from the point C with a radius equal to the diagonal AD (of the square representing the square of the base) describe arcs intersecting each other. The angle FDG will be the angle sought. We may suppose it taken along the line BC traced in fig. 487. 1202. In order to obtain the angles formed by the faces of an oblique pyramid (fig. 489.), through some point q of the axis draw the perpendicular mo, showing the base ogmq'o' of the right pyramid mpo, whose developement is shown in fig. 490., by the portion of the polygon a, b", c", e", d", a' F. 1203. By means of this base and the part developed, proceeding as we have already explained for the right pyramid, we shall find the angles formed by the meeting of the faces, and they will differ but little from those of the little polygon oqmq'o'. 1204. In respect to the angles formed by the faces inclined to the base, that of the face answering to the side De of the base is expressed by the angle ADP of the vertical projection, fig. 489. 1205. As to the other faces, for instance, that which corresponds to the side AE of the base (fig. 490.), through any point g draw gf perpendicular to it, meeting the line AF, showing the projection of one of the sides of the inclined face; upon the developement of this face, expressed by A"E'F, raise at the same distance from the point E' another perpendicular g'm', which will give the prolongation of the line shown on the base by Af. If we transfer A"m of the developement upon Am, which expresses the inclination of the arris represented by this line, we shall have the perpendicular height mf of the point m above the base, which, being transferred from fm" upon a perpendicular to gf, we shall have the two sides of a triangle whose hypothenuse gm" will give m'gf, the angle sought. 1206. In the oblique prism (fig. 491.), the angles of the faces are indicated by the plane of the section perpendicular to the axis, represented by the polygon hiklmn. 1207. Those of the sides perpendicular to the plan of the inclination of the axis are expressed by the angles Ddb, Abd of the profile in the figure last named. 1208. In order to obtain the angles formed with the other sides, for instance Cc Dd and Cc Ab, draw the perpendiculars csbt, whose projection in plan are indicated by s'c' and b't′, then upon fe, drawn aside, raise a perpendicular c'c"" equal to cs of the profile, fig. 491. Through the point c'" draw a line parallel to fc, upon which, having transferred c's' of the projection in plan (fig. 492.), draw the hypothenuse s"c", and it will give the angle s'e"ƒ formed by the face Cc Dd with the inferior base. 1209. To obtain the angles of the face Cc Ab, raise upon Fb", drawn on one side, a perpendicular b't"", equal to bt (fig. 492.), and drawing as before a parallel to b" through the point, transfer b't' of fig. 492. to t"t"; and drawing t"b", the angle t'b"F is that required. 1210. As the bases of this prism are parallel, these faces necessarily form the same angles with the superior base. 1211. An acquaintance with the angles of planes is of the greatest utility in the preparation of stone, and a thorough acquaintance with it will well repay the architectural student for the labour he may bestow on the subject. SECT. VII. MENSURATION. D 4 C 1212. The area of a plane figure is the measure of its surface or of the space contained within its extremities or boundaries, without regard to thickness. This area, or the content of the plane figure, is estimated by the number of small squares it contains, the sides of each whereof may be an inch, a foot, a yard, or any other fixed quantity. Hence the area is said to consist of so many square inches, feet, yards, &c., as the case may be. 1213. Thus if the rectangle to be measured be ABCD (fig. 512.), and the small square E, whose side we will suppose to be one inch, be the measuring unit proposed; then, as often as such small square is contained in the rectangle, so many square inches are said to be contained in the rectangle. Here it will be seen by inspection that the number is 12; that the side DC or AB, which is 4 times the length of the 3 measuring unit, multiplied by the number of times 3, which the length of the measuring unit is contained in AD or BC, will give 12 for the product. 1214. PROBLEM I. To find the area of a parallelogram, whether it be a square, a rectangle, a rhombus, or a rhomboid. Multiply the length by the perpendicular breadth or height, and the product will be the area. E Fig. 51. B Example 1. Required the area of a parallelogram whose length is 12.25 feet, and height 8.5 feet. 12.25 x 8.5=104·125 feet, or 104 feet 11 inches. Example 2. Required the content of a piece of land in the form of a rhombus whose length is 6 20 chains, and perpendicular height 5·45. Recollecting that 10 square chains are equal to a square acre, we have, 6.20 × 5·45=33.79 and 20 perches. 