MENSURATIONMENSURATION, or the art of measuring, involves the construction of measures, the methods of using them, and the investigation of rules by which magnitudes which it may be difficult or impossible to measure directly are calculated from the ascertained value of some associated magnitude. It is usual, however, to employ the term mensuration in the last of these senses; and we may therefore define it to be that department of mathematical science by which the various dimensions of bodies are calculated from the simplest possible measurements.

The determination of the lengths and directions of straight lines, including what are familiarly known as problems in heights and distances, generally depends on the solution of triangles, and will be discussed in the articles Trigonometry and Surveying. The remaining portions of the subject are the determinations of the lengths of curves, the areas of plane or other figures, and the volumes and surfaces of solids; and it is of mensuration as thus restricted that the present article will discuss some of the more important problems.

§ 1. Units of Length, Area, and Volume.— In measuring any magnitude we select some standard or “unit” to measure by. Thus in measuring length we take for unit an inch, a foot, or a yard. From the unit of length we derive the units of area and volume. Thus we define the unit of area to be the area of the square described upon the unit of length, and the unit of volume to be the volume of the cube whose edge is the unit of length or whose side is the unit of area. For example, if an inch be taken as the unit of length, the square whose side is 1 inch is the unit of area, and the cube whose edge is 1 inch is the unit of volume. The length of a line, the area of a surface, and the volume of a solid are then expressed by the numbers, whole or fractional, of units of length, area, and volume which they respectively contain. Hence, if I denote the linear unit, the length of a line which contains a units is al, or simply a since I is unity; similarly the area of a surface which contains b units of area is bm, or simply b, where m is the unit of area.

§ 2. Commensurable and Incommensurable Magnitudes.— When two magnitudes have a common measure, that is, when another magnitude can be found which is contained in each an exact number of times, they are said to be “commensurable.” Thus a line 4⅜ and another 3⅛ inches long are commensurable; for, if ⅜ inch be taken as unit of length, the former contains the unit nine times and the latter seven times. If no common measure can be found, the two magnitudes are said to be “incommensurable.” For instance, 1 and √2 have no common measure; for V2 = 1'4142 . . . an interminable decimal, and hence no unit, however small, can be found which will be contained in each an exact number of times. If, however, we take χ∕2 = Γ4, the error will be less than τ ⅛j if V2 = 1'414, the error will be less than τj ⅛ π , &c. Hence, by taking a sufficient number of figures, we can find a fraction which will differ from χ∕2 by less than any assignable quantity, and therefore we can always find two commensurable magnitudes that will represent two incommensurable ones to any degree of accuracy we please. In what follows we need therefore only consider commensurable lines.

§ 3. Area of a Rectangle.— Let the side AB (fig. 1) contain a units and the side BC b units of length. If we divide AB into a equal parts, each equal to the unit of length, and similarly BC into b equal parts, and if through the points of division we draw lines parallel to the sides of the rectangle, these lines will divide the rectangle into a series of rectangles, each of which is the unit of area, since each is a square whose sides are of unit length. As we have a rows of these rectangles, and b in each row, the whole number of rectangles will be ab. Therefore area of ABCD= ab units of area *=ab.

PART I.—PLANE FIGURES.

Section I.— Plane Figures contained by Straight Lines.

A. The Rectangle.

§ 4. Let ABCD (fig. 2) be a rectangle, and let AB = CD=α, BC = DA = 5, AC = c, and the angle BAC = a ; it is required to find its area. Since a rectangle is completely determined when two independent data, one of which at least is a length, are given connecting its parts, we can determine its area in the following cases.

(a) When its length a and its breadth b are given. — It has already been proved (§ 3) that area of ABCD = ab ; or the area of a rectangle is equal to its length multiplied by its breadth.

Example.— Let α = 12 feet 6 inches and δ=9 inches, then area of ABCD = 12'5 × '75 = 9'375 square feet.

If we make use of logarithms in the above calculation we have log area=logα + logδ. logα = logl2'5 =1'0969100 log b = log '75 = 1'8750613

therefore log area = '9719713; hence area-= 9 '375.

(3) When a side a and the diagonal c are given.— By Euclid i. 47 we have

5 2 = c 2 -α 2 , or δ = Vc 2 -a 2 , therefore area of ABCD=αδ=αVc 2 -α 2 5 or logarea=loga + ⅜log(c + a) + ∣log(c-a).

