rhombus contain, the base of which is 1490, and the dicular breadth 1280 links? Problem 2. To find the area of a Triangle. RULE. Multiply the base by the altitude, and half the product will be the area. (1) Required the number of square yards in a triangle, whose base is 49 feet, and height 25 feet. (2) What is the area of the gable of a house, the base or distance between the eaves being 22 feet 5 inches, and the perpendicular from the ridge to the middle of the base, 9 feet 4 inches? A Fig. 4. C perpen D B RULE 2. When the three sides only are given.-From half the sum of the sides subtract each side severally: multiply the half sum and the three remainders continually together; and the square root of their product will be the area. (3) The three sides of a triangular fish-pond, are 140, 336, and 415 yards respectively. What is the value of the land which it occupies, at £225. per acre? Problem 3. To find the area of a Trapezium, or a Trapezoid. RULE. Divide the trapezium into two triangles by a diagonal: multiply the diagonal by the sum of the two perpendiculars falling upon it; and half the product will be the area. That is, D A DE+ BF XAC 2 Fig. 5. F B the area. For a trapezoid. Multiply the sum of the two parallel sides by the perpendicular distance between them; and half the product will be the area. (1) How many square yards of paving are in the trapezium, whose diagonal is 65 feet, and the perpendiculars 28 and 33 feet? (2) Find the area of a trapezium whose south side is 2740 links, east side 3575, west side 4105, and north side 3755 links; and the diagonal from the south-west to the north-east angle 4835 links. (3) Required the area of a trapezoid whose parallel sides are 20 feet and 12 feet respectively; the perpendicular distance being 10 feet? (4) How many square feet are in a board, whose length is 12 feet, and the breadths of the two ends 15 inches and 11 inches respectively? Problem 4. To find the area of an Irregular Figure. RULE. Divide it, by drawing diagonals, into trapeziums and triangles. Find the area of each, and their sum will be the area of the whole. 1. Required the content of the irregular figure ABCDEFGA, in which are given the following diagonals and perpendiculars: namely, AC 5.5 FD = 5.2 Fig. 6. n C Problem 5. To find the area of a Regular Polygon. RULE 1. Multiply the perimeter (or sum of the sides) by the perpendicular drawn from the centre to one of the sides; and half the product will be the area. 1 RULE 2. Multiply the square of the side by the corresponding tabular area, or multiplier opposite to the name in the following table; and the product will be the area. (1) Required the area of a regular pentagon whose side is 25 feet. (2) Required the area of an octagon whose side is 20 feet. Problem 6. To find the Diameter or Circumference of a Circle the one from the other.* or, as 113 : 355+ or, as 1 : 3.1416§ diameter: the circumference; and reversing the terms will find the diameter. (1) Required the circumference of a circle whose diameter is 12.¶ To find the proportion which the circumference bears to the diameter, and thence to find the area of a circle, is a problem that has engaged the anxious attention of mathematicians of all ages. It is now generally considered impossible to determine it exactly; but various approximations have been found, some of which have been carried to so great a degree of accuracy, that in a circle as immense in magnitude as the orbit of the planet Saturn, the diameter of which is about 158 millions of miles, we are enabled to express the circumference (the diameter being given) so nearly approximating to the truth, as not to deviate from it so much as the breadth of a single hair. The three approximations in the Rule are those in general use. This is the ratio assigned by Archimedes, a celebrated philosopher of Syracuse, who flourished about two centuries before the Christian era. It answers the purpose sufficiently well when particular accuracy is not required. This was discovered by Metius, a Dutchman, about two centuries since. It is a very good approximation, agreeing with the truth to the sixth figure. § This is an abridgment of the celebrated Van Ceulen's proportion, who was nearly contemporary with Metius. By a patient and most laborious investigation, he determined it truly to 36 places of figures. (3.141598, &c.) But it has been since extended to considerably more than 100 places.-This proportion is extremely convenient, from the circumstance of the first term being unity; which saves the labour of division, in finding the circumference of any other circle whose diameter is given. It is not quite so accurate as the preceding, or, as 1 : 31416: 12: 3-1416 X 1237.6992 H (2) What is the circumference when the diameter is 45? (3) What is the diameter of a column whose circumference is 9 feet 6 inches? (4) If the circumference of a great circle of the earth (as the equator) were exactly 25000 miles, what would be the diameter ? Problem 7. To find the area of a Circle. RULE 1. The area is equal to a fourth part of the product of the circumference into the diameter; or, the product of half the circumference into half the diameter. Therefore, when the diameter is 1, the area = ='7854; whence we have 1 x 3.1416 4 RULE 2. Multiply the square of the diameter by 7854; and the product will be the area. RULE 3. Multiply the square of the circumference by 07958 for the area. (1) Required the area of the circle proposed in Example 1, Problem 6.* (2) Find the area of the circle proposed in Example 2, Problem 6. (3) What is the area of the end of a rolier whose diameter is 2 feet 3 inches? (4) Required the area when the circumference is 81 feet. Problem 8, To find the side of a square inscribed in a circle. RULE. Multiply the radius by 1-4142 (that is by √2) or, multiply the diameter by 7071.† (1) Find the side of the square inscribed in the circle whose diameter is 12. (2) What is the side of the square inscribed in a circle whose diameter is 6 feet 5 inches ? 37.6992 X 12 -= 37·6992 × 3 = 113·0976 the area. or, 122 X 7854 = 12 X 12 X 7854 113.0976 = + The following Rules exhibit the principal relations between the circle and its equal square, or inscribed square. 1. The diameter X.8862269 2. The circumfer. X.2820948 3. The diameter X.7071068 the side of an equal square. 4. The circumfer. X2250791 the side of the inscribed square. 5. The area X.6866197 Problem 9. To find the length of a circular arc NOTE. Half the chord of the whole arc, the chord of half the arc, and the versed sine, are sides of a right angled triangle; any two of which being given, the third may be found as directed in page 117. RULE 2. Multiply the number of degrees in the arc by the radius, and the product by 01745, for the length of the arc. (1) The chord of the whole arc is 30, and the versed sine 8: what is the length of the arc ? (2) What is the length of the arc when the chord of the half arc is 10-625, and its versed sine 5? (3) Required the length of an arc of 12° 10′, the radius being 10 feet. Problem 10. To find the area of a Sector of a circle. RULE 1. Multiply the length of the arc by the radius, and half the product will be the area. RULE 2. As 360°: the degrees in the arc :: the area of C the circle the area of the Fig. 9, Α Sector. B (1) Required the area of the sector, when the radius is 15, and the chord of the whole arc 18 feet. (2) What is the area of a sector whose arc is 147° 29′, and the radius 25? (3) Required the area of a sector whose radius is 20 feet, and the versed sine 1 foot 9 inches.* of its circumscribing circle. of an equal cir 6. The side of a square X1·414214 the diameter 7. The side of a square X4·442883=the circumf. 8. The side of a square X1·128379=the diameter 9. The side of a square X3.544908 the circumf. cle. * By the properties of the circle, the versed sine X the remaining part of the diameter the square of half the chord of the arc; whence all the requisites may be found. |