 7s. 5{d. is the loss the carpenter sustained by accepting the second proposal.

Again, by Mr. A. Nesbit, of Farnley.
Put Cn=x, then Fn=10—x, and by sim. triangles
CF:AB:: Cn:DE= Also A B_{DE=1-

, whence AG = 1 + 1, and A E’=AG? + GE = m _ 19.60396 x+ 100. But, by conics, AB* DE=.4x = sqr. of the conjugate, and the area of the

elliptic section being as the rectangle of these squares, · we have to? - 19.60396 x + 100 X.4x, or 4 x3 –

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