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. perpen Let each side of a rhombus be 15,5 feet, and the dicular 18,42 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 feet, what is the superficial conAns. 60,06 feet. tent? 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 $1,2 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 FIGURES. Irregular figures are all such as have more sides than four, and the sides and angles unequal. Ali 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. 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, 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? Ans. 9,330125. 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 circumference of the circle equal to the periphery of the polygon: therefore half the circumference, multiplied by half the diameter, gives 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 polygon 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. "This 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 equal. 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 never 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 36 places of decimals, which are engraven on his tombstone in St. Peter's church, 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 diameter, 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 the product {*}by {,51831} the quotient be be ÷ be is the circumference. ,886227 the product is the side of an equal by { 1,128379} the quotient ,707016 the product {} by {1,414218} the quotient S square. is the side of the square inscribed. be be be {x} If the square of the diameter of any circle be be by1,273241 the quotient is the area. If the circumference of any circle by {5,14159} the product is the diameter. quotient by {3,544907} the quotient is the side of the square equal If the circumference of any circle is the side of {} by {3,6275939} the quotient } lateral trian. inscribed. If the circumference of any circle ,225079 the product is the side of the square S inscribed. '. If the square of the circumference of any circle. {X} by {4,442877} the quotient ,079577525 4 by12,56636217 the product is the area. be If the area of any circle 1,273241 the product is the square of the ,785398 the quotient If the area of any circle diameter. (12,56636217 the product 2 is the square of the {X} by {12,579597525} the quotient circumference. When the 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 equa! 1,273241. If the diameter be 4, the circumference and area equal 12,566368. 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,,159151 by,5 and the product is ,079577 which is the area of a circle, whose circumference is 1. If the area of any circle be given to find the side of a square equal, you need 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 square within a circle be required, if you square the semidiameter, and double that square, and out of that sum extract the square |