Example.— Let a=238 and c=456, then loga-log238 = 2'3765770 ⅛log(c + a) = ⅜log694 = 1 '4206797 ⅛ log(c - a) - ⅛log218 = 1'1692282

therefore log area = 4 '9664849; hence area = 925/3'1.

(γ) When a side a and a its inclination to the diagonal are given. —Since = tan<x, 5 = αtanα, a and therefore area of ABCD = ab =α 2 tan a; or logarea=21ogα + Ltanα-10.

Example. — Let a = 36 and a = 32° 25' 15", then

21oga = 21og36= 3'1126050

Ltana=Ltan32° 25' 15" = 9'8028622

therefore logarea=12'9154672-10 = 2'9154672; hence area = 823'127.

(5) When the diagonal c and a its inclination to either of the sides are given. — W e have

α = ccosα, andδ=csina, therefore area of ABCD = ab =c 2 sin αoosa=⅛c 2 sin 2a; or 2area=c 2 sin2a, and hence log2 area=2loge+ Lsin2α -10.

§ 5. A square being a rectangle whose sides are equal, we can at once determine its area. When one datum, which must be a length, is given the square is completely determined, and hence we have only two cases to consider.

(a) When the side is given.— From § 4, a, we have at once area of square = ab =a × a=a?.

(3) When the diagonal c is given.— From § 4, 3, we have

α 2 + α 2 = c 2 , or a 2 = ⅜c 2 ; hence area of square=α 2 = ⅛c 2 , or 2 area = c 2 ; and therefore log 2 area = 2 log c.

B. Right-angled Triangles.

§ 6. The diagonal of every rectangle divides it into two equivalent right-angled triangles (Eucl. i. 34), and hence the area of the right-angled triangle ABC (fig. 2) is equal to half the area of the corresponding rectangle ABCD.

C. Triangles Generally.

§ 7. In every triangle there are six elements to be considered, namely, the three sides and the three angles. If any three of these six be given, provided one is a length, the triangle is completely determined, and hence its area can be found.

§ 8. Length of Perpendiculars of a Triangle.— In the triangle ABC (fig. 3) let BC = α, CA = 5, AB=c, AD the perpendicular from A on BC=⅛, BD=ae, and CD = y.

Since BDA and CDA are right angles, we have c 2 =aj 2 + 7ι 2 , and ⅛ 2 = y 2 + Λ 2 , and therefore

5 2 - c 2 =y 2 - aj 2 =(y+»)(y - x) ≈a(y - x);

δ 2 -c 2 whence y-x≈ . Cl

But y ÷α!=α, and, by solving these equations, we obtain

δ 2 + α 2 -c 2 y~ 2α -·

Again h*-V-y*≈V - ^ a +^ 2 -<∙ s y W-(⅛ 2 + α*-c 2 ) 2

(α + ⅜ + c)(⅛ + c - α)(c + « - b)(a + b-c )

4α 2 hence Λ=-A-√(α+5 + c)(5 + c-α)(c + α-δ)(α + b -c).

2α

Now let a + b + c≈2s, then 5 + c-α = 2(s-α), c + α-δ = 2(s-δ), and α + 5-c = 2(s-c).

Therefore, on substituting and reducing, we obtain

2

h = — √s(s - a)(s - b)(s - c).

Similarly the perpendiculars from B and C on the opposite sides are respectively

ο 2

y√s(s-α)(s-J)(s-c) , and —√i(s-α)(s-φ-c).

§ 9. We now proceed to investigate formulae for the area of a triangle in the following important cases.

(a) When the base a and the altitude h are given.— Since a triangle is equel to half a rectangle of the same base and altitude, we have at once area ABC —⅛α⅛.

Example.— Let α = 40 chains and A-14'52 chains, then area - ⅜ × 40 × 14 '52 = 290'4 square chains.

(3) When two sides a and c and the included angle B are given.— From fig. 3 — — sinB , and therefore Λ=csinB; hence area— ⅛ah- ⅛αcsinB; or log2area=logα + logc + LsinB-10.

Example.— Let a —40, c—30, and B —30°, then area = ⅛ac sin B — ⅜×40×30×⅛- 300.

