18. A DI-AG-O-NAL is a line joining two angles of a figure not next to each other. Thus, SU (Fig. Def. 14) is a diagonal of the square. A E 19. A CIRCLE is a figure bounded by a curve line, called the circumference, every part of which is equally distant from a point within, called the center. A DIAMETER is a straight line passing through the center and terminated both ways by the circumference. A RADIUS is Ta straight line drawn from the center to the circumference; it is half the diameter. D Thus, A DBE is the circumference; A B the diameter; A For B F the radius. 20. A TANGENT is a straight line which touches the circumference only in one point, called the point of contact. Thus, T P is a tangent. 21. An ARC of a circle is any part of the circumference, as AF. A CHORD is a straight line joining the extremities of an arc. 22. MENSURATION is the art of finding the surface, and also the solid contents of bodies. 23. The AREA of a figure is the surface which it contains. The quantity of this surface is denoted by the number of times it contains a given surface called the measuring unit. The MEASURING UNIT for surfaces is a square surface, whose side is some one of the common measures of length, such as a square inch, a square foot, &c. See Arts. 87, 88, and 89. MEASUREMENT OF SURFACES. ART. 311. To find the superficial contents or AREA of a Parallelogram, Rectangle, Square, or Rhombus, Rule.-Multiply the length by the perpendicular breadth, the product will be the area. NOTE. The learner must recollect (Art. 276) that feet in length multiplied by feet in breadth, produce square feet; and the same of the other denominations of lineal measure. 1. How many square feet in a floor 17 feet long and 15 feet wide? Ans. 255 sq. ft 2. Find the sq. ft. in a board 2 ft. 3 in. wide and 12 ft. 6 in. long. Ans. 28.125 28 sq. ft. 3. The sq. ft. in a board 15 in. wide and 16 ft. long. Ans. 20 sq. ft. 4. How many sq. ft. in a board 1 ft. 2 in. in mean breadth, 12 ft. 6 in. long? Ans. 14 sq. ft. 84 sq. in., or 147 sq. ft. 5. At $1.50 per sq. ft., what cost a marble slab; the length 5ft. 7 in.; breadth 1 ft. 10 in. ? Ans. $15.354+ . 6. How many acres of land in a parallelogram; the length, 120 rd.; the breadth, 84rd.? Ans. 63 A. 7. How many acres in a square field, each side of which is 65 rd.? Ans. 26 A. 1 R. 25 P. 8. How many acres in a field in the form of a rhombus; each side measures 35 rd.; the perpendicular distance across it, 16 rd.? Ans. 3 A. 2 R. 9. Each side of the base of a pyramid is 693 ft. long: how many acres does it cover? Ans. 11 A. 4P. 10. Find the difference between a floor 30 ft. sq., and two others each 15 ft. sq. Ans. 450 sq. ft. 11. If a room is 10 ft. long, how wide must it be to contain 80 sq. ft.? (See Art. 90.) Ans. 8 ft. Ans. 12 ft. 12. A board is 10 inches wide: what must be its length to contain 10 sq. ft.? 13. How many yd. of carpet 11 yd. wide, will cover a floor 6 yd. long, 5 yd. wide? Ans. 20 yd. 14. How many yd. of carpet 11 yd. wide, will cover a floor 21 ft. 3 in. long, 13 ft. 6 in. wide? Ans. 251yd. 15. How many yd. of flannel 3 yd. wide, will line 3 yd. of cloth, 1yd. wide? Ans. 6 yd. Plasterers', Pavers', Painters', and Carpenters' Work. ART. 312. Several kinds of artificers' work are measured by the preceding rule. Plasterers', Pavers', and Painters' work, is computed in sq. yards: Glaziers' work, by the sq. ft., or by the pane Carpenters' and Joiners' work, some parts by the sq. yard; other parts by the SQUARE, which contains 100 sq. ft. 1. How many square yards in a ceiling 25 ft. 9 in long, and 21 ft. 3 in. wide? Ans. 60 sq. yd. 7 sq. ft.+ 2. At 20 cts. a sq. yd., what will it cost to plaster a ceil ing 22 ft. 7 in. long, 13 ft. 11 in. wide? Ans. $6.984+ 3. A room is 20 ft. 6 in. long, 16 ft. 3 in. broad, 10 ft. 4 in. high: how many yd. of plastering in it, deducting a fireplace 6 ft. 3 in. by 4 ft 2 in.; a door 7 ft. by 4 ft. 2 in., and two windows, each 6 ft. by 3 ft. 3 in.? Ans. 110 sq. yd. 85 sq. ft. : 4. A room is 20 ft. long, 14 ft. 6 in. broad, and 10 ft. 4 in. high what will the coloring of the walls cost, at 27 cts. per sq. yd., deducting a fireplace 4 ft. by 4 ft. 4 in., and two windows, each 6 ft. by 3 ft. 2 in. ? Ans. $19.73 5. At 18 cts. per sq. yd., find the cost of paving a walk 35 ft. 4 in. long, 8 ft. 3 in. broad. *Ans. $5.83 6. What will it cost to pave a rectangular yard, 21 yd. long, and 15 yd. broad, in which a footpath, 5 ft. 3 in. wide, runs the whole length of