 | Edward Rutledge Robbins - Geometry, Plane - 1906 - 268 pages
...each other as the products of their bases by their altitudes. Proof: (?). 378. THEOREM. The area of a triangle is equal to half the product of its base by its altitude. Given : A ABC ; base = b \ altitude = h. To Prove : \ Area of A ABC = -£ b - h. Proof :... | |
 | Edward Rutledge Robbins - Geometry - 1907 - 428 pages
...each other as the products of their bases by their altitudes. Proof: (?). 378. THEOREM. The area of a triangle is equal to half the product of its base by its altitude. Given: A ABC; altitude = h. To Prove : Area of A ABC = % b • h. Proof : Through A draw... | |
 | George Albert Wentworth, David Eugene Smith - Geometry, Plane - 1910 - 287 pages
...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. o A b BX Given the triangle ABC, with altitude a and base b. To prove that the area of the... | |
 | United States. Office of Education - 1911 - 1154 pages
...between a secant from the same point and the external segment of the secant. 6. Prove that the area of a triangle is equal to half the product of Its base by Its altitude. Group III. 7. Find the area of a circle inscribed in an equilateral triangle whose side is... | |
 | David Eugene Smith - Geometry - 1911 - 358 pages
...an approximation. The cutting of the paper is in every way more satisfactory. THEOREM. The area of a triangle is equal to half the product of its base by its altitude. Of course, the Greeks would never have used the wording of either of these two propositions.... | |
 | Education - 1911 - 1030 pages
...between a secant from the same point and the external segment of the secant. 0. Prove that the area of a triangle is equal to half the product of Its base by its altitude. GROUP III. 7. Find the area of a circle inscribed in an equilateral triangle whose side is... | |
 | John William Hopkins, Patrick Healy Underwood - Arithmetic - 1912 - 406 pages
...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... | |
 | George Albert Wentworth, David Eugene Smith - Geometry - 1913 - 496 pages
...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 area of the... | |
 | John Wesley Young, Albert John Schwartz - Geometry, Modern - 1915 - 250 pages
...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.... | |
 | John H. Williams, Kenneth P. Williams - Geometry, Solid - 1916 - 184 pages
...altitude. 347. Parallelograms having equal bases and equal altitudes are equivalent. 349. The area of a triangle is equal to half the product of its base by its altitude. 352. The area of a trapezoid is equal to half the product of its altitude by the sum of its... | |
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The area of a triangle is equal to half the product of its base by its altitude. 