(γ) When the three sides a, b, c are given.— From § 8

⅛ = - Vs(s-α)(s-δ)(s-c) , and therefore area = ⅛ah - ⅜α × -∣- √s(s - a)(s - b)(s - c) - Vs(s - α)(s - b){s - c); or log area — ⅛ {logs + log (s - α) + log (s - b) + log (s - c)}.

Since 2s=a + b+c, we have area of triangle — ∣ V2(α 2 δ 2 + b i c i + c 2 a 2 ) - (a i + b i +c 4 ).

Example 1.—Leta = 13, 5 = 14, and c—15, then s-⅜(13 + 14 + 15)-21, s-α-21-13 = 8, s-b≈ 21-14 — 7, and s-c—21-15=6; therefore area = √21 × 8 × 7 × 6 = 84.

Example 2.—Let a —255, δ = 238, and c = 221, then logs-log 357= 2'5526682

log (s-α) = log 102— 2'0086002 log(s-δ) = logll9= 2Ό755470 log (s - c) = log 136 = 2Ί335389

therefore logarea —J(8'7703543) = ^'3851771; hence area = 24276.

(δ) When any two angles B and C and the adjacent side a are given. —Since c sinC αsinC

a sinA ’ sinA ’ and therefore (by 3)

area = ⅛αcsinB - ? 8smBsln Ç where A = 180°- (B + C),

2 sin A

or log 2 area=2 log a + Lsin B + I. sin C + I. cosec A - 30.

Since all the angles of a triangle are given when any two are given, we can find the area of a triangle when any two angles and any one side are given. Thus, when A, B, and c are given, we know C also, and the problem reduces to a case of the preceding.

(«) When the three medians a, β, y are given.— If a, b, c be the three sides of a triangle, and α, 3, y the three medians, i.e., the lines drawn from the angles to the middle points of the opposite sides, then by well-known geometrical propositions we have

4(α 2 + 3 2 + τ 2 ) = 3(α 2 + ⅛ 2 + c 2 ),

16(a 2 3 2 + 3 2 τ 2 + τ 2 a 2 ) = 9(a⅞ 2 + δ 2 c 2 + c 2 a 2 ), and 16(a 4 + 3 4 + γ 4 ) = 9(a 4 + 5 4 + c 4 ).

Now (§ 9, γ)

area of triangle — ∣ V2(α 2 6 2 + 5 2 c 2 + c 2 α 2 ) - (a* + b* + c 4 ), therefore = ⅛√2(α 2 j8 2 + 3 2 γ 2 + 7 2 a 2 ) - (a 4 + 3 4 + γ 4 ).

D. Parallelograms.

§ 10. The opposite sides and angles of a parallelogram being equal, three independent data, one of which at least is a length, are necessary and sufficient to determine it completely.

In theparallelogramABCD (fig. 4) let BC = DA = α, AB = CD = δ, AC=c, AE = ⅛, the angle ABC = α and AOD = 3∙

Since the diagonal AC divides the parallelogram into two equivalent triangles, we obtain

(a) area of ABCD —2 area of triangle ABC

= 2X⅛ a×h (§9, a)=ah;

(3) area of ABCD — 2area ABC-2 × ⅛α⅛sinα (§ 9, 3) = a5sina; or logarea = loga + logδ + Lsina - 10;

(7) area of ABCD = 2 area ABC = 2(ABO+ CBO)

= 2{⅛BO. A0sinA0B + ⅜BO.COsinCOB}-2{⅜BO.ACsin3}

= ⅜BD. AC sin 3 = ⅜c<7sin 3 , or Iog2area=logc + Iog<i + Lsin3-10.

§ 11. If the parallelogram be equiangular (a rectangle), c— d, and area — ⅜c 2 sin3∙ If it be equilateral (a rhombus), 3 = 90°, and area =⅜crf. If it be both equiangular and equilateral (a square), c— d and 3 = 90°, and area — ⅛c 2 as before (§ 5, 3).

E. Trapeziums.

§ 12. To determine a trapezium completely four data are necessary and sufficient.

In the trapezium ABCD (fig. 5)letBC = a, CD — b, DA=c, AB-ci, and AE perpendicular to BC — h, and draw AF parallel to CD, then

(a) area ABCD —area ABC + area ADC

= ⅛α⅛ + ⅛ch ≈⅛(a + c)h; or the area is equal to half the sum of the parallel sides multiplied by the perpendicular between them.

