=

And 24 half the perimeter, therefore 24 X 688191 1·720477 the area of the pentagon.

The area 1.720477 will serve as a multiplier for finding the content of any other regular pentagon whose side is given; Thus,

2. Suppose it is required to find the content of a pentagon whose side is 20: Then, similar plane figures being in the same proportion as the squares of their homologous sides (102), we have

12: 1.720477 :: 20: 400 x 1.720477 or 688.19 the area required.

3. If the side of a regular hexagon be 1, what is the area?

The hexagon is composed of 6 equilateral triangles, each side being 1.

Now 1 × × is the square of the area of one of the triangles (260); and the square root of is 433013 nearly, therefore 133013 × 6 = 2.598078, the content of the hexagon. And this area will be a multiplier for finding the content of any other regular hexagon whose side is given.

265. To find the area of an irregular Polygon.

DIVIDE the polygon into triangles, or into triangles and trapezoids; then their areas added together will be the content of the polygon.

In land-measuring, an instrument called the Cross-staff will be very useful for finding the points O, P, R, &c. where the perpendiculars IO, BP, HR, &c. fall from the corners of the Field upon the base line AE. Or the same thing may be done with a pocket Sextant, thus: Set the index to 90°, and as you walk along the line AE (if towards E) direct the sight to an object at E, then suppose you see the corner B (for example) by reflection when you are at P, the angle BPE will be a right one.

What is the superficial content of the section?

The perpendiculars divide the figure into 2 triangles, 4 trapezoids, and a rectangle; and their areas added together make 698 feet, the content required.

3. Let ABC be the profile or perpendicular section of a breast-work, and EP that of the ditch; now suppose the area of

the section ABC is 88 feet, the depth of the ditch RD = 6 feet, and ER3 feet; what is the breadth of the ditch at top when the sections of the ditch and breast-work are equal, or when the earth thrown out of the ditch is supposed to make the breast-work?

If the slope on each side of the ditch is the same, the areas of the triangles ERD, SPO together make 18 feet, which taken from 88 leaves 70, the area of the rectangle RP; this divided by the depth RD or SP gives RS, therefore EO the breadth of the ditch at top is 113+6 =

113 173fect.

4. Let the section of the breast-work ABC be as in the preceding example, and EO the breadth of the ditch at top = 20 feet; also suppose the slopes of the ditch are unequal according to the following proportions, ER: RD :: 2 : 3, and SO: SH :: 2 : 4; RD and SH being perpendicular to EO : Now what must be the depth of the ditch, if the earth when thrown out is also to form a glacis whose height is 3 feet, and base OG 14?

Then the triangles RDQ, SHO are similar, whence

:

HS SO: DR: RQ

or 4 : 2:: 3 : 14 RQ; therefore EQ2+1=35

And because the triangles EDQ, EVO are similar, we have

Therefore (262), EQ : RD :: triang. NVT: VK2

or 12 : 3 ::

623

107, and the square

10.3=

root is 103 nearly, VK; whence the depth LK = 174

What must be the breadth of the ditch so that its section ISOT shall be equal to the profile ABGC and TQV (the section of the glacis) together, when the talus BG and QV the exterior slope of the glacis are in the same plane; the slopes IS, TO being equal?

266.

Ans. IT breadth at top 26.67 feet.

Of the Circle.

we have

IF 00051132692 be multiplied by 12288 6.28318519296 the length of the perimeter of the inscribed polygon of 12288 sides when the radius of the circle is 1 (208). This perimeter must be very nearly equal to the circumference of the circle, but somewhat less. It may there. fore be worth while to calculate the perimeter of the circumscribing polygon of the same number of sides, because the

circumference of the circle must be greater than the one, but less than the other.

Let C be the centre of the circle, AB the side of the inscribed polygon, and GD that of the cor responding circumscribing one; and suppose CR is perpendicular to AB and GD.

R

G

Then OB is 0002556346 or half of 00051132692 : and CB being = 1, we get CO = 99999996728 nearly, (83, corol.).

And because the perimeters of similar plane figures are in the same proportion as their homologous sides (103) we have

CC CR perim. inscribed polyg. : perim. circums. polyg.

or 99999996728:1::628318519296: 6-2831854 nearly, the peri meter of the circumscribing polygon. We therefore conclude that the circumference of the circle

is less than 6.2831854

but greater than 6.2831852

consequently half their sum or 6-2831853 must be very nearly the cir cumference.

Now the diameter being 2, the circumference of a circle whose diameter is 1 will be half of 6.2831853, or 3.14159265; which is correct in the last decimal, and sufficiently near to give the circumference of the Earth true to less than 2 inches, supposing it globular and the diameter 8000 miles.

When much accuracy is not required, the proportion of the diameter to the circumference may be taken as 1 to 3 1416. Or that of 7 to 22 will serve for common purposes. The ratio of 113 to 355 is a nearer approximation than either.

267.

To find the area of a Circle.

1. MULTIPLY the radius or half the diameter by half the circumference, and the product will be the area (106, corol.).

II. Or the square of the diameter multiplied by 7854 gives the area.