pencil placed inside the string may be made to describe an ellipse, if the string is kept tightly and uniformly stretched while the pencil is in motion.

NOTE. This is the method commonly employed to lay off an ellipse upon the ground, as when making garden beds, etc.

The cord and straightedge methods are theoretically capable of describing a true ellipse. That is, the methods are founded on the mathematical principles governing the ellipse, but it is not generally possible to manipulate either the string or straightedge so that perfect results may be obtained.

MENSURATION OF SURFACES

TRIANGLES

A triangle is a plane surface bounded by three straight lines. Some of the different kinds of triangles are shown here.

A A L

m = distance from foot of

h to nearest vertex

a, b, c

sides opposite angles A, B, and C

P = perimeter

h = perpendicular upon base from vertex of angle opposite

Angles. A+B+C=180°. A = 180° (B+C), and similarly for B and C. A+A180°, and similarly for B+B' and C+C'. A'= 180°-A, and similarly for B' and C'. A'+B+C'=360°. A'=360° - (B'+C'), and similarly for B' and C'. A'=B+C; B'=A+C; C'=A+B.

Perimeter and Sides.-P=a+b+c. In all acute-angled triangles, including equilateral, isosceles, and right-angled triangles, any side a = √b2+c2=2bm. In an obtuse-angled triangle, the side opposite the obtuse angle a = √b2+c2+2bm. Altitude.—In any triangle, the altitude, or perpendicular distance from the base to the vertex of the opposite angle, is h= √c2 — m2.

Area. In any triangle, the area is equal to the product of the base by one-half the altitude, or A = Any side may be selected as the base.

bh
2

Thus,

if the base is the

bh in the obtuse-angled triangle, if the base is the side b, A = 2 ah' side a, A = If the length of the three sides is given, let pone-half the 2

sum of the three sides, or p=

a+b+c
2

and A

=

√p(p − a) (p − b) (p−c); that is,

the area is equal to the square root of one-half the sum of the sides multiplied by this one-half sum less each one of the sides, respectively.

Sid

Special Cases.-Equilateral Triangle.-Angle ABC-60°. Side a= = C. Altitude h=side X.866025. Side h÷.866025= h X 1.154701. =VareaX 1.51967. Area-side2X.443013. Length of side of square havin same area as an equilateral triangle = side of triangle X.658037. Diameter d circle of same area as an equilateral triangle-side of triangle +1.34677. Th perpendicular h bisects the angle B and the side b; and similarly for the othe angles and sides. This triangle is also known as the equiangular or 60° triangle Isosceles Triangle.-Angle A=C. B=90°-A or C. Side a=c. The per pendicular h bisects the angle B and the side b.

Right-Angled Triangle.-Angle A=90°. Angles B+C-90°. Side a2= b2+ c2, and a= √b2+c2, b = √a2— c2, c= √a2—b2.

For other properties and methods of solving triangles, see unde Trigonometry.

PARALLELOGRAMS

A parallelogram is a plane figure bounded by four straight lines which are parallel, two and two. Some of the different kinds of parallelograms are shown here.

Angles. The sum of the exterior angles the sum of the interior angles =A+B+C+D=A'+B'+C'+D'=360°. In the square and the rectangle, the four angles are equal and each is 90°; in the rhombus and rhomboid, A=C and B= D. In the square and the rhombus, the diagonals are perpendicular to one another and bisect one another and the angles at their opposite extremities.

Perimeter and Sides. Let the sides be a, b, c, and d, respectively, then the perimeter, P=a+b+c+d. In the square and rectangle, any side = area ÷ by an adjacent side. In the rhombus and rhomboid, a side area ÷ by the altitude h. In all four cases, the diagonal d= area ÷ by perpendicular p. Area. In all cases, the area Abh=dp.

