their sum by the perpendicular breadth or distance between them; and half the product will be the area, by Geometry, theorem 25.

Ex. 1. In a trapezoid the parallel sides are 750 and 1225, and the perpendicular distance between them 1540 links: to find the area.

1225 750

1975 × 770=152075 sq. links = 15 acres, 33 perches. Ex. 2. How many square feet are contained in the plank whose length is 12 feet 6 inches, the breadth at the greater end 15 inches, and at the less end 11 inches? Ans. 13 feet.

Ex. 3. In measuring along one side AB of a quadrangular field, that side and the two perpendiculars let fall on it from the two opposite corners, measured as below required the content.

D

Ans. 4 acres, 1 rood, 5·792 perches.

PROBLEM IV.

To find the area of any trapezium.

Divide the trapezium into two triangles by a diagonal: then find the areas of these triangles, and add them together.

Note. If two perpendiculars be let fall on the diagonal, from the other two opposite angles, the sum of these perpendiculars being multiplied by the diagonal, half the product will be the area of the trapezium.

Ex. 1. To find the area of the trapezium whose diagonal is 42, and the two perpendiculars on it 16. and 18.

Here

Then

16 +18= 34; its half is 17.

42 × 17 = 714, the area.

Ex. 2. How many square yards of paving are in

the trapezium whose diagonal is 65 feet, and the two perpendiculars let fall on it 28 and 331 feet? Ans. 222 yards.

Ex. 3. In the quadrangular field ABCD, on account of obstructions, there could only be taken the following measures, viz.: the two sides BC 265, and AD 220 yards, the diagonal AC 378, and the two distances of the perpendiculars from the ends of the diagonal, namely, AE 100, and CF 70 yards. Required the area in acres when 4840 square yards make an acre. Ans. 17 acres, 2 roods, 21 perches.

PROBLEM V.

To find the area of an irregular polygon. Draw diagonals dividing the proposed polygon into trapeziums and triangles. Then find the areas of all these separately, and add them together for the content of the whole polygon.

Ex. To find the content of the irregular figure ABCDEFGA, in which are given the following diagonals and perpendiculars,

namely:

AC 55

B

To find the area of a regular polygon.

RULE I. Multiply the perimeter of the polygon, or sum of its sides, by the perpendicular drawn from its center on one of its sides, and take half the product for the area.*

* The demonstration of this is given in th. 73.

Ex. 1. To find the area of the regular pentagon, each side being 25 feet, and the perpendicular from the center on each side 17·2047737.

Here
And

25 x 5125, is the perimeter. 17:2047737 × 125=2150.5967125.

Its half, 1075-298356, is the area sought.

RULE II. Square the side of the polygon; then multiply that square by the area or multiplier set against its name in the following table, and the product will be the area.*

Ex. Taking here the same example as before, namely, a pentagon, whose side is 25 feet. Then, 25' being = 625,

And the tabular area 1.7204774;

Therefore, 1.7204774 × 625 = 1075·298375, as before.

*This rule is founded on the property that regular polygons of the same number of sides, being similar figures, are as the squares of their sides. Now the multipliers in the table are the areas of the respective polygons to the side 1. Whence the rule is manifest.

Note. The areas in the table, to each side 1, may be computed in the following manner, with the aid of plane trigonometry: From the center C of the polygon draw lines to every angle, dividing the whole figure into as many equal triangles as the polygon has sides; and let ABC be one of those triangles, the perpendicular of which is CD. Divide 360 degrees by the number of sides in the polygon, the quotient gives the angle at the center ACB. The half of this gives the angle ACD; and this taken from 90°, leaves the angle CAD. Then, as radius is to AD, so is tangent angle CAD to the perpendicular CD. This, multiplied by AD, gives the area of the triangle ABC; which, being multiplied by the number of the triangles, or of the sides of the polygon, gives its whole area, as in the table.

A D B

Ex. 2. To find the area of the trigon, or equilateral triangle, whose side is 20. Ans. 173.20508. Ex. 3. To find the area of a hexagon whose side is 20. Ans. 1039-23048. Ex. 4. To find the area of an octagon whose side Ans. 1931-37084. Ex. 5. To find the area of a decagon whose side Ans. 3077-68352.

is 20.

is 20.

