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. AP = 110 links. AQ = 745 AB 1110 CP = 352 DQ = 595 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. 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 334 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. quired the area in acres, when 4840 square yards make an acre? Re 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. EXAM.—To find the content of the irregular figure ABCDEFGA, in which are given the following diagonals and perpendiculars; namely, AC 55 FD 52 GC 44 Gm 13 Bn 18 Go 12 Ер 8 Ans. 1878.5. K K B m n F E PROBLEM VI. To find the area of a regular polygon. RULE 1.-Multiply the perimeter of the polygon, or sum of its sides, by the perpendicular drawn from its centre on one of its sides, and take half the product for the area.* Ex. 1.—To find the area of the regular pentagon, each side being 25 feet, and the perpendicular from the centre on each side is 17.2047737. Here 25 X 5 = 125 is the perimeter. And 17.2047737 × 125 = 2150-5967125. 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, t EXAM.—Taking here the same example as before, namely, a pentagon, whose side is 25 feet. Then, 252 being = 625, And the tabular area 1·7204774; Therefore, 17204774 X 625 = 1075 298375, as before. Ex. 2.—To find the area of the trigon, or equilateral triangle, whose side is 20, Ans. 173-20508. *This is only in effect resolving the polygon into as many equal triangles as it has sides, by drawing lines from the centre to all the angles; then finding their areas, and adding them all together. This rule is founded on the property, that like polygons, being similar figures, are to one another as the squares of their like sides; which is proved in the Geomety, Theorem 89. pliers in the table, are the areas of the respective polygons to the side 1. manifest. Now, the multiWhence the rule is Note. The areas in the table, to each side 1, may be computed in the following manner: From the centre 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 numver of sides in the polygon, the quotient gives the angle at the centre 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. 3. To find the area of a hexagon, whose side is 20 Ans. 1039-23048. PROB. 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. Or, As I is to 3.1416, so is the diameter to the circumference.* Ex. 1. To find the circumference of the circle whose diameter is 20. By the first rule, as 7 : 22 :: 20: 624, the answer. Ex. 2. If the circumference of the earth be 25000 miles, what is its diameter? By the 2d rule, as 3·1416: 1 :: 25000: 7957, nearly the diameter. For, let ABCD be any circle, whose centre 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. 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, these being subtended by the equal arcs AB, BC; also the exterior angle FAB of the quadrangie ABCD, is equal the opposite interior angle at C; 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: 2 DE+DC. Then, by taking the rectangles of the it is DB2 2 DE2+DE. DC. extremes and means, E D 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 supplemental 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 of the circumference, and be successively bisected by the above theorem: thus, the chord AC, of of the circumference, is the side of the inscribed regular hexagon, and is therefore equal the radius AĚ or 1 ; hence, in the right-angle triangle ACD, it will be DC=√AD2—AC2 = √/22—12= √/3=1.7320508076, the supplemental chord of 16 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, 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. PROBLEM VIII. To find the length of any arc of a circle. Multiply the degrees in the given arc by the radius of the circle, and the product again by the decimal 01745, for the length of the arc.* Ex. 1.—To find the length of an arc of 30 degrees, the radius being 9 feet. Ans. 4.7115. Ex. 2.-To find the length of an arc of 12° 10′, or 12°, the radius being, 10 Ans. 2.1231. feet. PROBLEM IX. To find the area of a circle. + RULE I. -Multiply half the circumference by half the diameter. Cr multiply the whole circumference by the whole diameter, and take of the product. RULE II.-Square the diameter, and multiply that square by the decimal 7854, for the area. Ex. 1.-To find the area of a circle whose diameter is 10, and its circumference 31415. By Rule 1. By Rule 2. 