Square.-Diagonal d=side X 1.41421. Side = diagonalX.707107. The side of a square equal in area to a given circle = diameter of circle X.886277. The area of the largest square that may be inscribed in a circle = 2X radius of circle.2 It should be noted that all problems relating to parallelograms, as well as to trapezoids, trapeziums, and regular and irregular polygons, may be solved by resolving these figures into triangles.

TRAPEZOIDS

A trapezoid is a plane figure bounded by four straight lines, only two of which are parallel one to the other.

B

One is shown in the accompanying figure. Angles.-The sum of the interior angles the sum of the exterior angles 360°, or A+B+C+D=A' + B' + C' +D'=360°. A=B'; B=A'; C=D'; D =C'. A = 180°-A'; B = 180° - B'; and similarly for C and D. A+B = C' + D'; A+CB'+D'; A+D = B'+C'; B+C =A'+D'; and similarly for other combinations of A, B, C, and D.

Perimeter. The sides being a, b, c, and d, the perimeter P=a+b+c+d.

Diagonal. The diagonal d=2X area÷(p+b).

Area. Case I.-Given the two parallel sides a and c and the perpendicular

distance between them h, A= (a+c).

Case II.-Given the diagonal d and the perpendiculars upon it ♪ and p'.

d

A = (p+p').

2

Case III.-Given the two parallel sides a and c and the angles adjacent to one of them, A and D, the area,

c2-a2

2 (cot A+cot B)

(c-a) (c+a) sin A sin B

=

2 sin (A+B)

Given all four sides a, b, c, and d. Let c-a-ƒ, and } (b+d+f)

A=

Case IV.

=s, then area A =

(a+c) √s (s−d) (s—b)(s—f).

TRAPEZIUMS

A trapezium is a plane figure bounded by four straight lines, no two of which are parallel.

=

Angles. As with trapezoids, the sum of the interior angles the sum of the exterior angles=360°. Also, A = 180°-A', and similarly for B, C, and D. Perimeter and Diagonal.—The same relations prevail for the perimeter

and the diagonal as for trapezoids.

Area. In the trapezium shown in Fig. 1, A = } [b(h+h')+ah+ch']. d

In the trapezium shown in Fig. 2, A =1⁄2 (Þ+p').

POLYGONS

A polygon is a plane figure bounded by three or more straight lines. Some of the more common forms are shown in Fig. 1.

Pentagon

Hexagon

Heptagon

FIG. 1

Octagon

If all the sides and angles are equal, each to each, the figure is a regular polygon; otherwise it is not. Of the figures previously discussed, the equilateral triangle and the square are regular polygons; the others are irregular.

In any regular polygon, Fig. 2, let the central angle (AOB=BOC, etc.) = C; let the interior angle (ABC= BCD, etc.) = 1; let the exterior angle (A'AB = B'BC, etc.) E. Also let R= -radius of circumscribed circle, and let r

radius of inscribed circle=apothem. Likewise, let S-length of a side

as AB, BC, etc., and let N = number of sides.

360°

Central Angle.-The central angle is equal to the exterior angle and is equal to 360° divided by the number of sides in the polygon, or C=E= sum of either the central or the exterior angles of any polygon is 360°.

The

N

Interior Angle.-The interior angle is equal to 180° minus either the cer or the exterior angle, or I = 180°-C=180°-E. The sum of all the inte

-R

B

FIG. 2

angles of any polygon is equal to twice as m right angles as the polygon has sides, less right angles, or I = (2XNX 90°) — 360°.

Diagonals. The diagonals AD, BE, etc. a regular polygon bisect the interior angle: bisect one another, intersect at the center the inscribed and circumscribed circles, divide polygon into as many isosceles triangles as it sides; also, they are the diameters Ď of the Dcumscribed circle.

Apothems.-The apothems LO, etc., of regular polygon are perpendicular to the sid They bisect the sides, the central angles, a one another; divide the fundamental isosce triangles of the polygon into two equal rig angled triangles BLO+CLO= BOC; and

the radii r of the inscribed circle. Perimeter and Sides.

