 | Robert Gibson - Surveying - 1795 - 384 pages
...Sum of the Squares, Plate I. Squares, = BKLH+KCML, the Sum of the two Parallelograms or Square BCMH ; therefore the Sum of the Squares on AB and AC is equal to the Square on BC. Q^ ED Cor. i. Hence the Hypothenufe of a rightangled Triangle may be found by having the Legs ; thus,... | |
 | Robert Gibson - Surveying - 1806 - 486 pages
...ACGF the sum of the Plate I. squares = BELH + KCML, the sum of the two parallelograms or square BCMH ; therefore the sum of the squares on AB and AC is equal to the square on BC. QED Cor. Hence the hypothenuse of a right-angled triangle may be found by having the legs ; thus, the... | |
 | Robert Gibson - 1808 - 482 pages
...ACGF the sum of the Plate I. squares, = BKLH+KCML, the sum of the two parallelograms or square BCMH ; therefore the sum of the squares on AB and AC is equal to the square on BC. -QED Cor. 1. Hence the hypothenuse of a rightangled triangle may be found by having the legs ; thus,... | |
 | Robert Gibson - Surveying - 1811 - 580 pages
...So AB DE + ACGF the sum of the &quvces=BKLH+KCML, the sum of the two parallelograms or square BCMH ; therefore the sum of the squares on AB and AC is equal to the square on BC. 2. ED Cor. 1. Hence the hypothenuse of a right-angled triangle may Its found by having the sides ;... | |
 | Robert Gibson - Surveying - 1814 - 558 pages
...ABDE+ACGFûie sum of the squares= BKLH + KCML, the sum of the two parallelograms or square BCMH ; thei'efore the sum of the squares on AB and AC is equal to tiie square on BC. QED Cor. 1. Hence the hypothenuse of a right-angled triangle may be found by having... | |
 | Nathaniel Bowditch - Nautical astronomy - 1826 - 764 pages
...parallelograms BKLH and KCML : the sum of these parallelograms is equal to the square BCMH, therefore sum of the squares on AB and AC is equal to the square on BC. Cor. Hence in any right angled triangle, if we have the hypotenuse one of the legs, we may easily find the... | |
 | Nathaniel Bowditch - Nautical astronomy - 1826 - 732 pages
...BKLH and KCML ; but lhe. sum of these parallelograms is equal to the square BCMH, therefore the iyirn of the squares on AB and AC is equal to the square on BC. Cor. Hence in any right angled triangle, if we have the hypotenuse and one of the legs, we may easily find... | |
 | Robert Gibson - Surveying - 1832 - 290 pages
...ABDE+ACGF the sum of the squares —BKLH-\-KCML, the sum of the two parallelograms or square BCMH; therefore the sum of the squares on AB and AC is equal to the square on BC. QED* . Cor. 1. Hence the hypothenuse of a right-angled triangle may be found by having the sides :... | |
 | Robert Gibson, James Ryan - Surveying - 1839 - 452 pages
...the squares =BKLH+KCML, the sum of the two parallelograms or square BCMH; therefore the sum of ihe squares on AB and AC is equal to the square on BC. QED* Cor. 1. Hence the hypothenuso of a right-angled triangle may be found by having the sides i thus,... | |
 | Nathaniel Bowditch - 1846 - 854 pages
...JMrallelograms BKLH and KCML; but the sum of these parallelograms is equal to the square ВСЛ1Н ; therefore the sum of the squares on AB and AC is equal to the square on BC. Cor. Hence, in any right-angled triangle, if we have the hypotenuse and one of the legs, we may easily find... | |
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KCML, the sum of the two parallelograms or square BCMH ; therefore the sum of the squares on AB and AC is equal to the square on BC. 
