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solution in finite terms. The fluxion of every fluent wasfound to be always assignable, but the converse proposition,viz., of finding the fluent of a given fluxion, could only beeffected in particular cases, and amongst these exceptions, tothe great disappointment and regret of geometricians, wasincluded the case of the circle, with regard to all the formsof fluxions under which it could be obtained.

At length, all 1 ropes of accurately squaring the circle andsome other curves being abandoned, mathematicians began toapply themselves to finding the most convenient series forapproximating towards their true length and quadrature;and the theory of mensuration now began to make rapidprogress towards perfection. Many of the rules, however,were given in the transactions of learned societies, or in sepa-rate and detached works, till at length Dr. Hutton formedthem into a complete treatise, intitled, A Treatise on Mensu-ration , in which the several rules are all demonstrated, andsome new ones introduced. Mr. Bonnycastle also publisheda very neat work on this subject, intitled An Introduction toMensuration . Several books on the subject have been pub-lished since the above, which, notwithstanding, still maintaintheir high reputation.

Particular rules for measuring the various kinds of geome-trical figures and solids will be found under their respectiveheads ; but as a collection of examples of the mensuration ofdistances capable of application to the purposes of engineryand architecture is a disideratum in science, thelikely to occur in general practice, are here inserted.

Mensuration of Lines.

Problem I .—Any two sides of a right-angled triangle beinggiven, to find the third.

Case I.—When the two legs are given, to find the hypo-thenuse.

Rule. —Add the squares of the two legs together, and thesquare root of the sum will give the hypothenuse.

Example 1.—In the right-angled triangle, ABC, are giventhe base, ab, equal to 195 feet, and the perpendicular, b c,— 28 feet: what is the length of the hypothenuse, A c ?

195 28

195 28

following,

975

1755

195

784

(a b) s = 38025(b cf = 784

3,88,09 ( 197 = A 0

1

29 ) 288261

387 )

2709

2709

Example 2 .—If the span, A b, of a roof, be 24 feet, andthe height, DC, 5 feet, what should be the length of therafters, a c and C B.

This is resolved into two equal and similar right-angledtriangles, as follows :—

24 -T- 2 = 12, the half base.

(12 2 + 5 2 ) 2 = 13, the answer.

Case II.—When the hypothenuse and one of the legs aregiven, to find the other leg.

of

Rule. —From the square of the hypothenuse takethe given leg ; and the square root of the difference Winthe leg required. . 0

Example 1.—In the right-angled triangle, ABC, arethe hypothenuse B c, 601, the perpendicular AC, ‘ ’

required the base. ^

(601 2 —240 2 ) 2 = 551, the answer. _ , ggtExample 2.—In a roof, whose span, a b, is 45 feet 6 in® * ^and the rafters, AC or b c, 25 feet 5 inches ; requiredheight, d c, of the roof.

45 .. 6 25 .. 5

12 12

2)546

Rafter, 305 inches.

Base of each triangle, 273 inches.

. Inches. Feet. Inches.

Then will (305 2 — 273 2 ) 2 = 136 = 11 ..

Example 3.—In a roof, whose rafters are each gt

and the perpendicular height 10 feet, what is the s P aI1distance between the feet of the rafters ?

4 the answer-

26 ft*

(26 2 — 10 2 ) 2 = 24 feet, half the span. ,

C ah Ot

Consequently, 48 feet is the distance between the i ethe rafters.

Examples for Practice.

1. Given the hypothenuse, 1625 yards, and the perp ecular, 400 ; required the base.— Answer, 1575 yards. t j )g

2. Wanted to prop a building with raking-shores & g fheight of 25 feet from the ground, having several P* eC ”j, 9 e>wood of equal length which might be used for the p ur P peeach 30 feet long ; how far must the bottom of the sh<W eplaced from the base of the building ?— Answer, 16.5 “„ s led

The following rule for the construction of right

triangles, the sides of which shall be commensura

ble

jstru®'

each other, may be found useful in the practice of c° niing roofs. _ t ))e a

Rule. —Take any two square numbers at pleasure »their sum, their difference, and the double product oroots, will give the sides of a right-angled triangleshall be commensurable with each other. , 4 b 0

Example 1.—Let the two square numbers 1 aDtaken: the roots of which are 1 and 2.

Then 4 +_ 1 =5, the hypothenuse.

4 — 1 = 3, one of the legs, . 0 pr®'

and 2 X 2 X 1 = 4, the other leg, equal to double gl0 \)^duct of the roots ; consequently, 3, 4, 5, are therequired, and are the least numbers by which a rig* 11triangle can be constructed. a0 ,j 2 J-

Example 2. —Let the two square numbers be 14 4Then 144 + 25 = 169, the hypothenuse.

144 — 25 = 119, one of the legs.

2X 12 X 5 = 120, the other leg. fouo^ ^

In this manner an infinite variety of ratios may bethe sides of right-angled triangles. •thn^'fl’

The above rule may be found in some books of ar jjp ( W’but having obtained the three sides, suppose 169> nstr aC ‘Vthe following progressive table of ratios may be co 0 pp°_by adding 10 continually to the fourth column, and c ol a,J) ^site number of the vertical arithmetical progress* ge cO aon the left hand to each horizontal number in t e( ji» te ^and third columns, which will generate those >“* ^

below. j 30: ^[,0

Example. —To generate the numbers 194, 144» 0 f ,25 to 169, and the sum will be 194, the hypothec leg 9triangle. In like manner, 25 + 119= 144, one 0

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