The perimeter of any polygon is equal to the su of the lengths of all its sides. The perimeter of a regular polygon, P=N C C P Any side, S= 2√R2-r2=2R sin =2r tan

=

2 N'

Area. In any polygon, the area is equal to the sum of the areas of t triangles into which it is divided by its diagonals. The area of a regul polygon is equal to the area of one of the fundamental triangles, as AO multiplied by the number of sides. Likewise,

C

A=Nr2 tan}NR2 sin C = NSr = }Pr.

The accompanying table gives, for the more important regular polygon the number of sides; the name; the central angle, which equals the exteri angle; the interior angle; the length of the side, in terms of the radius of bot the circumscribed and inscribed circles R and r; and the area, in terms of th side S and in terms of the radius of the circumscribed and inscribed circles and r, respectively.

NAMES AND RELATIONS OF REGULAR POLYGONS

EXAMPLE 1.-What is the length of the side of a triangle inscribed in a circle of 2 in. radius?

SOLUTION.-Here R=2 and the side S=2X1.73205=3.46410.

EXAMPLE 2. What is the length of the side of a pentagon circumscribed about a circle of 4 in. radius?

SOLUTION.-Here r=4 and side S-4X1.45308-5.81232 in.

EXAMPLE 3.-What is the area of a hexagon whose side is 2 in. long? SOLUTION.-Here S=2 and area=S2 or 4X2.59808-10.39232 sq. in. EXAMPLE 4.-What is the area of a dodecagon that may be inscribed in a circle of 3 in. radius?

SOLUTION.-In this case, R=3 and area =R2 or 9X3.000000=27.000000

sq. in.

EXAMPLE 5.-What is the area of a decagon that may be circumscribed about a circle of 4 in. radius?

SOLUTION. In this case r=4 and area=r2 or 16X3.24921-51.9874 sq. in. EXAMPLE 6.-What is the radius of the circle that may be circumscribed about a square whose side is 2 in.?

SOLUTION.-Here R=2 and it is required to find S. S=2÷1.41421

= 1.41421.

Area of Irregular Polygons.-If the figure is bounded by straight lines, to find its area divide it into triangles; the sum of the areas of these will be equal to that of the irregular polygon. This method is commonly used by engineers as a useful check upon the accuracy of the areas obtained by calculation. In many cases the results thus obtained answer every purpose.

If one or more of the boundaries is an irregular line, as would be that of the bank of a river, the area may be found by one of several methods.

First Method.-Let A and E, Fig. 3,

be two corners on the river or on any other irregular boundary. Draw the A lines AF and FE to include as much water on the land side as they exclude land on the water side. If the map has been platted by means of coordi- B nates, those of F may be found by measurement, and the area can then be calculated by double latitudes. Or the entire figure may be divided into a series of triangles ás shown and the sum of their areas taken as that of the property.

FIG. 3

E

Second Method: By Selected Ordinates.-Draw perpendiculars on AB, Fig. 4, from the points of the curve at which its direction changes appreciably, and consider the portion of the curve between two consecutive perpendiculars to be a straight line. The figure is then treated as if divided into a number of trapezoids, whose areas can be computed by the rules already given.

Third Method: Trapezoidal Rule.-The ordinates are measured at regular intervals d along the line, as shown in Fig. 5. If the end ordinates are a and n

respectively, the area is A = (a++h)d in which Σh is the sum of all the

intermediate ordinates.

EXAMPLE. If the ordinates from the straight line AB to the curved boundary DC, are 19, 18, 14, 12, 13, 17, and 23 li., respectively, and are at equal distances of 50 li., what is the area included between the curved boundary and the straight line?

19+23

2

+18+14+12+13+17) × 50 = 4,750

sq. li.

Fourth Method: Simpson's Rule. The base line must be divided into an even number of equal parts, as shown in Fig. 6. The area is then equal to A

=(a+n+4Σh2+2h), in which a+n is the sum of the end ordinates; 4h2 is four times the sum of all intermediate even-numbered ordinates; and 2Σhs is