 | Webster Wells - Geometry - 1899 - 424 pages
...The base of an isosceles triangle is 56, and each of the equal sides is 53 ; find its area. 30. The area of a triangle is equal to one-half the product of its perimeter by the radius of the inscribed circle. B 31. The area of an isosceles right triangle is equal... | |
 | Floyd Davis - Mining engineering - 1900 - 148 pages
...the altitude of the rhomboid. 21 LESSON VI. Q. 83. How is the area of a triangle determined? A. The area of a triangle is equal to one-half the product of its base and altitude. Q. 84. What is the area of a scalene triangle whose base is 20 feet and altitude 12 feet and 9 inches?... | |
 | International Correspondence Schools - Sheet-metal work - 1901 - 578 pages
...figures. SOLUTION. — - = .375. O Therefore, 4 = A/7375 = .61. Ans. GEOMETRICAL APPLICATIONS. 62. Area of a Triangle. — The area of a triangle is...to one-half the product of its base and altitude. Let ABC, Fig. 7, be a triangle whose base is BC and whose altitude is A D. Through A draw EF parallel... | |
 | Metal-work - 1901 - 548 pages
...D SOLUTION. — = = .375. Therefore. | = yC375 = .61. Ans. GTCOMKTRK'AL APPLICATIONS. 62. Area of n Triangle. — The area of a triangle is equal to one-half the product of its base and altitude. A~ ..JF Let ABC, Fig. 7, be a triangle whose base is BC and whose altitude is A D. Through A draw EF... | |
 | Arthur Schultze, Frank Louis Sevenoak - Geometry - 1901 - 394 pages
...area. Ex. 759. To divide a parallelogram into three equivalent parts. PROPOSITION V. THEOREM 346- The area of a triangle is equal to one-half the product of its base and altitude. AB Hyp. In A ABC, the base is b, and the altitude h. To prove area of ABC = £ b x h. Proof. Construct... | |
 | Arthur Schultze, Frank Louis Sevenoak - Geometry - 1902 - 394 pages
...area. Ex. 759. To divide a parallelogram into three equivalent parts. PROPOSITION V. TIIEOREM 346. The area of a triangle is equal to one-half the product of its base and altitude. AB Hyp. In A ABC, the base is b, and the altitude h. To prove area of ABC = \ bx /<. Proof. Construct... | |
 | Arthur Schultze - 1901 - 260 pages
...area. Ex. 759. To divide a parallelogram into three equivalent parts. PROPOSITION V. THEOREM 346. The area of a triangle is equal to one-half the product of its base and altitude. AB Hyp. In A ABC, the base is b, and the altitude h. To prove area of ABC = | bx h. Proof. Construct... | |
 | Edward Brooks - Geometry, Modern - 1901 - 278 pages
...; base 21 in. THEOREMS FOR DEMONSTRATION. (A Huh- more difficult than the preceding list.) 1 . The area of a triangle is equal to one-half the product of its perimeter by the radius of the inscribed circle. 2. The four lines joining the middle points of the... | |
 | George Albert Wentworth - Trigonometry - 1902 - 256 pages
...by logarithms as the value of 6 given under Case IV. See also p. 23. SECTION XIV AREA OF THE RIGHT TRIANGLE The area of a triangle is equal to one-half the product of the base by the altitude ; therefore, if a and b denote the legs of a right triangle, and F the area,... | |
 | George Albert Wentworth - Plane trigonometry - 1902 - 286 pages
...logarithms as the value of 6 given under Case IV. 8ee also p. 23. f, л8ECTION XIV AREA OF THE RIGHT TRIANGLE The area of a triangle is equal to one-half the product of the base by the altitude ; therefore, if a and Ь denote the legs of a right triangle, and F the area,... | |
| |
The area of a triangle is equal to one-half the product of its base and altitude. 
