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" The square described on the hypothenuse of a rightangled triangle is equal to the sum of the squares described on the other two sides. "
Elements of Drawing and Mensuration Applied to the Mechanic Arts: A Book for ... - Page 23
by Charles Davies - 1846 - 240 pages
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Elements of Geometry and Trigonometry

Adrien Marie Legendre - Geometry - 1852 - 436 pages
...right-angled triangle, right-angled at A : then will the square described on the hypothenuse BC be equivalent to the sum of the squares described on the other two sides, BA, AC. FGI H D Haying described a square on each of the three sides, let fall from A, on the hypothenuse,...
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A Theoretical and Practical Arithmetic: Designed for Common Schools and ...

Daniel Leach - Arithmetic - 1853 - 622 pages
...sides the base and perpendicular. 293. The square described on the hypothenuse, or longest side, is equal to the sum of the squares described on the other two sides. Thus, suppose the longest side is 10 ft., the base 6 ft., and the perpendicular 8 ft. 10a=100. 6a=36. 82+36=100....
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Elements of Geometry and Trigonometry from the Works of A.M. Legendre ...

Charles Davies - Geometry - 1854 - 436 pages
...triangle, right•angled at A : then will the square described on the hypothenuse BC be equivalent to the sum of the squares described on the other two sides, BA, AC. 1 GEOMETRY. Having described a square on each of the three sides, let fall from A, on the hy•...
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A Treatise on Arithmetic, Combining Analysis and Synthesis: Adapted to the ...

James Stewart Eaton - Arithmetic - 1857 - 376 pages
...or circumferences. FIG. 12. 6. The square described on the hypothenuse of a right-angled triangle is equal to the sum of the squares described on the other two sides. This will be seen by counting the small squares in the square of the hypothenuse and those in the squares...
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The Mathematical Monthly, Volume 1

John Daniel Runkle - Mathematics - 1859 - 478 pages
...BT JAMES IIIUVAIIII OLIVER. The square described on t/te hypothenusc of a right-angled triangle is equal to the sum of the squares described on the other two sides. Drop a perpendicular from the right angle to the hypothenuse, and prove in the usual way that the two...
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The Pasha papers, epistles of Mohammed Pasha, tr. into Anglo-Amerian [really ...

William Wirt Howe - 1859 - 324 pages
...confusion to the fact that the square described on the hypothenuse of a right-angled triangle is equivalent to the sum of the squares described on the other two sides ; or the able Editors should denounce the incoming flow of a spring tide as an altogether unprecedented...
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Higher Arithmetic : Or, The Science and Application of Numbers: Combining ...

James Bates Thomson - Arithmetic - 1860 - 440 pages
...29. 30. 207*?. 34967 A371 578. The square described on the hypothenu.se of a rightangled triangle, is equal to the sum of the squares described on the other two sides. (Thomson's Legendre, B. IV. 11, Euc. I. 47.) The Irii/h of Ms principle may be seen from, the following...
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Iconographic Encyclopaedia of Science, Literature & Art, Volume 1

Johann Georg Heck - Encyclopedias and dictionaries - 1860 - 332 pages
...this proposition is known as the Pythagorean: the square described upon the hypothenuse is equivalent to the sum of the squares described on the other two sides. As the unit of measure for the determination of the superficial relations of figures, we use a square...
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Davies' University Arithmetic: Embracing the Answers, and a Full Analysis ...

Charles Davies - Arithmetic - 1861 - 496 pages
...angles to each other. 384. In a right-angled triangle the square described on thr Base. hypothenuse is equal to the sum of the squares described on the other two sides. Thus, if ACB be a right-angled triangle, right-angled at C, -then will the large square, D, described in the...
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Sabbath Evening Readings on the New Testament, Volume 12

John Cumming - 1861 - 540 pages
...first book of Euclid, that the square described on the hypothenuse of any right-angled triangle is equal to the sum of the squares described on the other two sides — I remember I could prove that step by step ; but I have been so much out of the way of mathematics...
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