Multiply the transverse by the conjugate diameter, and the product by 7854 for the area.

Let ABCD be an ellipsis, whereof the transverse diameter AB is 45 inches, and the conjugate diameter CD, 32 inches; the area is required.

If the square of the mean proportional between the diameters be multiplied by the constant decimal ·7854, the product will in like manner be the area; for every ellipsis is a geometrical mean between its inscribed and circumscribing circles.

BY THE SLIDING RULE.

Set 45 on C, to the same number on D, and against 32 on C, will be found 38 on D, for the mean proportional between the diameters.

Then set 7854 on C, to unity on D, and you will have, for the area, 1130-98 on C, opposite to 38 on D.

PROBLEM XI.

To find the Area of a regular Polygon.

RULE.

Multiply the square of the side by the number corresponding to the figure in the following Table.

TABLE OF REGULAR POLYGONS.

EXAMPLE.

Let ABCDEF be a hexagon, whereof the side is 30 inches; the area is required.

900 square of the side

2.598 tabular number for hexagons

2338.2 square inches, area required.

BY THE SLIDING RULE.

Set the tabular number 2.598 on C, to 1 on D, and opposite to 30 on D, look for the area on C, which you will find to be 2338-2 square inches, as before.

It is evident that the area of any polygon may be found by dividing the figure into triangles, and then letting fall perpendiculars, as already directed for right-lined multilateral plane figures.

PROBLEM XII.

To find the Area of a Parabola.

RULE.

Multiply the base or ordinate by the perpendicular

or axis, and two-thirds of the product will be the area : for every parabola is equal to two-thirds of its circumscribing rectangle, and on this property is the Rule founded.

Let ABCD be a parabola, whereof the base BD measures 40.5 inches, and the axis AC, 34 inches; the area is required.

Set 1 on B to 27 (two-thirds of the base) on A, and opposite to 34 (the altitude) on B, may be seen 918 for the area on A.

PROBLEM XIII.

To find the Area of a Sector.

RULE.

Multiply the radius by half the length of the arc, and the product will be the area.

Let ABDC be a sector, whereof the radius AB measures 17 inches, and the arc BDC, 26 inches; the area is required.

B

SOLUTION.

17 radius

13 half the arc

51 17

221 square inches, area of the sector.

When the chord of the arc, and chord of half the arc are given, the length of the arc may be found nearly, though not exactly, as follows:

Multiply the chord of half the arc by 8, and from the product subtract the chord of the whole arc; then divide the remainder by 3, and the quotient will be the arc nearly.