PROBLEM VII.

To find the diameter and circumference of any circle, the one from the other.

This may be done nearly by either of the two following proportions, viz. :

As 7 is to 22, so is the diameter to the circumfer

ence.

Or, as 1 is to 3.1416, so is the diameter to the circumference.*

* For, let ABCD be any circle whose center is E, and let AB, BC be any two equal arcs. Draw the several chords as in the figure, and join BE; also, draw the diameter DA, which produce to F,* till BF be equal to the chord BD.

E

F

B

Then the two isosceles triangles DEB, DBF are equiangular, because they have the angle at D common; consequently, DE: DB::DB: DF. But the two triangles AFB, DCB are identical, or equal in all respects, because they have the angle F= the angle BDC, being each equal the angle ADB (see th. 39, cor. 1); also, the exterior angle FAB of the quadrangle ABCD equal the opposite interior angle at C (exercise 32, p. 48); and the two triangles have, also, the side BF = the side BD; therefore, the side AF is also equal the side DC. Hence the proportion above, viz., DE: DB::DB: DF =DA+AF, becomes DE: DB::DB: 2DE + DC. Then, by taking the rectangles of the extremes and means, it is DB2 = 2DE2 + DE.

DC.

Now if the radius DE be taken 1, this expression becomes DB2 =2+ DC; and hence DB=√2+ DC. That is, if the measure of the supplementalt chord of any arc be increased by the number 2, the square root of the sum will be the supplemental chord of half that

arc.

Now, to apply this to the calculation of the circumference of the circle, let the arc AC be taken equal to one sixth of the circumference, *The point F may be found by describing an arc with B as center, and radius. =BD. The supplemental chord is the chord of the supplement.

Ex. 1. To find the circumference of the circle whose diameter is 20.

and be successively bisected by the above theorem: thus, the chord AC of one sixth of the circumference is the side of the inscribed regular hexagon, and is, therefore, equal the radius AE or 1; hence, in the right-angled triangle ACD, we shall have DC=√AD2 — AC2 = √22 — 1o = √/ 3 = 1·7320508076, the supplemental chord of one sixth of the periphery.

Then, by the foregoing theorem, by always bisecting the arcs, and adding 2 to the last square root, there will be found the supplemental chords of the 12th, the 24th, the 48th, the 96th, &c., parts of the periphery; thus,

=

9 P

Since, then, it is found that 3.9999832669 is the square of the supplemental chord of the 1536th part of the periphery, let this number be taken from 4, the square of the diameter, and the remainder 0.0000167331 will be the square of the chord of the said 1536th part of the periphery, and, consequently, the root 0.0000167331 0-0040906112 is the length of that chord; this number, then, being multiplied by 1536, gives 6.2831788 for the perimeter of a regular polygon of 1536 sides inscribed in the circle; which, as the sides of the polygon nearly coincide with the circumference of the circle, must also express the length of the circumference itself, very nearly. But now, to show how near this determination is to the SRT truth, let AQP = 0.0040906112 represent one side of such a regular polygon of 1536 sides, and SRT a side of another A similar polygon described about the circle; and from the center E let the perpendicular EQR be drawn, bisecting AP and ST in Q and R. Then, since AQ is = {AP = 0-0020453056, and EA=1, therefore EQ2 = EA2 — AQ2 =9999958167, and, consequently, its root gives EQ= -9999979084; then, because of the parallels AP, ST, we have the proportion EQ: ER::AP:ST:: the whole inscribed perimeter: the circumscribed one; that is, as -9999979084:1::6-2831788: 6-2831920, the perimeter of the circumscribed polygon. But the circumference of the circle being greater than the perimeter of the inner polygon, and less than that of the outer, it must, consequently, be greater than 6.2831788, but less than 6.2831920,

E

and must, therefore, be nearly equal half their sum or a mean between them, or 6-2831854, which, in fact, is true to the last figure, which should be a 3 instead of the 4.