| Archibald Patoun - 1734 - 414 sider
...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 - 319 sider
...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 - 452 sider
...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 - 508 sider
...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 - 508 sider
...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 - 478 sider
...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 - 544 sider
...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 - 617 sider
...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 - 617 sider
...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 - 348 sider
...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.... | |
| |