Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

3d. OF A RHOMBUS. A rhombus is a figure representing a quarry, of glass, having four equal sides, the opposites thereof being equal, two angles being obtuse, and two acute. To find the superficial content thereof, this is the

RULE. Multiply one of the sides by a perpendicular let fall from one of the obtuse angles, to the opposite side, and the product is the content.

Let each side of a rhombus be 15,5 feet, and the perpendicular 13,42 feet; required the superficial content?

Ans. 208,01 feet.

4th. OF A RHOMBOIDES. A rhomboides is a figure having four sides, the opposite whereof is equal, and parallel; also four angles, the opposite whereof is equal. To find the superficial content thereof,

Rule. Multiply one of the longest sides thereof by a perpendicular let fall from one of the obtuse angles, to one of the longest sides, and the product is the content.

Let the longest sides of a rhomboides be 19,5 feet, and the perpendicular let fall be 10,2 feet; required the superficial content?

Ans. 198,9 feet.

5th. OF A TRIANGLE, A triangle is a figure having three sides and three angles. Triangles are either right-angled, or oblique-angled. Rightangled triangles are such as have one right angle; obliqueangled triangles are such as have their angles either acute or obtuse. An obtuse angle is greater than a right angle, that is, it is more than 90 degrees; and an acute angle is less than a right angle. To find the superficial content thereof, this is the

Rule. Let the triangle be of what kind soever, multiply the base by half the perpendicular, or half the base by the whole perpendicular; or multiply the whole base by the whole perpendicular, and take half the product. Any of these ways will give the content.

Let the base of a right-angled triangle be 14,1 feet, and the perpendicular 12 feet; required the content?

Ans. 84,6 feet. Let the base of an oblique-angled triangle be 15,4 feet, and the perpendicular 7,8 feety what is the superficial content?

Ans. 60,06 feet.

To find the area of any plain triangle, by having the three

sides given, without the help of a perpendicular. RULE. Add the three sides together, and take half that sum; then subtract each side severally from that half sum, which done, multiply that half sum and the three differences continually, and out of the last product extract the square root, which square root shall be the area of the triangle sought.

Let a triangle be given, whose three sides are as follows, viz. 43,3 20,5 and 31,2 feet; required the area ?

Ans. 296,31 feet.

6th. OF A TRAPEZIUM.

A trapezium is a figure having four unequal sides, and oblique angles. To find the area, or superficial content thereof, this is the

RULE. Add the two perpendiculars together, and take half the sum, and multiply that half sum by the diagonal; or multiply the whole sum by half the diagonal, and the product is the area. Or you may find the areas of the two triangles, add these two areas together, and the sum will be the area of the trapezium.

Let a trapezium be given, the diagonal whereof is 80,5 and the perpendiculars severally 30,1 and 24,5 feet; required the area !

Ans. 2197,65 feet.

7th. OF IRREGULAR FIGURE Irregular figures are all such as have more sides than four, and the sides and angles unequal. All such figures may be divided into as many triangles as there are sides, wanting two. To find the area of such figures, they must be divided into trapeziums and triangles, by lines drawn from one angle to another, and to find the areas of the trapeziums and triangles severally, and then add all the areas together, and you will have the area of the whole figure.

NOTE. This figure being composed of triangles and trapeziums, and these being shewn in the 5th and 6th problems aforegoing, it will be needless to mention any thing more in this place.

8th. OF REGULAR POLYGONS. Regular polygons are all such figures as have more than four sides, all the sides and angles thereof being equal.

P

Polygons are denominated from the number of their sides and angles.. To find the area, or superficial content of any regular polygon, this is the

Rule. Multiply the whole perimeter, or sum of the sides, by half the perpendicular let fall from the centre to the middle of one of the sides; or multiply the half perimeter by the whole perpendicular, and the product is the area.

If the figure consists of
5 equal sides and angles, it is called a regular Pentagon.
6 equal sides and angles, it is called a regular Hexagon.
7 equal sides and angles, it is called a regular Heplagon.
8 equal sides and angles, it is called a regular Octagon.

equal sides and angles, it is called a regular Enneagon.
10 equal sides and angles, it is called a regular Decagon.
11 equal sides and angles, it is called a regular Endecagon.

12 equal sides and angles, it is called a regular Dodecagon. Let a regular hexagon be given, whose sides are severally 14,6 feet, and the sum of all the sides 87,6 feet, and the perpendicular 12,64 feet: required the area, or superficial content?

Ans. 553,632 feet.

A Table for the more ready finding the area of a polygon.

RULE. Multiply the square of the side by the tabular number, and the product is the area of the polygon,

[blocks in formation]

How to find these tabular numbers. These numbers are found by trigonometry, thus: Find the angle at the centre of the polygon, by dividing 360 degrees by the number of sides of the polygon.

