 | Robert Louis Short, William Harris Elson - Mathematics - 1911 - 216 pages
...find that in volume theorems the ratios involved the products of three lines. THEOREM LIII 207. Tlie area of a triangle is equal to one half the product of the base and altitude. Draw A ABC. Through C draw II to AB. Through B draw II to AC. A parallelogram... | |
 | Clara Avis Hart, Daniel D. Feldman, Virgil Snyder - Geometry, Solid - 1912 - 216 pages
...bases. 485. The area of a triangle 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... | |
 | Clara Avis Hart, Daniel D. Feldman - Geometry - 1912 - 504 pages
...area of triangle ABC ; the altitude upon b ; the altitude upoii c. PROPOSITION V. THEOREM 491. The area of a triangle is equal to one half the product of its perimeter and the radius of the inscribed circle. Given A ABC, with area '/; sides a, b, and c, and... | |
 | Walter Burton Ford, Charles Ammerman - Geometry, Solid - 1913 - 184 pages
...two parallelograms that have equal bases are to each other as their altitudes. 189. Theorem II. The area of a triangle is equal to one half the product of its base by its altitude. 190. Corollary 1. (a) Two triangles are to each other as the products of their bases and... | |
 | Walter Burton Ford, Earle Raymond Hedrick - Geometry, Modern - 1913 - 272 pages
...greatest area obtainable and what the least, provided that AB= 6 in. and BC=4: in. ? 189. Theorem U. The area of a triangle is equal to one half the product of its base and its altitude. FIG. l.'il Given the triangle ABC, having the base b and the altitude h. To prove... | |
 | Arthur Schultze, Frank Louis Sevenoak - Geometry - 1913 - 486 pages
...and 5 in. and the angle included by them equal 45°. Find the area. PROPOSITION V. THEOREM 355. The area of a triangle is equal to one half the product of its base and altitude. Proof. Construct O ABDC. The diagonal of a parallelogram divides it into two equal triangles.... | |
 | Walter Burton Ford, Charles Ammerman - Geometry, Plane - 1913 - 378 pages
...greatest area obtainable and what the least, provided that AB= 6 in. and BC = 4 in. ? 189. Theorem II. The area of a triangle is equal to one half the product of its base and its altitude. Given the triangle ABC, having the base b and the altitude h. To prove that the area... | |
 | Arthur Schultze, Frank Louis Sevenoak - Geometry, Plane - 1913 - 328 pages
...and 5 in. and the angle included by them equal 45°. Find the area. PEOPOSITION V. THEOEEM 355. The area of a triangle is equal to one half the product of its base and altitude. Proof. Construct £7 ABDC. The diagonal of a parallelogram divides it into two equal... | |
 | William Benjamin Fite - Algebra - 1913 - 304 pages
...theorems from geometry : 1. The area of a square is equal to the square of the length of one side. 2. The area of a triangle is equal to one half the product of its base and altitude. 4. The area of a circle is equal to the square of its radius multiplied by IT (=3| approximately).... | |
 | William Benjamin Fite - Algebra - 1913 - 368 pages
...theorems from geometry : 1. The area of a square is equal to the square of the length of one side. 2. The area of a triangle is equal to one half the product of its base and altitude. 3. The square of the hypotenuse of a right triangle is equal to the sum of the squares... | |
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The area of a triangle is equal to one half 'the product of its base and altitude. 
