2

A C2 — A g2 = C g2, and C g + B C = B g. Ag2+Cg2= A C2.

Twice the area of any triangle, divided by the base, equals the perpendicular to that base.

Half the side of any triangle, multiplied by the perpendicular to that side, equals the area of that triangle.

NOTE. If a triangle have one obtuse-angle, a perpendicular dropped from either of its acute-angles will fall outside the figure.

In an equilateral triangle, a perpendicular dropped from either angle will bisect the opposite side or base, and the triangle will be divided into two equal right-angled triangles. In an isosceles triangle, a perpendicular dropped from the angle included between the two equal sides, will bisect the side opposite that angle; and the triangle will be divided into two equal right-angled triangles.

A perpendicular dropped from the angle opposite the longest side of any triangle, will fall within the figure, and will be shorter than either of the sides.

To divide a circle into any number of concentric circles of equal areas.

RULE.-1. Multiply the square of the radius of the given circle by the number of concentric circles less 1 required, and divide the product by the number of concentric circles required; the square root of the quotient will be the radius of the given circle, less the breadth of the outermost concentric circle.

2. Multiply the square of the radius of the given circle by the number of concentric circles less 2 required, and divide the product by the number of concentric circles required; the square root of the quotient will be the radius of the given circle, less the sum of the breadths of the two outermost concentric circles, &c.

EXAMPLE.-Required to divide a circle of 20 inches radius into two concentric circles of equal areas.

202 ÷ 2 = √200 = 14.142 inches, radius of inner circle.

The following diagram illustrates the principle of the foregoing Rule, and exhibits the mechanical method of solving the problem. It is perhaps unnecessary to add that it is wholly immaterial into how many concentric circles of equal areas the given circle is to be divided, or whether the concentric circles are to have equal areas or otherwise. The radius of the given circle has only to be divided into the required number of aliquot parts, or proportional parts, the parallel lines struck, the curves drawn, and the division is accomplished.

✔(1524) = 7.5, and 10.606-7.5 3.106 inches, -C.

=

15 — (2.01 +2.384 + 3.106) = 7.5 inches, — D.

NOTE.The square root of the quotient obtained by dividing the square of the radius of the given circle by the number of concentric circles required, is equal to the radius of the innermost concentric circle.

OF THE REGULAR BODIES.

The Regular Bodies are five in number, viz. :

The Tetrahedron or equilateral triangle; a solid bounded by four equilateral triangles.

The Hexahedron or cube; a solid bounded by six equal squares. The Octrahedron; a solid bounded by eight equilateral triangles. The Dodecahedron; a solid bounded by twelve regular and equal pentagons.

The Icosahedron; a solid bounded by twenty equilateral triangles. The following TABLE shows the superficies and solidity of each of the regular bodies, the linear edge of each being 1.

To find the superficies of any of the regular bodies, by help of the fore

going table.

RULE. - Multiply the tabular number in the column of superficies by the square of the linear edge, and the product will be the surface.

EXAMPLE. The linear edge of a tetrahedron is 3 feet; required the superficies.

1.732 X 32=15.588 square feet. Ans.

To find the solidity of any of the regular bodies, by help of the fore

RULE.

going table.

Multiply the tabular number in the column of solidities by the cube of the linear edge, and the product will be the solidity or contents.

EXAMPLE. The linear edge of an octahedron is 2 feet; required the solidity.

.4714 X 2.5 X 2.5 × 2.5 = 7.3656 cubic feet. Ans.

OF THE CIRCLE AND ITS SECTIONS.

Relationship of the sectional lines of the circle, one with another, whereby, any two being given, any other may be found.

To furnish several examples, and to exhibit the proof with the operation, suppose the lines a b and p d given, and let a b= 20, inches, feet, yards- any linear measure, - and let pd=3.51.

ab

P

10, sine of half the arc, or half the chord of the segment.

ар

202 =

a p2

d2 100+12.323.51

p p d 2 ca

=

32,

diameter.

12.49, cosine of half the arc.

ca ср p d

1612.49-3.51, versed sine of half the arc, or height of segment.

ср

pd ca

12.49 +3.51 = 16, radius.

Radius and height of segment given, to find chord of segment. ca-pd=cp, and c a2 chord, a b. Ans.

— c p2 = √ a p2 = a p, and a p × 2=2 ap=

Chord and height of segment given, to find diameter.

§ a b2+p d2 = a d2, and a d2 ÷ p d = 2 c a. Ans.

Radius and height of segment given, to find chord of half the arc. ca-pdcp, and ca2 — cp2 = a p2, and a p2 + p d2 =➡√ad2=ad.

Ans.

Radius and sine given, to find versed sine, or height.

c a2 — a p2 = c p2, and ca—c p= pd.

Ans.

Radius and cosine given, to find chord of half the arc. c a2 — c p2 = a p2, ca — c p =p d, and a p2+p d2 √ a ď2= aà.

Ans.

Radius and chord of half the arc given, to find sine of half the arc. ad2 2 ca= pd, and a d2 pd2 = √ a p2 = ap. Ans. Chord of half the arc and sine of half the arc given, to find radius. a d2 — a p2 = pd2, and a d2 ÷ 2 p d = ca. Ans.

Chord of half the arc and sine of half the arc given, to find cosine. a p2= pd2, and a d2 2 p d =ca, and ca2

a d2

Ncp2=cp. Ans.

Radius and sine of half the arc given, to find versed sine. ca2- a p2=cp2, and ca cp=pd. Ans.

a p2:

Sine and versed sine given, to find cosine, and, thereby, radius. a p2 + p d2 = a d2, and a ď÷ 2 p d=c a, radius, and c a

-

=

・pd=cp. Ans. Having the circumference of a circle given, to find the diameter. RULE. Divide the circumference by 3.1416, (3.141592) and the quotient is the diameter; or, multiply the circumference by .31831, and the product is the diameter; or, multiply the circumference by 7 and divide the product by 22. More correctly,— multiply the circumference by 113 and divide the product by 355, and the quotient will be the diameter.

Having the diameter of a circle given, to find the circumference. RULE. Multiply the diameter by 3.1416, and the product is the circumference; or, multiply the diameter by 22 and divide the product by 7; or, (and nearer than either the foregoing to the truth - the nearest known rule that can be expressed in small numbers) — multiply the diameter by 355 and divide the product by 113, and the quotient will be the circumference.

Having the area of a circle given, to find the diameter or circumference.

RULE.-Multiply the square root of the area by 1.12838, and the product is the diameter; or, multiply the square root of the area by 3.5449, and the product is the circumference.

EXAMPLE. What is the diameter of a circle that contains an area of 784 square feet?

✓78428 X 1.12837 = 31.594+ feet. Ans.