 | Charles Hutton - 1811 - 406 σελίδες
...— BC = AB.AD + BD in the 1st figure. THEOREM XXXVI. IN any Obtuse-angled Triangle, the Square of the Side subtending the Obtuse Angle, is Greater than the Sum of the Squares of the other two Sides, by Twice the Rectangle of the Base and the Distance of the Perpendicular from... | |
 | Charles Hutton - 1812 - 620 σελίδες
...— BC =AB . AD + BD in the 1st figure. THEOREM XXXVI. IN any Obtuse-angled Triangle, the Square of the Side subtending the Obtuse Angle, is Greater than the Sum of the Squares of the other two Sides, by Twice the Rectangle of the Base and the Distance of the Perpendicular from... | |
 | Charles Butler - 1814 - 528 σελίδες
...from the former,) -- Therefore AB>* —~Ub\* + ~*Tl1= ...... 2 bx * * That is, the square of ^B/the side subtending the obtuse angle,, is greater than the sum of the squares of CB and AC, the sides containing the obtuse angle, by (2 bx) twice the rectangle BC, CD. QED 175.... | |
 | Charles Hutton - 1822 - 616 σελίδες
...BC = AB . AB -f- BB in the 1st figure. THEOREM XXXVI. IN any Obtuse-angled Triangle, the Square of the Side subtending the Obtuse Angle, is Greater than the Sum of the Squares of the other two Sides, by Twice the Rectangle of the Base and the Distance of the Perpendicular from... | |
 | John Playfair - 1829 - 210 σελίδες
...perpendicular be drawn from either of the two acute angles to the opposite side produced, the square of the side subtending the obtuse angle is greater than the sum of the squares of the sides containing it by twice the rectangle under the base and the distance between the perpendicular... | |
 | George Roberts Perkins - 1847 - 326 σελίδες
...remainders, and it will be the area. PEOPOSITION X. THEOREM. In any obtuse-angled triangle, the square of the side subtending the obtuse angle is greater than the sum of the squares of the other two sides, by twice the rectangle of the base and the distance of the perpendicular from... | |
 | Charles William Hackley - 1847 - 248 σελίδες
...- BC) = AB . (AD + BD) in the 1st fig. THEOREM XXVJU. In any obtuse-angled triangle, the square of the side subtending the obtuse angle is greater than the sum of the squares of the other two sides, by twice the rectangle of one of the sides containing the obtuse angle and... | |
 | Thomas Tate (mathematical master.) - 1848 - 284 σελίδες
...5, will have the same property. 49. THEOREM. In an obtuse-angled triangle, the square of the side AB subtending the obtuse angle, is greater than the sum of the squares of the other two sides, BC and AC, by twice the rectangle of the base BC and the B cb distance CD of... | |
 | George Roberts Perkins - 1850 - 332 σελίδες
...remainders, and it will be the area. PROPOSITION X. THEOREM. In any obtuse-angled triangle, the square of the side subtending the obtuse angle is greater than the sum of the squares of the other two sides, by twice the rectangle of the base and the distance of the perpendicular from... | |
 | Great Britain. Committee on Education - 1850 - 942 σελίδες
...same base, and between the same parallels, are equal. 3. In an obtuse angled triangle, the square of the side subtending, the obtuse angle is greater than the sum of the squares of the sides containing it by twice the rectangle, contained by the base and the straight line intercepted... | |
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In obtuse-angled triangles, if a perpendicular be drawn from either of the acute angles to the opposite side produced, the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle, by twice the... 