Again, area of ABF -⅜BF × AE (§ 9, α) = ⅜(α- c)h, also area of ABF = Vs(s-AB)(s-BF)(a-FA), where 2s—AB + BF + FA; hence Λ=^-^√s(s-AB)(s-BF)(s-FA), therefore

(3) area of ABCD = ⅛(α + c)A=^√ s (s - AB)(s - BF)(s - FA)

(- α +δ+ c + i ≈0(α ÷δ-c-ti)(α + δ-c + ci)(α-5-c + ci) l since AB=ci, BF— a-c, and FA = CD — b.

Thus we can find the area of a trapezium in terms of its sides.

§ 13. If c—0, ABCD becomes a triangle, and its area

= ∣∖∕( -α+5+ci)(α+δ-ci)(α + ⅛ + d)(α-5 + ci).

Again, if c=α, then also b=d, and ABCD becomes a parallelogram, and its area takes the indeterminate form-^-, as it should do, since four sides do not completely determine a parallelogram.

F. Quadrilaterals Generally.

§ 14. A quadrilateral is completely determined when five independent data are given. We consider the following cases.

(a) When any diagonal and the perpendiculars on it from the opposite vertices are given.

The quadrilateral ABCD (fig. 6) = ABD + BCD

= ⅛BD.AE + ⅜BD.CF

= ∣BD(AE + CF)j or the area is equal to half the product of the diagonal and the sum of the perpendiculars.

If the diagonal BD fall without the figure, as in the concave quadrilateral ABCD (fig. 7), then it is clear that area ABCD = ⅛BD( AE - CF).

(3) When the diagonals and their included angle are given.— In the quadrilateral ABCD (fig. 8, p. 16) let BD — h, AC = h, and angle DEA — a, then

ABCD-ABD+BCD

= ⅛BD.AEsinα + iBD.CEsinα (§ 10, 7)

= ⅛Λ(AE + CE)sina

— ⅜λλ'sina; or the area is equal to the product of the diagonals and the sine of their contained angle.

The same result holds when one of the diagonals falls without the quadrilateral, as in fig. 7, as the reader can easily verify.

(7) When the four sides and the angle between the diagonals arc given.— If a, b, c, d be the sides and a the angle between the diagonals it can easily be shown that area of quadrilateral = ∣(α 2 - b 2 + c i - cP) tan a.

(δ) When the four sides are given and the opposite angles are supplementary.— In fig. 8 let AB = α BC = δ, CD—c, DA—d, AC = ½, angle ABC = α, angle CDA = 0, and let α + 3 = 180°, then area of ABCD = ABC + ADC

= ⅛ab sine + ⅜c<i sin β.

But sin j8 — sin (180°- α)—sin α, therefore area of ABCD = ⅛(ab + cd) sin α.

This gives us the area of the quadrilateral in terms of the four sides and one angle.

Again we have

α 2 + ⅛ 2 - 2ab cos a=h? ≈ c 2 + cP - 2cd cos β = c 2 + cf 2 + 2cd cos a,

., , α 2 + ⅛ 2 -c 2 -d 2 , , therefore cosα= —· .-. - κ — , and hence

2(αo + cd)

(α + δ + c-cf)(α + 5-c + d) , l + cosα∙=- ,'∖ ' and

2(α⅛ + cd) ’ 1 (c + d+α-δ)(c+<Z-α + δ)

X I.·UOU — — ·

2(ab + cd)

From this we obtain sin 2 α= (1 +cosa)(l - cos a)

(δ + c + d-a)(c + d + a-δ)(rf + a + ⅛-c)(a + δ + c-<Z) 4(ab+cd)*

Now let a + b+c+d=2s, then ⅜ (ab + cd) sin a = √(s - a)(s -b)(sc)(s - d); therefore area of ABCD = V(s - α)(s - b)(s - c)(s - d), or log area = ∣{log(s-α) + log(s-δ) + log(s-c) + log(s-iZ)} .

If d—0, the qua drilateral becomes a triangle, and its area is V (s - a)(s - b)(s - c)s as before. In extracting the square root of sin a we take the positive sign, since the angle a is less than two right angles.

G. Regular Polygons.

§ 15. Since a regular polygon is both equilateral and equiangular, a circle can be inscribed within it and also described about it, and thus the n straight lines drawn from the common centre of the two circles to the n vertices of the polygon divide it into n triangles equal in every respect. Therefore the area of the polygon is equal to n times the area of any one of these triangles.

