 | Sir Henry Edward Landor Thuillier - 1851 - 826 páginas
...of the base and perpendicular, will be the hypothenuse. Cor. 2. Having the hypothenuse and one side given to find the other; the square root of the difference of the squares of the hypothenuse and given side will be the required side. THEO. XIV. In all circles the chord of 60 degrees is always equal in... | |
 | Francis Walkingame - 1855 - 186 páginas
...top of the wall. Ans. 39-05 fed. fhc hypotfienuse and perpendicular given, to find tlie base. EULE. The square root of the difference of the squares of the hypothenuse and perpendicular is the length of the base. The base and hypotlienuse given, to find the perpendicular.... | |
 | Silas Lawrence Loomis - 1859 - 324 páginas
...difference of the squares of the Jiypothenuse and base. PROB. CUI. — To FIND THE BASE. RULE. — The square root of the difference of the squares of the hypothenuse and perpendicular. PROB. CIV. — To FIND THE AREA OF A TRIANGLE, THE BASE AND ALTITUDE BEING GIVEN. RULE.... | |
 | Francis Walkingame - 1865 - 222 páginas
...top of the wall? Ans. 39-05 feet. The hypothenuse and perpendicular given to find the base. RULE. — The square root of the difference of the squares of the hypothenuse and perpendicular is the length of the base. The base and hypothenuse given to find the perpendicular.... | |
 | Richard Nelson - 1867 - 304 páginas
...square root of the sum of the squares of the base and perpendicular will give the hypothenuse. And the square root of the difference ,of the squares of the hypothenuse and either of the two sides will give the third side. 22. What length of a ladder will reach across a 15-foot... | |
 | Francis Walkingame - 1868 - 154 páginas
...wall. Ans. 29-05 feet, and 975 rem. The Hypoihenuse and Perpendicular given to find the Base. RULE. — The square root of the difference of the squares of the hypothenuse and perpendicular is the length of the base. (8) There is a tower of 103 feet in height, standing close... | |
 | Henry Bartlett Maglathlin - 1869 - 332 páginas
...the other two sides; and . 3. Hither of the two shorter sides of a right-angled triangle is equal to the square root of the difference of the squares of the hypothenuse and the other side. Exercises. 1. If the base of a right-angled triangle is 60 feet, and the perpendicular... | |
 | David Munn - 1873 - 160 páginas
...-AC" 2 (2) = v/(AB +AC)(AB-AC) (3). Interpreting formula (2), we find the remaining side, by taking the square root of the difference of the squares of the hypothenuse and given side. Interpreting formula (3), we find the remaining side, by taking the square root of the product... | |
 | Henry Bartlett Maglathlin - 1873 - 362 páginas
...the other two sides; and 3. Either of the two shorter sides of a right-angled triangle is equal to the square root of the difference of the squares of the hypothenuse and the other side. Exercises. 1. If the base of a right-angled triangle is 60 feet, and the perpendicular... | |
 | William Edward Armytage Axon - 1875 - 504 páginas
...number of times the base line is divided into, which will give the exact length of the hypothenuse. Then the square root of the difference of the squares of the hypothenuse and the base is the height of the perpendicular. The sketch F of the sliding staff will explain itself.... | |
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The hypothenuse of a right-angled triangle may be found by having the other two sides ; thus, the square root of the sum of the squares of the base and perpendicular, will be the hypothenuse. Cor. 2. Having the hypothenuse and one side given to find the... 
