 | American Oriental Society - Oriental philology - 1860 - 632 pages
...the gnomon is found by the converse process. This is simply an application of the familiar rule, that in a rightangled triangle the square of the hypothenuse is equal to the sum of the squares of the other two sides, to the triangle produced by the gnomon as perpendicular, the shadow... | |
 | Benjamin Greenleaf - 1851 - 332 pages
...perpendicular, the side AC the hypolhenuse, and the angle at B is a right angle. ART. 272. In every right angled triangle the square of the hypothenuse is equal to the sum of the squares of the base and perpendicular, as shown by the following diagram. It will be seen by examining... | |
 | Charles William Hackley - Trigonometry - 1851 - 536 pages
...and a very simple formula depending upon the well-known property of the right angled triangle, that the square of the hypothenuse is equal to the sum of the squares of the other two sides, a formula expressing the value of the sine of half an arc in terms... | |
 | Horace Mann - 1851 - 384 pages
...lines 10ft. apart. This mode of operation is founded on the property of right-angled triangles, that the square of the hypothenuse is equal to the sum of the squares of the other two sides. A roof is said to have a true pitch, when the length of each rafter... | |
 | Edward Deering Mansfield - Education - 1851 - 348 pages
...in the year five hundred and ninety before Christ, who discovered the fundamental proposition that the square of the hypothenuse is equal to the sum of the squares of the other two sides. Euclid appeared in the year three hundred BC His object was to systematize... | |
 | Jeremiah Day - Geometry - 1851 - 418 pages
...referred to. 94. Other relations of the sine, tangent, &c., may be derived from the proposition, that the square of the hypothenuse is equal to the sum of the squares of the perpendicular sides. (Euc. 47. 1.) In the right angled triangles CBG, CAD, and CHP,... | |
 | William Smyth - Algebra - 1851 - 272 pages
...the other two sides ? NOTE. In solving this and other similar questions, it will be recollected that the square of the hypothenuse is equal to the sum of the squares of the other two sides, and the area is equal to one half the product of these sideS. ANS.... | |
 | Vasiliĭ Mikhaĭlovich Golovnin - Great Britain - 1852 - 308 pages
...Japanese knew how to demonstrate geometrical truths, I asked whether they were perfectly convinced that in a right-angled triangle the square of the hypothenuse is equal to the squares of the other two sides ? He answered in^he affirmative. I then asked how they were certain... | |
 | Nicholas Patrick Wiseman - 1852 - 892 pages
...Japanese knew how to demonstrate geometrical truths, I asked whether they were perfectly convinced that in a right-angled triangle the square of the hypothenuse is equal to the squares of the other two sides ? He answered in the affirmative. I then asked how they were certain... | |
 | Thomas Kentish - Geometrical drawing - 1852 - 258 pages
...29, and raise a perpendicular BC = 17. Join AB; apply it to the scale, and it will be found 33.6. For the square of the hypothenuse is equal to the sum of the squares of the base and perpendicular. It is required to find the diameter of a copper, that, being... | |
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The square of the hypothenuse is equal to the sum of the squares of the other two sides ; as, 5033 402+302. 
