Application of the Table.

1. The radius of a circle being 6 feet, required the side of the greatest heptagon that may be inscribed therein.

•867 X 6.5=5.6355, or 5 feet 7 inches nearly.

2. Each side of a pentagon is required to be 9 feet required the radius of circumscribing circle.

•8529=7-668, or 7 feet 8 inches.

3. A perpendicular from the centre to either side of an octagon is required to be 12 feet; what must be the radius of circumscribing circle?

1.08 X 12 12.96, or 12 feet 11 inches.

4. Each side of a hexagon is 4 yards; required its superficial contents.

4122.59852.6095 square yards.

THE CIRCLE AND ITS SECTIONS.

Observations and Definitions.

1. The circle contains a greater area than any other plane figure bounded by the same perimeter or outline.

2. The areas of circles are to each other as the squares of their diameters; any circle twice the diameter of another contains four times the area of the other.

3. The radius of a circle is a straight line drawn from the centre to the circumference, as O B.

4. The diameter of a circle is a straight line drawn through the centre, and terminated both ways at the circumference, as CO A.

C

B

A

G

F

E

5. A chord is a straight line joining any two points

of the circumference, as D F.

6. The versed sine is a straight line joining the chord and circumference, as E G.

7. An arc is any part of the circumference, as CDE. 8. A semicircle is half the circumference cut off by a diameter, as CEA.

9. A segment is any portion of a circle cut off by a chord, as DEF.

10. A sector is a part of a circle cut off by two radii, as A OB.

General Rules in Relation to the Circle.

1. Multiply the diameter by 3-1416, the product is the circumference.

2. Multiply the circumference by 31831, the product Is the diameter.

3. Multiply the square of the diameter by 7854, the product is the area.

4. Multiply the square root of the area by 1·12837, the product is the diameter.

5. Multiply the diameter by 8862, the product is the side of a square of equal area.

6. Multiply the side of a squarè by 1-128, the product is the diameter of a circle of equal area.

Application of the Rules as to Purposes of Practice. 1. The diameter of a circle being 7 inches, required its circumference.

7.1875 X 3.1416=22-58025 inches, the circumference. Or, the diameter being 30 feet, required the circumference.

3.1416 30.595-8188 feet, the circumference.

2. A straight line, or the circumference of a circle, being 274-89 inches, required the circle's diameter corresponding thereto.

274-89 X-31831 87.5 inches diameter.

=

Or, what is the diameter of a circle, when the circumference is 39 feet?

31831 3912-41409 feet, and 41409 × 12 = 4.96908 inches, or 12 feet 5 inches, very nearly the diameter.

3. The diameter of a circle is 33 inches; what is its area in square inches?

3-75214-06257854 11.044, &c., inches area.

=

Or, suppose the diameter of a circle 25 feet 6 inches, required the area.

25.52650-25X7854510-706, &c., feet, the area.

4. What must the diameter of a circle be, to contain an area equal to 706-86 square inches?

V706-8626·586 × 1·12837 = 29.998 or 30 inches, the diam eter required.

5. The diameter of a circle is 144 inches; what must I make each side of a square, to be equal in area to the given circle?

14.25 X 8862= 12.62835 inches, length of side required.

Any chord and versed sine of a circle being given, to find the diameter.

Rule. Divide the sum of the squares of the chord and versed sine by the versed sine, the quotient is A the diameter of corresponding circle. 1. The chord of a circle A B equal 6 feet, and the versed sine CD equal 2 feet, required the circle's diameter.

D

6.52+22=46.252=23·125 feet, the diameter.

2. In a curve of a railway, I stretched a line 72 feet in length, and the distance from the line to the curve 1 found to be 14 ft.; required the radius of the curve.

722+1·2525185.5625, and

5185-5625 1.25 X 2

= 2074-225 ft., the radius.

To find the length of any given arc of a circle.

Rule. From eight times the chord of half the arc subtract the chord of the whole arc, and one third of the remainder is equal the length of the arc.

Required the length of the arc ABC, the chord A B of half the

arc being 4 feet 3 inches,

A

B

and chord A C of the whole arc 8 feet 4 inches.

To find the area of the sector of a circle.

Rule. Multiply the length of the arc by its radius,

and half the product is the area.

The length of the arc A CB, equal 93 feet, and the radii F A, FB, equal each 7 feet, required the

area.

A

e

B

9.5X765.5-232-75, the area. Note.-The most simple means whereby to find the area of the segment of a circle is, to first find the area of a sector whose arc is equal to that of the given segment; and if it be less than a semicircle, subtract the area of the triangle formed by the chord of the segment and radii of its extremities; but if more than a semicircle, add the area of the triangle to the area of the sector, and the remainder, or sum, is the area of the segment.

Thus, suppose the area of the segment A C B e is required, and that the length of the arc A CB equal 9 feet, FA and FB each equal 7 feet, and the chord A B equal 8 feet 4 inches, also the perpendicular e F equal 3 feet.

9.75 X 7

2

=34-125 feet, the area of the sector.

8.333 x 3.75

15.624 feet, area of the triangle.

And 34-125-15-624 18-501 feet, the area of the segment.

To find the area of the space contained between two concentric circles.

Rule.-Multiply the sum of the inside and outside diameters by their difference, and by 7854, the product

is the area.

1. Suppose the external circle A B equal 32 inches, and internal circle C D equal 28 inches; required the area of the space con- A tained between them.

32+2860, and 32-28-4, hence 60 x 4 x 7854-188-496 in., the area.

D

2. The exterior diameter of the fly-wheel of a steam engine is 20 feet, and the interior diameter 18 feet; required the area of the surface or rim of the wheel.

2018-538-5 and 20-18.5=1.5, hence 38.5 x 1.5 × 7854 45.35, &c., feet, the area.

To find the area of an ellipsis or oval.

Rule.

Multiply the longest diameter by the shortest, and the product by 7854; the result is the area. An oval is 25 inches by 16-5; what are its superficial contents?

25 x 16.5=412.5 × 7854323-9775 inches, the area.

Note.-Multiply half the sum of the two diameters by 3.1416, and the product is the circumference of the oval or ellipsis.

To find the area of a parabola, or its segment.

Rule. Multiply the base by the perpendicular height, and two thirds of the product is the area. What is the area of a parabola whose base is 20 feet and height 12?

Note. Although the whole of the preceding practical applications or examples are given in measures of feet or inches, these being considered as the most generally familiar, yet the rules are equally applicable to any other unit of measurement whatever, as yards. chains, acres, &c. &c. &c.