 | Francis Holliday - 1749 - 360 pàgines
...lejjer circle mujl be lefs than half the radius of the greater circle. DE ( "9 ) • DEMONSTRATION.' The area of any triangle is equal to the product of the radius of its infcrib'd circle into half the perimeter ; therefore, if this prop'jfition be true, the... | |
 | Edinburgh Mathematical Society - 1887 - 316 pàgines
...eccentric circles, the radius of the lesser circle must be less than half the radius of the greater circle. The area of any triangle is equal to the product of the radius of its inscribed circle into half the perimeter ; therefore, if this proposition be true, the... | |
 | George D. Pettee - 1896 - 272 pàgines
...equilateral triangle to the three sides is equal to the altitude of the triangle. 295. The area of a triangle is equal to the product of the semiperimeter and the radius of the inscribed circle. 296. If an angle of a triangle is 30°, prove that the area of the triangle equals one-fourth the product... | |
 | Alfred Cook - 1904 - 408 pàgines
...cylinders and of cones are subject to a system of conceptions. Take, for example, that simple case that the area of any triangle is equal to the product of the base into half the altitude. What an importance must attach to it! Take also the case that the surface... | |
 | George William Myers - 1910 - 304 pàgines
...F.) a \J Hence from (1) and (2) ABC=-=~ , and we have thus proved — PROPOSITION VI 190. Theorem: The area of any triangle is equal to the product of the three sides divided by twice the diameter of the circumscribed circle. Give a complete proof. EXERCISES... | |
 | Alfred Monroe Kenyon, Louis Ingold - 1919 - 306 pàgines
...product of its area into the diameter of its circumscribed circle. See §§ 40 and 52. 33. Prove that the area of any triangle is equal to the product of the radii of ite inscribed and circumscribed circles into the sum of the sines of its angles. See §§... | |
 | Edinburgh Mathematical Society - 1887 - 162 pàgines
...eccentric circles, the radius of the leaner circle must be less than half the radius of the yreater circle. The area of any triangle is equal to the product of the radius of ita inscribed circle into half the perimeter ; therefore, if thia proposition be true, the... | |
 | 1898 - 734 pàgines
...(«) If A + B + C=180°, prove that tan. A + tan. B + tan. C = tan. A, tan. B, tan. C. (/) Prove that the area of any triangle is equal to the product of the radius of the inscribed circle and the semi-perimeter of the triangle. ELEMENTARY SPHERICAL TRIGONOMETRY.... | |
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The area of any triangle is equal to the product of the radius of... 