§ 16. Radius of Inscribed and Circumscribed Circles.— Let AB (fig. 9)=αbea side of a regular polygon of n sides; let C be the centre of the inscribed and circumscribed circles, CD=r the radius of the former, and CE=R the radius of the latter. The angle ACB is evidently equal to the nth part of four right angles, that is

ACB = - , and ACD = ⅛ACB = 18 -. η n

Now AD=-^--CDtanACD = rtanl^-,

2 n and AD = -^- = ACsin ACD = Rsinl^-;

2 n therefore r=α×icot- , n and R=αx ⅜ cosec— θ.

n

§ 17. Perimeter of Polygon.— The perimeter of the polygon of 180° Ί R∏ o n sides is na, i,e., 2 nr tan f or 2πRsin-—

η n

From this it follows that the perimeters of the inscribed and circumscribed regular polygons of n sides of a circle of radius r are o · 180 ° 4 « t 180° i . 1

2nrsιn-^- and 2nrtan - — respectively.

§ 18. Area of Polygon.

(a) In terms of r.—The area of polygon = nACB≈nAD.CD = n× r 2 tanl 8 θl .

n (β) In terms ofR.— The triangle ACB

= ⅜AC. CBsinACB = ⅛R 2 sinθθθl, n and therefore area of polygon ≈∣nR 2 sin -θθ-.

(γ) In terms of a.— The triangle ACB

up nn a a i , 18° 0 α 2 180° = ⅜AB. CD = -g-×r=-× ⅛αcot-—ψcot~^- , and therefore area of polygon =α 2 × — cot^^

4 n or log4area=logn + Lcotl^- + 21ogα-10.

n

From α and β it follows that the areas of the inscribed and circumscribed regular polygons of n sides of a circle of radius r are 360° 180°

⅜nr 2 sin —— and nr 2 tan-—respectively.

§ 19. In the formula (§ 18, γ) for the area of a polygon, the factor ∣cot ι 8 θ - has a definite value for every value of n, and hence, if we find its value once for all for a large number of values of n, and tabulate the results, we can find the area of a regular polygon of n sides by multiplying the square of its side by the appropriate tabular value.

Again, if α = l we have r=¾coti^- and R =⅜cosec^-! η n ’ and thus we obtain in a similar manner the radius of the inscribed and circumscribed circles by multiplying the side by the appro- 180° 180°

priate tabular value of ⅛cot-— and ⅛cosec-respectively.

§ 20. The following table contains the values of — cot^^- and 4 n

180° 180°

their logarithms, and the values of ⅛cot — and ⅛ cosec—— for all η n values of n from 3 to 12. [table]

Let A denote the area of a polygon of n sides and A , the corresponding tabular value of—cot ί— , then

4 n A = α 2 A', or log A = 2 log a + log A'.

H.— Length of the Radius of the In&cribcd, Escribed, and Circumscribed Circles of a Triangle.

§ 21. (a) Radius of Inscribed Circle.— LetO (fig. 10) be the centre of the circle inscribed in the triangle ABC and touching the sides in D, E, and F. Join OA, OB, and OC. The angles at D, E, F are right angles (Euch iii. 18). Let BC = α, CA = b, AB—c, and OD = OE = OF-r.

Now A B C - B O C +COA+AOB

= ⅜αr + ⅛δr + ⅛cr=⅜(α + ⅛ + c)r=rs; whence — area ^BC ∖∕s(s -a)(sb)(s -^c) ’ = s” s (β) Radius of Escribed Circles.— Let CD = OE = OF = r α , then ABC=ACO+ABO-BOC

= ⅛br a + ⅛cr a - ⅜ara = ⅜(⅛ + c-a)ra = r a (s -a),

, area of ABC ∖∕s(s - a)(s - b)(s - c) ∕s(s-b)(s-c) and rα = = __ = λ ∕ .

Similarly ∕⅜ t -< t )(,→) and rι _ ∕≠-α)(>-⅜). ∖∕ s — b ∖∕ s — c (γ) Radius of Circumscribed Circle.— Let AD (fig. 12) =7? the perpendicular from A on the side BC, and AE = 2R the diameter of the circle, then (Encl. vi. C) we have

2R ×p=b×c, therefore 2R × ap=dbc. Now ap≈2∆, where Δ denotes the area of ABC; hence R = —=—. a bc .

