Problem 90.

When the length of the arc is given.

Rule 2. Multiply the radius of the circle by half the length of the arc of the sector, and the product will be the area.

Example 2.

BO 2.50 × A B C 60°

15000 × .0174524 = 2.61786 length of arc (See Prob. 66) 2.61786

= 1.30893 half the arc

Then 1.30893 × 2.50 =

Another method:

3.2723 the area required

Rule. Multiply the diameter of the circle by the length of the arc of the sector; and one-fourth of the product will be the

area.

Ex. 5.00 x 2.61786 3.2723 the area
Problem 91.

To find the area of a segment of a circle, Fig. 74, Plate 5. Rule 1. By the help of the Table of circular segments, No. 15, divide the versed sine or height C D by the diameter; with the quotient thence arising from the table take out the segment area opposite the versed sine; multiply this segment area by the square of the diameter of the circle, the product will be the area of the segment.

Example 1. Given the versed sine 7, and diameter 35, to find the area.

"735.2 the segment area = .11182 × 352 = 136.989 the area

Rule 2. Multiply the versed sine CD by the decimal .626; and to the square of the product add the square of half the chord A B; multiply twice the square root of the same by twothirds of the versed sine, the product will be the area.

Example 2. Required the area of a segment whose chord A B = 52.5, and versed sine C D = 11.5.

11.5 × 626 = 7.19902 = 740.89810100

Then ✓740.89810100 = 27.2194 × 2 = 54.4388 × ver. sine 7.7 = 419.168 the area

Or the area of a segment may be found by first finding the area of the sector, having the same radius as the segment, and then deducting the area of the triangle, leaves the area of the segment.

Problem 92.

To find the area of a circular ring or space included between two concentric circles, Fig. 75, Plate 5.

Rule. Add the inside and outside diameters together, multiply the product by their difference and by .7854; the result will be the area.

Example. The diameters of two concentric circles A B = 25 and CD=15; required the area of the ring or space contained between them.

25+15= 40 × 10 400 × .7854 314.1600 the area

Problem 93.

To find the area of an ellipse, Fig. 57, Plate 3.

Rule. Multiply the transverse diameter by the conjugate diameter, and then by .7854; the product will be the area.

Or multiply the two diameters together, divide that product by 4, which multiply by 3.1416 will be the area.

Required the area of an ellipse, the transverse diameter A B = 680, and the conjugate diameter C D = 420.

To reduce a trapezium to a plane triangle, Fig. 76, Plate 5. Draw the diagonal A C, parallel to it draw the line D E, draw the line A E, then will A B E be the triangle required.

Examples for Practice.

Required the area of the triangle A B C, also of the trapezium A B C D; by the rule, Problems 79, 80.

F

Problem 95.

To reduce a parallelogram to a square of equal area.

Fig. 77. Rule. Multiply the length by the breadth, and the square root of the product will be the side of the square equal to the parallelogram.

Note. The square root of the product of any two numbers is the geometrical mean proportion of those numbers.

To construct the same geometrically.

Continue the line A B to E equal to B C; bisect A E at F, with the radius A F or FE describe the semicircle A G E, erect a perpendicular from B intersecting the semicircle at G, then will B G be the side of the square B G I H equal to the parallelogram A B C D.

Example. What is the side of a square whose area shall be equal to the parallelogram, the length A B=440, the breadth BC=140?

440 × 140 = √61600 the area = 248.19 the side of the square

For Practice.

The sides of a parallelogram are 600 and 300, what is the side of a square of equal area?

Problem 96.

To reduce a figure of five sides to a triangle, Fig. 78, Plate 5. Draw the line B D, and parallel to it draw the line C F, from F draw the line B F, which forms one side of the triangle; in like manner draw B E, and parallel to it draw the line A G, from G draw the line G B the other side of the triangle; then will G B F be the triangle required.

Let these be cast by the rule as before, by the scale, Fig. 1, Plate 39; if the quantity of the triangle G B F does not

agree with the quantities of A B D E and B C D taken together, repeat the operation.

Examples for Practice.

Required the area of Figs. 71 and 72, Plate 4.

Both these figures are similar; the dotted lines show the triangles and trapeziums each figure is divided into for calculation as follows:

Fig. 71. Two triangles and one trapezium.
Fig. 72. Two trapeziums and one triangle.
Note.-If correct, both quantities should agree within a few decimals.

Problem 97.

Fig. 84, Plate 6, represents all the reduced lines of fences to Plan, Fig. 85, Plate 6.

Calculate these four pieces by the scale as before; then calculate Fig. 85, Plate 6, by the improved computation scale, as described in Part IV.

Note.-If there be any trifling difference in the decimals in the different modes of calculating them, take the mean between the two for the actual quantity.

Problem 98.

Examples for reducing crooked fences to straight lines by the parallel rule, Fig. 79, Plate 5.

As before stated, all crooked fences have to be reduced to straight lines to form the sides of the regular figures for computing the quantities as shown by the last example.

The following system is decidedly a most perfect one, although tedious, and was, until lately, generally adopted.

The irregular line from A to B represents the side of a field, which has to be reduced to a straight line that shall equalise the quantities on both sides, as shown by the shaded part on one side the line, and plain on the other side of the line A C.

First draw the line A B, and perpendicular to it the line BC; then lay the edge of the parallel rule at the points B and 2 on the fence, move the upper part of the rule to 1 on the

fence, with a fine needle fix it on the line B C as marked 1; holding the needle firm at that point, bring the edge of the parallel rule to the needle and to the point 3 on the fence, then move the upper part of the rule to the point 2 on the fence, and the needle on the line BC at the mark 2; holding the needle firm at that point, place the edge of the rule against the needle and point 4 on the fence line, move the rule to 3 on the fence line and fix the needle at 3 on B C; bring the rule to the needle and at 5 on the fence line, move the rule to 4 on the fence and place the needle at 4 on B C; bring the rule to 6 on the fence and against the needle, move the rule to 5 and fix the needle at 5 on B C; bring the rule to that point and 6 on the fence, move the rule to 5 on the fence, and the needle to C on the line BC; draw the line A C, it will be the equalised line of the fence from A to B, forming one side of a triangle for calculation.

Note. This method may be applied to equalise a fence between two fields, giving to each an equal quantity.

Another method is with a piece of transparent horn perfectly straight on the edges. Place the edge of the horn over the irregular fence, shifting the horn until by the eye there appears to be an equal portion on both sides of the edge of the horn; from the point A draw the line A C; then ABC will be a triangle, and the line A B one side of a trapezium.

Proceed in like manner for practice with Figs. 80 and 81, Plate 5, as shown by the dotted lines; the quantities are marked in each figure, and collected together for the total quantity as shown.

To find the area of a field, Fig. 82, Plate 6, by the computing scales, Fig. 1, Plate 39 (which may be obtained at any of the mathematical instrument-makers).

A full description of these scales is given in Part IV.

The parallel lines drawn over the plot are supposed to be lines drawn on the tracing paper to the same scale as the plan, and laid over the piece to be calculated, Fig. 83, Plate 6.