§ 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...