4∆ 4 √ s (s _ β ξ s _ jξ β _ c )

Example.— Let a = 13, δ = 14, and c=15; then r will be found to be 4, r a 10⅜, n 12, r c 14, and R 8j.

Section II.— Plane Figures contained by Curved Lines.

A. The Circle.

§22. Circumference of a Circle.— If we inscribe in any circle a regular polygon of n sides, and also circumscribe a regular polygon of the same number of sides, it is clear that the perimeter of the circle is intermediate between the perimeters of the inscribed and circumscribed polygons, and that the difference between the perimeters of the inscribed and circumscribed polygons can be made as small as we please by sufficiently increasing n. A similar statement holds with reference to the areas of the circle and the inscribed and circumscribed polygons. With the above assumptions it is easily proved that the circumference of a circle bears a constant ratio to its diameter. Hence we have

Circumference=C = constant × radius = constant × r.

It is usual to denote this constant by 2τr, and therefore

C = 2πr= πd, where d is the diameter of the circle.

§ 23. Numerical Value of τr.—The constant π being, as can be easily proved, an interminable decimal, we can only approximate to its value, but this we can do to any degree of accuracy we please.

If s and σ denote respectively a side of the inscribed and circumscribed polygons of n sides, and s' and σ' a side of the inscribed and circumscribed polygons of 2n sides, it can easily be shown that ( α ) σ= √T 2 '^- ⅛s)a > (3) s' 2 = 2r j r - √r 2 - (Js)≡ j ,

(γ)∙σ' —,

√r 2 -(⅛s') 2 where r is the radius of the circle.

If we take r=⅜ we find, by means of these formulae, and by assuming the value of s when n=6, that the perimeter of inscribed polygon of 96 sides = 3 Ί40 .... , and the perimeter of circumscribed polygon of 96 sides =3 ’ 142. . . .

From this we learn that the circumference of the circle, in this oase π, is greater than 3140 .... and less than 3142 . . . . , and therefore as far as the second place of decimals

τr=314.

By taking greater and greater values of n we obtain closer and closer approximations to τr.

The preceding method for approximating to the value of π is the simplest afforded by elementary geometry, and is also the oldest; but better and more rapid methods are furnished by the higher mathematics. The calculation of π has been carried to 707 places of decimals, the following being the first 20 figures in the result:—

314159265358979323846.

For all practical purposes it is sufficient to take

π = 314159or=⅞∣.

§ 24. The following table contains the functions of π that are of most frequent occurrence in mensuration :— [table]

§25. Units of Angular Measurement.— In measuring lines we select some line of constant length as the standard or unit; similarly in measuring angles we require to take some angle of constant magnitude as unit angle. The right angle is by its nature the simplest unit angle, but, for convenience, we take the ⅛th part of a right angle for unit, and call it a degree, which is subdivided into sixty equal parts called minutes, and these again into sixty equal parts called seconds. For theoretical purposes we define the unit angle to be the angle subtended at the centre of a circle by an arc equal to the radius. This angle we call a “radian.” In many treatises the radian measure of an angle is called the circular measure.

§ 26. The radian is a constant angle.— Let OA (fig. 13) = arc AB = r, then A0B = radian, and let AOD = 90°j then arc AD = ⅛ × 2τr= ⅞τrr; and, since angles at the centre of a circle are proportional to the arcs on which they stand (Euch vi. 33), number of degrees in radian AOB AB_ r _ 2 number of degrees in AOD AD ⅛πr π ’ therefore number of degrees in radian

2

= 90° × — = 57° ∙ 2957795 = constant.

7Γ

§ 27. Number of Radians in any Angle.— Let AOC (fig. 13) be any angle, AOB the radian, and AC=s; then number of radians in AOC AC s .

one radian AB r ’ therefore number of radians in AOC=- .

r

If AOC=90°, then s=⅛πr, and number of radians = ⅜ιr; there are thus 7r radians in two and 2π in four right angles.

When r≈l we have number of radians=s, and hence in some treatises for the number of radians in an angle we find the length of the arc given.

§ 28. To transfer from degrees to radians and conversely.— Let x denote the number of degrees in an angle, and θ the number of radians in the same, then, since »

<∙>'-⅛>

§ 29. The following table contains the values of θ for values of x up to 180°, and also for minutes and seconds. [table]

As an example of the use of this table we proceed to find the value of θ when ic=68° 45' 17"’8.