33.79 3.379 acres, which are equal to 3 acres, 1 rood, Example 3. Required the number of square yards in a rhomboid whose length is 37 feet, and breadth 5 feet 3 inches (= 5·25 feet). Recollecting that 9 square feet are equal to 1 square yard, then we have 1215. PROBLEM II. To find the area of a triangle. area. =21584 yards. Rule 1. Multiply the base by the perpendicular height, and take half the product for the Or multiply either of these dimensions by half the other. The truth of this rule is evident, because all triangles are equal to one half of a parallelogram of equal base and altitude. (See Geometry, 904.) Example 1. To find the area of a triangle whose base is 625 feet, and its perpendicular height 520 feet. Here, 625 x 260=162500 feet, the area of the triangle. Rule 2. When two sides and their contained angle are given: multiply the two given sides together, and take half their product; then say, as radius is to the sine of the given angle, so is half that product to the area of the triangle. Or multiply that half product by the natural sine of the said angle for the area. This rule is founded on proofs which will be found in Chap. IV., on which it is unnecessary here to say more. Example. Required the area of a triangle whose sides are 30 and 40 feet respectively, and their contained angles 28° 57'. Rule 3. When the three sides are given, take half the sum of the three sides added together. Then subtract each side severally from such half sum, by which three remainders will be obtained. Multiply such half sum and the three remainders together, and extract the square root of the last product, which is the area of the triangle. This rule is founded on one of the theorems in Trigonometry to be found in the section relating to that branch of the subject. Example. Required the area of a triangle whose three sides are 20, 30, and 40. 20+30+40=90, whose half sum is 45. 45-20=25, first remainder; 45-30=15, second remainder; 45-40=5, third remainder. Then, 45 x 25 x 15 x 5 =84375, whose root is 290-4737, area required. 1216. PROBLEM III. To find the area of a trapezoid. Add together the parallel sides, multiply their sum by the perpendicular breadth or distance between them, and half the product is the area. (See Geometry, 932.) Example 1. Required the area of a trapezoid whose parallel sides are 750 and 1225, and their vertical distance from each other 1540. 1217. PROBLEM IV. To find the area of any trapezium. Divide the trapezium into two triangles by a diagonal; then find the areas of the two triangles, and their sum is the area. Observation. If two perpendiculars be let fall on the diagonal from the other two opposite angles, then add these two perpendiculars together, and multiply that sum by the diagonal. Half the product is the area of the trapezium. Example. Required the area of a trapezium whose diagonal is 42, and the two perpendiculars on it 16 and 18. Here, 16+18=34, whose half = 17; Then, 42 x 17-714, the area. 1218. PROBLEM V. To find the area of an irregular polygon. Draw diagonals dividing the proposed polygon into trapezia and triangles. Then, having found the areas of all these separately, their sum will be the content required of the whole polygon. Example. Required the content of the irregular AC=55, GC-44, Bn=18, Ep=8, And 55 x 9 =495 55 x 6.5 =357.5 44 x 11.5=506 A B Rule 1. To find the area of a regular polygon. Multiply the perimeter of the polygon, or sum of its sides, by the perpendicu lar drawn from its centre on one of its sides, and take half Example. Required the area of the regular pentagon ABCDE Here 25x5 2 =62.5=half the perimeter, and 62.5 × 17.2=1075 D P Fig. 515. Rule 2. Square the side of the polygon, and multiply the square by the tabular area or multiplier set against its name in the following table, and the product will be the area. This rule is founded on the property, that like polygons, being similar figures, are to one another as the squares of their like sides. Now the multipliers of the table are the respective areas of the respective polygons to a side =1; whence the rule is evident. In the table is added the angle OBP in the several polygons. Example. Required the area of an octagon whose side is 20 feet. Here 202-400, and the tabular area 4.8284271; Therefore 4.8284271 × 400=1931.37084 feet, area required. 1220. PROBLEM VII. To find the diameter and circumference of any circle, either from the other. Rule 1. As 7 is to 22, or as 1 is to 3·1416, so is the diameter to the circumference. as 22 is to 7, so is the circumference to the diameter. Example. Required the circumference of a circle whose diameter is 9. 22x9 Here 7 22::9: 284; or, 7 Required the diameter of a circle whose circumference is 36. 36×7 Here 22: 7:36:1119; or, =119, the diameter required. 