 | Thomas Keith - 1817 - 306 páginas
...annual reiit. PROBLEM VI. Any two Sides of a right angled Triangle being given,, tofind the third Side, RULE 1. To the square of the base add the square of...perpendicular, the square root of the sum will give the hypothcnuse, or longest side. 2. Multiply the sum of the hypothenuse and one side, by their difference... | |
 | Thomas Keith - 1822 - 354 páginas
...The base and perpendicular of a right angled triangle being given, to find the hypothenuse. C Rule. To the square of the base add the square of the perpendicular, the square-root of the sum will give the hypothenuse. BASE. B Prop. 7. Given the hypothenuse, or longest... | |
 | Thomas Keith - 1825 - 360 páginas
...6. The bate and perpendicular of a rightangled triangle being given, to find the hypothenuse. Rule. To the square of the base add the square of the perpendicular, the square root of the sum will invi> tin' ]l vn/itliiinn square root o te su give the hypothenuse. ,A BASE. B I'roji. 7. Given Ike... | |
 | William Templeton (engineer.) - 1833 - 224 páginas
...8 12 16 10 6 2 and v'ISx 10x6 x 2 = 46.47 the area. PROBLEM IT. If any two sides of a right.angled Triangle be given, the third side may be found by the following rules : — 2. — Multiply the sum of the hypotenuse, and one side by their difference; and the square... | |
 | Nathan Scholfield - 1845 - 894 páginas
...find the third side. RULE. I. — When the base and perpendicular are given, to find the liypothennse. To the square of the base, add the square of the perpendicular;...square root of the sum will give the hypothenuse, or the longest side, (Prop. XXIV. B. IV.) Ei. Required the hypothenuse AC of a right angled triangle,... | |
 | William Templeton (engineer.) - 1846 - 236 páginas
...two sides of a right-angled triangle be given, the third side may be found by the following rules. 1 . — To the square of the base add the square of the perpendicular ; and the square root of the sum will be the hypotenuse or longest side. 2. — Multiply the sum of... | |
 | Charles Haslett - 1855 - 482 páginas
...2=46-47, the area. PROBLEM IV. If any two aides of a right-angled triangle be given, the third tide may be found by the following rule». 1. — To the...square of the base add the square of the perpendicular : and the square root of the sum will be the hypothenuse or longest side. 2. — Multiply the sum of... | |
 | Charles W. Hackley - 1856 - 530 páginas
...10 6 18 18 2 and y 18 x 10 x 6 x 2=46 '47, the area. PROBLEM IY. Jf any two sides of a right-angled triangle be given, the third side may be found by the following rules. 1. — To the square of the base add the square of the perpendicular ; and the square root of... | |
 | Charles Haslett - 1855 - 544 páginas
...12 18 18 6 2 and \f 18 x 10 x 6 x 2=46-47, the area. PROBLEM IV. If any Iwo rides of a right-angled triangle be given, the third side may be found by the following rules. 1. — To the square of the base add the square of the perpendicular : and the square root of... | |
 | Joseph Ray - 1857 - 348 páginas
...square of the side A C. RULE FOB FINDING THE HYPOTENUSE WHEN TUB BASE AND PERPENDICULAR ARE KKQWS. To the square of the base, add the square of the perpendicular ; the square root of the sum will give the hypotenuse. FOR FINDING EITHER SIDB WHEN THE HYPOTENUSE AND THE OTHER SIDE ARE ENOWN. I \ * From the... | |
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To the square of the base add the square of the perpendicular ; and the square root of the sum will give the hypothenuse (Prop. 
