 | Roswell Chamberlain Smith - Arithmetic - 1827 - 216 pages
...legues ; the oilier due east 30 leagues ; don- far are they apart ? A. 50 leagues. Port. PROBLEM V. — To find the area of a piece of land in the form of a triangle. RULE — Add together the three sides ; from their half sura subtract each side and note... | |
 | B[enjamin] Franklin Callender - Measurement - 1836 - 226 pages
...the area of a semicircle, the diameter of which is 25 inches '>. ANS. 245.4375 sq. in . 4. What is the area of a piece of land in the form of a semicircle, the straight side of which is 40 feet 1 ANS. 628.32 sq. feet. 5. If a garden, in the form... | |
 | Charles Davies - Navigation - 1837 - 342 pages
...log 1217 . . 3.085291 2)13.578489 Area in square links, 6155225 .. . 6.789244 2R 8P. PROBLEM III. 1 10. To find the area of a piece of land in the form of n trapezoid. Measure the two parallel sides, and also the perpendicular distance between them. Add... | |
 | Charles Davies - Surveying - 1839 - 376 pages
...3.085991 fc 2)13.578489 Area in square links, 6155225 . . 6.789244 Ans. 6l.# 2R 8P. PROBLEM III. 110. To find the area of a piece of land in the form of a trapezoid. Measure the two parallel sides, and also the perpendicular distance between them. Add the... | |
 | Charles Davies - Surveying - 1839 - 376 pages
...are 20 and 32 ch, and the perpendicular distance between them 88 chAns. 67^# 2R 16P. PROBLEM IV. 111. To find the area of a piece of land in the form of a tfuadrilateral. (wo triangles, in both of which all the sides will be known. Then, find the areas of... | |
 | Charles Davies - Navigation - 1854 - 446 pages
...3.085291 2)13.578489 Area in square l;uks, 6155225 . . . 6.789244. Ans. 61A. 2R 8P. PEOBLEM III. 9. To find the area of a piece of land in the form of & trapezoid. Measure the two parallel sides, and also the perpendicular distance between them. Add... | |
 | Benjamin Greenleaf - Geometry - 1866 - 328 pages
...as its base 354 feet ; what is the altitude ? 67,968 -=- 354 = 192 feet, the altitude required. 2. The area of a piece of land in the form of a rhombus is 69,452 square feet, and the perpendicular distance between two of its opposite sides is... | |
 | Charles Davies - Leveling - 1871 - 448 pages
...3.085291 2)13.578489 Area in square links, 6155225 . . . 6.789244 Ans. 61 A. 2Jt. 8 P. PROBLEM III. 55* To find the area of a piece of land in the form of a trapezoid. RULE. — Measure the two parallel sides, and also the perpendicular distance between them.... | |
 | Benjamin Greenleaf - Geometry - 1873 - 202 pages
...taken as its base 354 feet; what is thcf altitude ? 67,968 -h 354 = 192 feet, the altitude required. 2. The area of a piece of land in the form of a rhombus is 69,452 square feet, and the perpendicular distance between two of its opposite sides is... | |
 | Benjamin Greenleaf - Geometry - 1874 - 206 pages
...as its base 354 feet ; what is tho altitude ? 67,968 -:- 354 = 192 feet, the altitude required. 2. The area of a piece of land in the form of a rhombus is 69,452 square feet, and the perpendicular distance between two of its opposite sides is... | |
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What must be the nominal value of 4% bonds that will yield to their owner an annual income of $720 ? 7. A owns $6000 of 5% bonds; B owns $8000 of 4£% bonds. How much greater is the annual income from B's bonds than from A's ? 8. Find the area of a piece... 