1221. PROBLEM VIII. To find the length of any arc of a circle. Or Rule 1. Multiply the decimal 01745 by the number of degrees in the given arc, and that by the radius of the circle; then the last product will be the length of the arc. This rule is founded on the circumference of a circle being 6.2831854 when the diameter is 2, or 3.1415927 when the diameter is 1. The length of the whole circumference then being divided into 360 degrees, we have 360°: 6-2831854 ::1° 01745. Example. Rule 2. Required the length of an arc of 30 degrees, the radius being 9 feet. Here 01745 x 30 x 9=4·7115, the length of the arc. From 8 times the chord of half the arc subtract the chord of the whole arc, and one third of the remainder will be the length of the arc nearly. Example. Required the length of an arc DCE (fig. 516.) whose chord DE is 48, and versed sine 18. 1222. PROBLEM IX. To find the area of a circle. Rule 1. Multiply half the circumference by half the diame ter. Or multiply the whole circumference by the whole Rule 2. Square the diameter, and multiply such square by 7854. Rule 3. Square the circumference, and multiply that square by the decimal 07958. Example. Required the area of a circle whose diameter is 10, and its circumference By rule 2., 10° x 7854100 × ·7854 =78·54. So that by the three rules the area is 78.54. 1223. PROBLEM X. To find the area of a circular ring, or of the space included between the circumferences of two circles, the one being contained within the other. Rule. The difference between the areas of the two circles will be the area of the ring. Or, multiply the sum of the diameters by their difference, and this product again by 7854, and it will give the area required. Example. 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. 1224. PROBLEM XI. To find the area of the sector of a circle. area. Rule 1. Multiply the radius, or half the diameter, by half the arc of the sector for the Or multiply the whole diameter by the whole arc of the sector, and take of the product. This rule is founded on the same basis as that to Problem IX. Rule 2. 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 manifest, because it is proportional to the length of the arc. Example. Required the area of a circular sector whose arc contains 18 degrees, the diameter being 3 feet. By the first rule, 3·1416 × 3=9'4248, the circumference. •47124 × 3÷4=1·41372÷4=35343, the area of the sector. By the second rule, 7854 x 32=7·0686, area of the whole circle. 1225. PROBLEM XII. To find the area of a segment of a circle. Rule 1. Find the area of the sector having the same arc with the segment by the last problem. Then find the area of the triangle formed by the chord of the segment Take the sum of these two for the answer when and the two radii of the sector. the segment is greater than a semicircle, and their difference when less than a semicircle. Example. Required the area of the segment ACBDA (fig. 517.), its chord AB being 12, and the radius AE As AE sin. 4 D 90° :: AD; sin. 36° : 52}=36.87 degrees Their double 73-74 degrees in are ACB. Now, 7854 x 400314·16, the area of the whole circle. Again, AE-AD2/100-36=√64=8=DE. Therefore, AD × DE=6 × 8=48, the area of the triangle B Hence the sector ACBE (64.350), less triangle AEB (48) =16·3504, area of segment ACBDA. Fig. 517. Rule 2. Divide the height of the segment by the diameter, and find the quotient in the column of heights in the following table. Take out the corresponding area in the next column on the right hand, and multiply it by the square of the circle's diameter for the area of the segment. This rule is founded on the principle of similar plane figures being to one another as the squares of their like lineal dimensions. The segments in the table are those of a circle whose diameter is 1. In the first column is contained the versed sines divided by the diameter. Hence the area of the similar segment taken from the table and multiplied by the square of the diameter gives the area of the segment to such diameter. When the quotient is not found exactly in the table, a proportion is used between the next less and greater area. Example. As before, let the chord AB be 12, and the radius 10 or diameter 20. Hence Having found as above DE 8: then CE-DE-CD=10-8=2. AREAS OF THE SEGMENTS OF A CIRCLE WHOSE DIAMETER, UNITY, IS SUPPOSED TO BE Hght. Area Seg. Hght. Area Seg. Hght. Area Seg. Hght. Area Seg. Hght. Area Seg. Hght. Area Seg. 001 000042 022 004322 043 002 032186 106 044522 032745|·107 045139 033307 108 045759 033872·109 046381 011734064 021168 085 000119023 004618 044 0121 42 065 021659] 086 *003 000219 024 004921045 012554 066 022154 087 004 000337 | 025 005230046012971 067 022652 088 005 000470026 005546 | 047 | 013392 068 023154 089 034441110 047005 | *006 000618027 005867048 013818 069 023659 090 035011|·111 ·047632| 007 000779 028 006194049 01 4247 070 024168091|·035585112 048262 008 000951029 006527050 014681 071 024680 092 036162·113 048894 009 001135 030 006865 051 015119 072 025195093 036741|·114 049528 010 001329 031 007209 052 011001533032 007558 053 012 001746033 007913 054 016 015561 073 025714 094 013 001968 034 008273 055 016911 076 027289 097 021 037323·115·050165 037909 116 050804 038496117 051 446| 039087|·118 052090 039680119052736| 029435 101 040276·120 053385 040875121 | ·054036 041476122 054689 042080123 055345 042687124 056003 043296125 056663 004031042 011330063 020680084 031629105 043908126 057326 |