| John Martin Frederick Wright - Astronomy - 1831 - 282 pages
...of the squares be joined ; the sum of the squares of the sides of the hexagonal figure thus formed is equal to four times the sum of the squares of the sides of the triangle. 14,. If all the angular points of a regular polygon of n sides be joined, and... | |
| Samuel Earnshaw - Statics - 1845 - 290 pages
...+ BC* + BD* + CD1 - *(AG* + BGF + CG2 + Hence the sum of the squares of the six edges of a pyramid is equal to four times the sum of the squares of the distances of its angular points from its centre of gravity. 168. When a system of bodies is in equilibrium... | |
| Alexander Jamieson - Mechanics - 1845 - 572 pages
...(a* + J'+c'+^ + S' + S'2), that 18 In any triangular pyramid, the sum of the squares of its six edges is equal to four times the sum of the squares of the distances of the centre of gravity from each angle of the figure. This property of the triangular pyramid... | |
| Euclides - 1845 - 546 pages
...of the squares be joined ; the sum of the squares of the sides of the hexagonal figure thus formed is equal to four times the sum of the squares of the sides of the triangle. GEOMETRICAL EXERCISES ON BOOK III. THEOREM I. If AB, CD be chords of a circle... | |
| John Narrien - Conic sections - 1846 - 252 pages
...proposition it may be proved, that the sum of the squares of the four diagonals of the parallelepiped BN is equal to four times the sum of the squares of the three edges about one of the solid angles ; that is, We have already, as above, the value of OP2. Now... | |
| 1847 - 364 pages
...with much regret. LXXXVI. By Amicus. In every tetrahedron, the sum of the squares of the six edges is equal to four times the sum of the squares of the lines which join the middles of the opposite edges. [FntST SOLUTION. Mr. JW Elliott, Greatham.~\ Let ABCD... | |
| Royal Military Academy, Woolwich - Mathematics - 1853 - 400 pages
...together with double the square on half the base. 8. Three times the sum of the squares on the three sides of a triangle is equal to four times the sum of the squares of the Hues drawn from the angles to bisect the opposite sides. 9. The difference between the squares on two... | |
| John Hind - Trigonometry - 1855 - 546 pages
...of the squares be joined : the sum of the squares of the sides of the hexagonal figure thus formed, is equal to four times the sum of the squares of the sides of the triangle. 23. If all the angular points of a regular polygon be joined, and r be the radius... | |
| 1857 - 486 pages
...triangle, and the angular points be joined, the sum of the squares of the hexagonal figure thus formed, is equal to four times the sum of the squares of the sides of the triangle. 5. If from a point P, in the diameter AB of a circle, PQ, PR be drawn to the... | |
| Isaac Todhunter - Spherical trigonometry - 1859 - 156 pages
...be inscribed in the two solids. 10. The sum of the squares of the four diagonals of a parallelopiped is equal to four times the sum of the squares of the edges. 11. If with each angular point of any parallelopiped as centres equal spheres be described,... | |
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