 | Archibald Patoun - 1734 - Liczba stron: 568
...the Parallelograms BKLH and KCML-, but the Sum of thefe Parallelograms is equal to the Square B CMH, therefore the Sum of the Squares on AB and AC is equal to the Square on BC. Cor. i. Hence in a rightangled Triangle, the Hypothenufe and one of the Legs being given, we may eafily... | |
 | Robert Gibson - 1795 - Liczba stron: 384
...proved, that the Sqv.are ACGF, is equal to the Parallelogram KCLM. So AEDE.f ACGF the Sum of the Squares, Plate I. Squares, = BKLH+KCML, the Sum of the two...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 - 1806 - Liczba stron: 486
...ACGF, is equal to the parallelogram KCLM. So ABDE + ACGF the sum of the Plate I. squares = BELH + KCML, the sum of the two parallelograms or square BCMH ;...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 square... | |
 | Robert Gibson - 1811 - Liczba stron: 580
...square ACGF, is equal to the parallelogram KCLM. So AB DE + ACGF the sum of the &quvces=BKLH+KCML, the sum of the two parallelograms or square BCMH ;...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 - 1814 - Liczba stron: 558
...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... | |
 | Robert Gibson - 1818 - Liczba stron: 502
...may be proved, that the square ACGF, is equal to the parallelogram KCLM. So ABDE+ACGKF the sum of the squares, =BKLH+KCML, the sum of the two parallelograms...on AB and AC is equal to the square on BC. QED Cor. i. Hence the hypothenuse of a right-angled triangle may be found by having the legs ; thus, the square... | |
 | Robert Gibson - 1821 - Liczba stron: 594
...may be proved, that the square ACGF, is equal to the parallelogram KCLM So ABDE+ACGF the sum of the squares =* BKLH+KCML, the sum of the two parallelograms...square BCMH ; therefore the sum of the squares on A Band AC is equal to the square on BC. QE D* Cor. 1. Hence the hypothenuse of a right-angled triangle... | |
 | Nathaniel Bowditch - 1826 - Liczba stron: 764
...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... | |
 | Nathaniel Bowditch - 1826 - Liczba stron: 732
...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... | |
 | Robert Gibson - 1832 - Liczba stron: 290
...may be proved that the square ACGF is equal to the parallelogram KCLM. So ABDE+ACGF the sum of the squares —BKLH-\-KCML, the sum of the two parallelograms...a right-angled triangle may be found by having the sides : thus, the square root of the sum of the squares of the base and perpendicular will be the hypothenuse.... | |
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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. 
