 | Nathan Daboll - 1839 - 220 pages
...acres.] Ans. 26 acres 1 r. 25 rods. PROB. ii. — To find the area of a parallelogram, or long square. RULE. Multiply the length by the breadth, and the product will be the area. EXAMPLES. 1. How many square yards of ground are contained in a garden which is 126 feet long and 65... | |
 | Robert Goodacre - 1839 - 320 pages
...Example 1 29 Miscellaneous Questions. CASE 1. — To find the area of any rectangular superficies. Multiply the length by the breadth, and the product will be the area in square measure. In measuring superficies, in which the dimensions vary, it is common to take the... | |
 | Jason M. Mahan - Arithmetic - 1839 - 312 pages
...given, to find the area ; or the area and one side given, to find the length of the other aide. RULE. 1. Multiply the length by the breadth, and the product will be the area. 2. Divide the area by one of the sides, and the quotient will be the adjacent side. Examples. 1. What... | |
 | Joseph Stockton - Arithmetic - 1839 - 216 pages
...32J perches. 3<2. To find the content of an oblong square piece of ground, called a parallelogram. RULE. Multiply the length by the breadth, and the product will be the answer. EXAMPLE. i 1. There is an oblong square piece of ground, A, B, C, D, the longest sides of which... | |
 | Charles Davies - Geometrical drawing - 1840 - 262 pages
...the more accurate methods by means of figures. PROBLEM I. 2. To find the area of a board or plank. RULE. Multiply the length by the breadth, and the product will be the content required.' Q.OEsT. — 1. What methods of measuring timber have already been explained ? 2.... | |
 | William Ruger - Arithmetic - 1841 - 268 pages
...w'de, and 20ft. long? Ans. 80yds. NOTE. — To obtain the surface of a plane, as a floor, Jield, if-e. Multiply the length by the breadth, and the product will be the surface ; then as the number of square feet in a yard of the carpeting which is 4,5 feet, is to one... | |
 | Nathan Daboll - Arithmetic - 1843 - 254 pages
...acres.] Ans. 26 acres I r. 25 rods. PROB. ii. — To find the area of a parallelogram, or long square. RULE. Multiply the length by the breadth, and the product will be the area. EXAMPLES. 1. How many square yards of ground are contained in a garden which is 126 feet long and 65... | |
 | William Watson (of Beverley.) - 1845 - 188 pages
...MENSURATION OF SURFACES. PROBLEM 1. — To find the area of a square, rectangle, parallelogram, SfC. * RULE. — Multiply the length by the breadth, and the product will be the area, or multiply half the sum of the parallel sides by their perpendicular distance. EXAMPLES. 1. What is... | |
 | Almon Ticknor - Arithmetic - 1846 - 274 pages
...sides are perpendicular to each other (right-angled), but the adjacent sides are longer and parallel. RULE. Multiply the length by the breadth, and the product will be the answer. 1 . A garden is 76 feet in length, and 42 feet in width ; how many square feet of ground are... | |
 | Almon Ticknor - Arithmetic - 1846 - 276 pages
...sides are perpendicular to each other (right-angled), but the adjacent side's are longer and parallel. RULE. Multiply the length by the breadth, and the product will be thejanswer. 1. A. garden is 76 feet in length, and 42 feet in width ; how many square feet of ground... | |
| |
To measure a Parallelogram, or long square. RULE. Multiply the length by the breadth, and the product will be the area or superficial content. 