100 10 78.54 But now, to show how near this determination is to the truth, let AQP 0.0040906112 represent one side of such a regular polygon of 1536 sides, and SRT a side of another similar polygon described about the circle; and from the centre 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 EQ' EA2- AQ' 9099958167, and consequently its root gives EQ 9999979084; then, because of the parallels AP, ST, it is 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, and must therefore be nearly equal their sum, figure, which should be a 3 instead of the 4. or 6.2831854, which in fact is true to the last Hence, the circumference being 6'2831854 when the diameter is 2, it will be the half of that, or 3.1415927, when the diameter is 1, to which the ratio in the rule, viz. 1 to 3·1416 is very near. Also the other ratio in the rule 7 to 22 or 1 to 3 = 31428, &c., is another near approximation. * It having been found, in the demonstration of the foregoing problem, that when the radius of a circle is 1, the length of the whole circumference is 6.2831854, which consists of 360 degrees; therefore, as 360°: 6-2831854 :: 1°: 01745, &c., the length of the arc of 1 degree. Hence, the number 01745, multiplied by any number of degrees, will give the length of the arc of those degrees. And, because the circumferences, and arcs, are as the diameters, or radii of the circles; therefore, as the radius 1 is to any other radius r, so is the length of the arc above mentioned to r X 01745 X degrees in the arc, which is the length of that arc as in the rule. This first rule is proved in the Geometry, Theor. 94. And the second rule is deduced from the first in this manner: It appears by the demonstration of Problem 7, that when the diameter of a circle is 1, its circumference is 3.1415927, or nearly 31416; then, by the first rule, 1 X 3141647854, which is therefore the area of the circle whose diameter is 1. But the areas of different circles are to each other as the square of their diameters, by Geometry, Theor. 93; therefore, as 1' :d :: 7854: 7854 d2, the area of the circle whose diameter is d, as in the second rule. ence 22. Ex. 2.—To find the area of a circle, whose diameter is 7, and circumferAns. 38. Ex. 3.--How many square yards are in a circle, whose diameter is 3 feet? Ans. 1.069. PROBLEM X. To find the area of a circular ring, or space included between two concentric circles. Take the difference between the areas of the two circles, as found by the last problem. Or, which is the same thing, subtract the square of the less diameter from the square of the greater, and multiply their difference by 7854.—Or, lastly, multiply the sum of the diameters by the difference of the same, and that product by 7854; which is still the same thing, because the product of the sum and difference of any two quantities, is equal to the difference of their squares. Ex. 1.-The diameters of two concentric circles being 10 and 6, required the area of the ring contained between their circumferences. Here 10+ 6 = 16 the sum, and 10 — 6 = 4 the difference, Therefore, 7854 X 16 X 4 = ·7854 × 64 = 50·2656, the area. Ex. 2.—What is the area of the ring, the diameters of whose bounding circles are 10 and 20? Ans. 235.62. PROBLEM XI. To find the area of the sector of a circle. RULE I-Multiply the radius, or half the diameter, by half the arc of the sector, for the area. Or, multiply the whole diameter by the whole arc of the sector, and take of the product. The reason of which is the same as for the first rule to problem 9. RULE II.—AS 360 is to the degrees in the arc of the sector, so is the area of the whole circle, to the area of the sector. This is evident, because the sector is proportional to the length of the arc, or to the degrees contained in it. Ex. 1.-To find the area of a circular sector, whose are contains 18 degrees; the diameter being 3 feet. 1. By the 1st Rule. First, 3-1416 × 3 = 9.4248, the circumference. And 360 18:: 9-4248 : ·47124, the length of the arc. Then, 47124 X 3 ÷ 4 = ∙11781 × 3 = ·35343, the area. 2. By the 2d Rule. First, 7854 X 370686, the area of the whole circle. Then, as 360: 18 :: 7·0686 : ·35343, the area of the sector. Ex. 2. To find the area of a sector, whose radius is 10, and arc 20. Ans. 100. Ex. 3.-Required the area of a sector, whose radius is 25, and its arc contairing 147° 29'. PROBLEM XII. To find the area of a segment of a circle. Ans. 804 4017. RULE 1.-Find the area of the sector having the same arc with the segment. by the last problem. |