Whenx=68° 0 = 1’1868239, X = 40' θ= Ό116355, x- 5' θ= Ό014544, x =10" θ= Ό000485, x= 7" ∙0000339, and when x=O'' ’ 8 0 = ’0000039, therefore when x=68° 45' 17"’8 0 = 1’2000001.

§ 30. Combining the results of §§ 27 and 28 we obtain r ∖ n s j 180 s . v r π r W r ’.I»» θ 7Γ X

ω s . rβ .i. ra .

§ 31. Length of Arcs of Circles.— The following are the more important cases:—

(a) In terms of the chord of the arc and the radius of the circle.— Let AB (fig. 14) = 2c, AC = r, and AEB=s, then

AD = ⅛AB=c=rsin⅜C, whence C is known, and therefore the arc s is found (§ 30, γ).

(3) In terms of the height of the arc and the radius of the circle.— Let DE =h = height of arc, then

CD = CE-DE = r-Λ,

, ιr , CD r-h and cos⅛C=-—= ,

AC r whence C is found, and therefore s.

§ 32. Huygens's Approximation to the Length of a Circular Arc.— Let AB (fig. 15) ≈p be the chord of the arc AEB, and AE = EB = g ’ that of half the arc, then the arc AEB = ⅛ (85 - p) approximately.

For, let r denote the radius, s the arc AEB, and 20 the angle ACB, then 0 = ^ 7 . Again, AB=p = 2AD = 2rsin0=2rsin^-; and similarly q =2r sin — .

4r

0 3 0 5

Now sin0 = 0-∣-g-+ ∣-g--&c.; ., f n ∖ s 1 I s ∖ 8 1 I s ∖ s „ ) therefore p = 2r < r -( ττ-I + rH 1 - &c. > p I 2r ∣3^∖2r∕ ^∣5∖2r∕ )

s 8 , s 5 =S ~2 .3 . 4 . r 2 + 2 . 3 . 4 . 5 . 16 . r 4

Stately 8 f -lβr j ⅛- j ∣(⅛)∙ +l ⅜(Λ)∙ - llc . j s 8 s 5 = 4s - 2 . 3 . 4 . r 2 + 2 . 3 . 4 . 5 . 64 . r 4 ^ &C ‘

Hence, neglecting powers of — be- r yond the fourth, we obtain fcp=√ 1 _ s4 Λ= s

3 Λ 1 7680r 4 ∕ approximately.

In practice it is sometimes more convenient to use the equivalent form s = 2q + ⅛(2q-p').

§33. Area of Sector of a Circle.— Let the sector be OAB (fig. 16). Divide the arc AB into n equal parts, and draw the chords of these. Let P denote the perimeter of the broken line AB, A the area of the polygon AOB, and p the altitude of any one of the n equal triangles of which this polygon is made up.

Now in the limit, when n is indefinitely increased, P becomes the arc AB=s, a result which we symbolize thus—

T j P = arcAB=s.

j -'n=oo

Similarly I j a - a0 P “radius= r.

Again, the area of the sector is equal to the area of the polygon when the broken line AB becomes the arc AB, that is, sector=I. w=x A = I. n = αo ∖P × P = ⅛I. n ≈ 00 P × I. a=00 P · = ⅛rs.

§ 34. Let 8 denote the area of a sector of a circle, then, by means of the above result and § 27, we have

(a) S = ⅜sr, (β) S = ⅛r0 . r = ⅛r 2 0.

§ 35. We proceed to find the area of a sector of a circle in the following additional cases :—

(a) When the chord of the sector and the radius of the circle are given.— In fig. 14 let AB = 2c, and let AC = r, then we have

. ACB AD c sm ^-=-rττ=—;

2 AC r whence ACB and therefore 0 is known, and S can be found by § 34. ACB has two values, the smaller one giving the area of the minor, and the larger that of the major conjugate sector.

(3) When the chord and height of the chord are given.— Let DE (fig. 14) =Λ and AB = 2c, then

AC 2 = r 2 = AD 2 + DC 2 =c 2 + (r-Λ) 2 , whence r= c -^-, and therefore by previous case the area can be found, (γ) When the chord and angle subtended at the centre are given.