the yard; the path paved with flags, at 36 cts. per sq. yd., and the rest with bricks, at 24 cts. per sq. yd.? Ans. $80.01 7. At 10 cts. a sq. yd., what the cost to paint the walls of a room 75 ft. 6 in. in compass, 12 ft. 6 in. high? Ans. $10.486+ 8. A house has 3 tiers of windows, 7 in a tier: the height of the first tier is 6 ft. 11 in.; of the 2d, 5 ft. 4 in.; the 3d, 4 ft. 3 in.; each window is 3 ft. 6 in. wide: what cost the glazing, at 16 cts. per sq. ft. ? Ans. $64.68 9. A floor is 36 ft. 3 in. long, 16 ft. 6 in. wide: what will it cost to lay it, at $3 a square? Ans. $17.943+ 10. A room is 35 ft. long, and 30 ft. wide: what will the flooring cost, at $5 per square, deducting a fireplace 6 ft. by 4 ft. 6 in., and a stairway, 8 ft. by 10 ft. 6 in.? Ans. $46.95 11. At $3.50 per square, what cost a roof 40 ft. long, the rafters on each side 18 ft. 6 in. long? Ans. $51.80 ART. 313. TO FIND THE AREA OF A TRIANGLE. Rule.-Multiply the base by the perpendicular hight, and take half the product for the area. Or, when the sides are given, the following RULE : 1st. Add the three sides together, and take half the sum. 2d. From the half sum take the 3 sides severally. 3d. Multiply the half sum and the 3 remainders together, and extract the square root of the product, which gives the area. 1. Find the area of the triangle, E F G H, the base, FH, is 15 feet; the perpendicular height, G E, 12 feet. Ans. 90 sq. ft. 2. The contents of a triangular space, the base 44 rd., perpendicular height 18 rd. Ans. 2 A. 1 R. 36 P. 3. How many acres in a triangular field; perpendicular height, 67 rd. ? the base 80 rd.; Ans. 16A. 3 R. NOTE. The area of any field or piece of land may be found by dividing it into triangles, and measuring the base and perpendicular height of each triangle thus formed. 4. What cost the glazing of a triangular skylight, at 12 cts. per sq. ft., the base, 12 ft. 6 in., the perpendicular height, 16 ft. 9 in.? Ans. $12.561 1 5. Find the area of a triangle, the sides being 13, 14, and 15 ft. Ans. 84 sq. ft. 6. The area of a triangle, the sides 2, 3, and 4 feet respectively. Ans. 2.9047375+sq. ft. ART. 314. TO FIND THE AREA OF A TRAPEZOID. Rule.-Multiply the sum of the parallel sides by the perpendicular breadth; take half the product. 1. The parallel sides of a trapezoid, F C G D, are 35 and 26 inches; its breadth 11 in.; required the area Ans. 335 sq. in. G 2. A field is the form of a trapezoid; one of the parallel sides is 25 rd., the other 19 rd.; the width 32 rd.: how many acres in it? Ans. 4A. 1 R. 24 P. ART. 315. TO FIND THE CIRCUMFERENCE OF A CIRCLE, WHEN THE DIAMETER IS GIVEN. Rule.-Multiply the diameter by 3.1416, the product will be the circumference. 1. The diameter A B of the circle A DBE is 48 feet: what is the circumference? Ans. 150.7968 ft. 2. The diameter of a wheel is 4 feet: find the circumference. Ans. 12 ft. 6.7968 in. E B 3 3. What is the circumference of the earth, the mean diameter being 7912.4 mi. ? Ans. 24857.59584 mi. ART. 316. TO FIND THE DIAMETER OF A CIRCLE, WHEN THE CIRCUMFERENCE IS GIVEN. Rule.-Divide the circumference by 3.1416, the quotient will be the diameter. 1. The circumference of a circle is 15 feet: what is the diameter ? Ans. 4 ft. 9.295+in. 2. If the girt of a tree is 12 feet 5 inches, what its thickness or diameter? Ans. 3 ft. 11.428+in. ART. 317. TO FIND THE AREA OF A CIRCLE. Rule.-Multiply the diameter by the circumference, and take one-fourth of the product. Or, Multiply the square of the diameter by .7854; or, for greater accuracy, by .785398 Or, Multiply the square of the radius by 3.1416 1. Find the area of a circle, the diameter being 42 feet. Ans. 1385.4456 sq. ft. 2. Find the area of a space on which a horse may graze, when confined by a cord 7 rods long, one of its ends being fixed at a certain point. Ans. 1 A. 16.715P. ART. 318. TO FIND THE DIAMETER OF A CIRCLE, WHEN Rule.-Divide the area by .7854; the square root of the quotient will be the diameter. 1. The area of a circle is 962.115: what its diameter and circumference? Ans. diam. 35: circum. 109.956 2. What length of halter will fasten a horse to a post in the center of an acre of grass, so that he can graze upon the 1 A. and no more? Ans. 7.1364+rd., or 117 ft. 9+in. ART. 319. MEASUREMENT OF BODIES OR SOLIDS. DEFINITIONS.-1. A BODY or SOLID, has length, breadth, and thickness or depth. 2. A PRISM is a solid whose ends, or bases, are parallel; its sides, parallelograms. Such a body is termed a RIGHT P |