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 Bøker Bok 1–10 av 22 på ABDE+ACGF the sum of the squares —BKLH-\-KCML, the sum of the two parallelograms...
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.
A Treatise of Practical Surveying, ... - Side 37
av Robert Gibson - 1808 - 440 sider
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## A Compleat Treatise of Practical Navigation Demonstrated from It's First ...

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...
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## A Treatise of Practical Surveying: Which is Demonstrated from Its First ...

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,...
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## A Treatise of Practical Surveying: Which is Demonstrated from Its First ...

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...
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## The Theory and Practice of Surveying: Containing All the Instructions ...

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 ;...
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## The Theory and Practice of Surveying: Containing All the Instructions ...

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...
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## A Treatise on Practical Surveying: Which is Demonstrated from Its First ...

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...
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## The Theory and Practice of Surveying: Containing All the Instructions ...

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...
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## The New American Practical Navigator: Being an Epitome of Navigation ...

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...
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## The New American Practical Navigator: Being an Epitome of Navigation ...

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...
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## The Theory and Practice of Surveying: Containing All the Instructions ...

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....
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