—Let AB (fig. 14) =2c and ACB = 0, then c . ACB c

— = sm — s - , or r = - , ,, r 2 sm⅛0 therefore area of sector = ⅜r 2 0=⅜ ( . c , I ×0.

∖sm⅛0∕

§ 36. Area of a Circle.— The circle being a sector whose arc is tho whole circumference we obtain at once area of circle = ⅛r × s =⅜r × 2πr = τrr 2 .

An independent proof of this proposition might be given by means of the inscribed and circumscribed polygons, and from the area of a circle the area of a sector can be deduced. The infinitesimal calculus affords a simple and elegant proof (see § 44).

§ 37. If A denote the area, r the radius, d=2r the diameter, and C the circumference of a circle, we have

(a) A = πr 3 ,

(0) A≈⅜×2πrxr=⅜Cr,

. . . 4ιr 2 r 2 C 2 ω a -ι√--s∙

(∙)

Whence we see that the area of a circle is obtained by multiplying (a) the square of its radius by π = 3Ι4159, (3) the radius by half the circumference, (γ) the square of the circumference by -I. = Ό7957, (δ) the square of the diameter by ∣π = - 78539.

§ 38. Again, from the above formulae we deduce

(a) r— A =’5641896 × A,

Vιr

(3) d —La=Γ1283792×A,

^ s ∕π

(γ) c=2√πA = 3’5449077 ×A, thus obtaining radius, diameter, and circumference from area.

§ 39. Area of a Circular Ring. — Let r and r 1 denote the radii of the outer and inner circles respect-

ively (fig. 17), then the area of the space between them

= πr 2 - πr J = π(r+r 1 )(r - r 1 ).

The circles need not be concentric, and the reader should note that the area of the ring is equal to the area of an ellipse whose major and minor axes are r + r 1 and r-r l (see § 51).

§ 40. Area of the Sector of an Annulus.— Let angle ACB = 0 in fig. 17, then the area of ABED

= sector ACB - sector DCE

= ir 2 0-⅛r 2 0, = i0(r + r 1 )(r-r 1 ).

Again, let AB — I, DE =» l 1 , and CA - CD — r - r 1 — ħ, then I , Z 1 .. .

r-- 3 - and r 1 = -±- ; therefore σ σ

r+*ι~∣∙(*+⅞λ and area of sector =≈⅛0(r + r 1 )(r-rι)-⅛0 = JA(Z + Z 1 ).

§41. Area of a Segment of aCircle.— (a) When the radius and the angle subtended at the centre are given.— In fig. 14, let AEB be a segment of a circle, then its area

— sector ACB - triangle ACB. - ⅛r 2 θ - ⅜r 2 sin θ (§§ 9, 34) — ∣r 2 (θ-sin0).

If the segment be greater than a semicircle sinβ is negative and the formula becomes

⅜r 2 (0 + sin0), as is also geometrically evident.

We might in a similar manner find the area of a segment of a circle (0) when the chord and radius are given, (γ) when the chord and its height are given, (δ) when the radius and height of the chord are given, (e) when the chord and angle subtended by the chord are given. In all these cases the method of proceeding is obvious, a segment being the difference between a sector and a triangle.

§ 42. Area of a Lune.— Let ADB and ACB (fig. 18) be two segments of circles, then the area of the lune ADBC

—segment ADB - segment ACB.

Hence if we so choose our data that we can determine the areas of the two segments we have only to take their difference to find the area of the lune.

§ 43. Area of a Circular Zone.— Let AB and CD (fig. 19) be two parallel chords, then the area of the zone ABCD

— circle - segment AHB - segment DFC; or — segment AED + trapezium ABCD + segment BGC — 2 segment AED + trapezium ABCD.

§ 44. The Infinitesimal Calculus (q. v.) furnishes a simple and elegant proof of the formulae for the areas of a circle and a sector. If y=-φ(x) be the equation to a plane curve referred to rectangular axes, then the area between the curve, the axis of x, and two ordinates corresponding to the abscissae a and b is represented by the integral

∕* i ∕ φ{x}dx. β ∕α

Let x and y be the coordinates of P (fig. 20), then if OP=r the equation to the circle is y 2 =r 2 -x 2 , and therefore area of quadrant AOB=∕ ∖∣ r t -a^dbx Jo and therefore area of whole circle — ∙nr 3 .

§ 45.