 | John William Hopkins, Patrick Healy Underwood - 1912 - 406 sider
...parallelogram ABCD. Therefore, the area of the triangle ABOia equal to half the product of AO by BH. The area of a triangle is equal to half the product of its base by its altitude. Thus, the area of a triangle, whose base is 12 ft. and altitude 5 ft., equals one half of... | |
 | William Herschel Bruce, Claude Carr Cody - 1912 - 134 sider
...base and altitude. 421. Parallelograms having equal bases and equal altitudes are equivalent. 425. The area of a triangle is equal to half the product of its base and altitude. 430. Triangles of equal bases and altitudes are equivalent. 4.34. The areas of two triangles,... | |
 | William Betz, Harrison Emmett Webb, Percey Franklyn Smith - 1912 - 360 sider
...street 100 ft. wide, how many miles of the street would be covered? PROPOSITION- II. THEOREM 333. The area of a triangle is equal to half the product of its base and altitude. Given the triangle ABC, with the base b and the altitude h. To prove that the area of... | |
 | William Betz, Harrison Emmett Webb - 1912 - 368 sider
...street 100 ft. wide, how many miles of the street would be covered? PROPOSITION II. THEOREM 333. The area of a triangle is equal to half the product of its base and altitude. Dr Given the triangle ABC, with the base b and the altitude h. To prove that the area... | |
 | George Albert Wentworth, David Eugene Smith - 1913 - 496 sider
...opposite sides cut one another into segments that are reciprocally proportional to each other. 53. The area of a triangle is equal to half the product of its perimeter by the radius of the inscribed circle. 54. The perimeter of a triangle is to one side as... | |
 | William James Milne - 1914 - 364 sider
...shaded triangle is one half of a parallelogram of the same base and altitude as the triangle. Hence, The area of a triangle is equal to half the product of its base and altitude, expressed in like units. Written Exercises Find the area of each of the following triangles... | |
 | Charles Ernest Chadsey - 1914 - 274 sider
...its altitude. This is different from its slant height. The line of breadth is called its base. The area of a triangle is equal to half the product of its altitude by its base. The first figure is called a right-angled triangle, because its base is perpendicular... | |
 | Edward Rutledge Robbins - 1915 - 282 sider
...having equal bases are to each other as their altitudes. Proof : (?). PROPOSITION V. THEOREM 364. The area of a triangle is equal to half the product of its base by its altitude. Given: A ABC, with base b and altitude h. To Prove : Area of A ABC = \ b • h. Proof : Through... | |
 | John Wesley Young, Albert John Schwartz - 1915 - 248 sider
...called the base of the triangle with reference to the altitude drawn to that side. 352. THEOREM. The area of a triangle is equal to half the product of its base by its altitude. Given the triangle ABC, with base b and altitude h. To prove that the area of A ABC= % bh.... | |
 | Claude Irwin Palmer, Daniel Pomeroy Taylor - 1915 - 296 sider
...parallelogram construct a rhombus equal in area to the parallelogram. THE TRIANGLE 362. Theorem. The area of a triangle is equal to half the product of its base and its altitude. A = \bh. Given AACE with base b and altitude h. To prove area of AACE = ^bh. Proof.... | |
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The area of a triangle is equal to half the product of its base by its altitude. 