Let each side of a dodecagon be 1; required the area thereof?

Ang. 9,380125.

9th. OF A CIRCLE. A circle is a plain figure, contained under one line, which is called a circumference, unto which all lines, drawn from

a point in the middle of the figure, called the centre, and falling upon the circumference thereof, are all equal the one to the other. The circle contains more space than any plain figure of equal compass.

Demonstration.—Every circle may be conceived to be a polygon of an infinite number of sides; and the semidiameter must be equal to the perpendicular of such a polygon, and the circamference of the circle equal to the periphery of the polygon : therefore half the circumference, multiplied by half the diameter, gires the area.

F. Ignat and G. Pardie says "Every circle is equal to a rectangle triangle, one of whose legs is the radius, and the other a right line, equal to the circumference of the circle; for such a triangle will be greater than any polygon inscribed, and less than any polygon circumscribed, by the 24th, 25th, 26th and 27th articles of the fourth book of his Elements of Geometry, and therefore must be equal to the circle; for should it be greater than the circle, be the excess as little as it will, a polygon may be circumscribed, whose difference from the circle shall be yet less than the difference between that circle and the rectangle triangle; and that that polygou will be less than the triangle, is absurd. If it be said that this rectangled triangle is less than the circle, an inscribed polygon may be made, which shall be greater than that triangle, which is impossible.

"l'his cannot but be admitted as a principle--that if two determinate quantities are such, that if every imaginable quantity, which is greater or less than one, is also greater or less than the other, these two quantities must be eqnal. And this principle being granted, which is in a manner self-evident, it may directly be proved, that the triangle before mentioned is equal to the circle, because every imaginable inscribed figure, which is less than the circle, is also less than the triangle; and every circumscribed figure, greater than the circle, is also greater than the triangle.

" The proportion of the diameter of a circle to the circumference was pever yet exactly found out, notwithstanding many eminent learned men have laboured very far therein, among whom the excellent Van Culen hath hitherto outdone all, in his having calculated the said proportion to 38 places of decimals, which are engraven on his tombstone in St. Peter's charch, in Leyden, which numbers are these :

Diameter, 1,00000,00000,00000,00000,00000,00000,00000.

Circumference, 3,14159,26535,89793,23846,26433,83279,50288. “Of which large number, these six places, 3,14159, answering to the di. ameter, 1,00000, may be sufficient.”

I will give many proportions, but for the most part that of Van Culen, as being most exact

If the diameter of any circle

product be by

is the circumference, be sx ,8862272 the product is the side of an equal

square. be Х ,707016 2 the product is the side of the square If the square of the diameter of

inscribed.

any

circle

be

If the circumference of any circle

be

De

{}by{1,295241} the quotient}

785598 ? the prodiert} is the area. {:}{

$5:3183! } the quodient } is the diameter. {*}{

vy ;quotient} {5} by{ } }

{ 1,472877} the quotient} spate the square by {} }

If the circumference of any circle
,282094 2 the product is the side of the square

equal
If the circumference of any circle

,2756646 2 the product ? is the side of the equi

be

<

be

be

If the circumference of any circle
x
} by

4,442877 ) the quotient) inscribed.'
If the square of the circumference of any circle

,079577525 3 the product is the area. by

If the area of any circle
Х

1,273241 2 the product is the square of the
,7

diameter. If the area of any

circle х

5 12,56636217 2 the product is the square of the

,079577525 S the quotient circumference.

be

be

:

When tbe diameter of a circle is 1, and the diameter of any other is 2, the circumference of the first circle is equal to the area of the second, 3,141592. If the circumference be 4, the diameter and area equal 1,273241. If the diameter be 4, the circumference and area equal 12,566358. If the diameter of one circle be double to that of another, the area of the first circle will be 4 times the area of the second. If the circumference of a circle be 1, the area will be ,079577. If the area be required, when the circumference is 1, first find what the diameter will be; thus,

cir. diam. cir. As 3,1416 : 1 :: 1 ,318309 which is the diameter, when the circumference is 1. Then multiply half ,318309 by half 1, that is, ,15915% by ,5 and the product is ,079577 which is the area of a circle, whose cir. cumference is 1.

If the area of any circle be given to find the side of a square equal, you peed but extract the square root of the area given, and it is done. So the square root of ,7854 is , 88,62 which is the side of a square equal, when the diameter is 1. And if you extract the square root of ,079577 it will be ,2821 which is the side of the square, equal to the circle whose circumference is 1.

Again, if the side of a equare within a circle be required, if you square the semidiameter, and double that square, and out of that sum extract the square

« ΠροηγούμενηΣυνέχεια »