 | Clara Avis Hart, Daniel D. Feldman, Virgil Snyder - Geometry, Solid - 1912 - 216 pages
...equals one half the product of its base and its altitude. 491. The area of a triangle is equal to one half the product of its perimeter and the radius of the inscribed circle. 492. The area of any polygon circumscribed about a circle is equal to one half its perimeter multiplied... | |
 | William Betz, Harrison Emmett Webb - Geometry, Modern - 1912 - 368 pages
...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... | |
 | John William Hopkins, Patrick Healy Underwood - Arithmetic - 1912 - 406 pages
...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... | |
 | William Herschel Bruce, Claude Carr Cody - Geometry, Solid - 1912 - 134 pages
...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... | |
 | William Betz, Harrison Emmett Webb, Percey Franklyn Smith - Geometry, Plane - 1912 - 360 pages
...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... | |
 | George Albert Wentworth, David Eugene Smith - Geometry - 1913 - 496 pages
...are to each other as the products of their bases by their altitudes. PROPOSITION V. THEOREM 325. The area of a triangle is equal to half the product of its ~base by its altitude. A b B x Given the triangle ABC, with altitude a and base b. To prove that the... | |
 | Charles Ernest Chadsey - 1914 - 274 pages
...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... | |
 | William James Milne - Arithmetic - 1914 - 364 pages
...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... | |
 | John Wesley Young, Albert John Schwartz - Geometry, Modern - 1915 - 248 pages
...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... | |
 | Edward Rutledge Robbins - Geometry, Plane - 1915 - 282 pages
...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 •... | |
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The area of a triangle is equal to half the product of its base by its altitude. 
