 | John Bonnycastle - Geometry - 1848 - 320 pages
...6 inches 1 fe. in. pa. Ans. 27 1 6 11. To find the area of a rectangular board, whose length is 12J feet, and breadth 9 inches. Ans. 9f feet. PROBLEM...Note. — The perpendicular height of the triangle is equa. to twice the area divided by the base. EXAMPLES. 1. Required the area of the triangle ABC, whose... | |
 | Thomas Tate (mathematical master.) - 1848 - 284 pages
...sq. yds. 2. PROBLEM. To find the area of a triangle when its base and perpendicular height are given. RULE. Multiply the base by the perpendicular height, and half the product will be the area. See Geo. p. 35. EXAMPLES. 1. Required the number of acres in the triangular field ABC, whose base AB... | |
 | Peter Nicholson, Joseph Gwilt - Architectural drawing Technique - 1848 - 750 pages
...Ans. 89 4 6 89 4 6 f. S'J-375 f. i ii f. i ii PROBLEM II. PLATE 50. To find the area of a triangle. Multiply the base by the perpendicular height, and half the product will be the area. EXAMPLE I. What is the area of a triangle ABC, the base AB being 12f. 3i, and the height вс 8f. 6i?... | |
 | Almon Ticknor - Measurement - 1849 - 156 pages
...— THE TRIANGLE. To find the area of a triangle, when the base and perpendicular height are given. RULE. — Multiply the base by the perpendicular height, and half the product will be the area. Or, which is the same, Multiply the base by half the altitude, and the product will be the area, 1.... | |
 | J. M. Scribner - Mechanical engineering - 1849 - 286 pages
...or height ; that is, equal to the length multiplied by the height, which is the rule. OF TRIANGLES. PROBLEM II To find the Area of a Triangle. Rule. — Multiply the length of one of the sides by the perpendicular falling upon it, and half the product will be the area.... | |
 | Ramchundra - Algebra - 1850 - 222 pages
...— -2— ^ = 1 equation to the Sphere. (5.) TO FIND THE AREA OF A TRIANGLE. (Fig. 4.) Rule 1st. — Multiply the base by the perpendicular height, and half the product will be the area. The truth of this rule is evident, because any triangle is the half of a parallelogram of equal base... | |
 | Benjamin Greenleaf - Arithmetic - 1850 - 368 pages
...base of a rhombus being 12 feet, and its height 8 feet, required the area. Ans. 96 feet. PROBLEM m. To find the area of a triangle. RULE. — Multiply the base by half the perpendicular height; or, add the three sides together ; then take half of that sum, and out... | |
 | Oliver Byrne - Engineering - 1851 - 310 pages
...— ^ — = 8-02676 acres = 8 ac. 10 0 ro. <ipo. area required. A To find the area of a triangle. — Multiply the base by the perpendicular height, and half the product will be the area. The perpendicular height of the triangle is equal to twice the area divided by the base. Required the... | |
 | Joseph Bateman - Excise tax - 1852 - 376 pages
...Acuteangled, aecording as it has one of its angles obtuse, or one of them acute. To find the area, It — Multiply the base by the perpendicular height, and half the product will be the area. Or, (5.) Lct the base of a rhomboid BC be 73-6, and perpendicular height AE 56-2 ; required the area... | |
 | Ezra S. Winslow - Business mathematics - 1853 - 264 pages
...distance from the side FG to the side HI being 8 feet? 12 X 8 = 96 square feet. Ans. OF TRIANGLES. To find the area of a triangle. RULE. — Multiply the base by half the perpendicular height, or the perpendicular height by half the base, and the product is the... | |
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To find the Area of a Triangle. Rule ] . Multiply the base by the perpendicular height, and half the product will be the area. 
