Elements of GeometryGinn and Heath, 1881 - 250 σελίδες |
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Σελίδα 6
... Equivalent Figures . Figures which have the same form and extent are called Equal Figures . • Opiem . ON STRAIGHT LINES . 18. If the direction of a straight line and a point in the line be known , the position of the line is known ...
... Equivalent Figures . Figures which have the same form and extent are called Equal Figures . • Opiem . ON STRAIGHT LINES . 18. If the direction of a straight line and a point in the line be known , the position of the line is known ...
Σελίδα 171
... equivalents in the last proportion . АН : АВ :: НВ : АН . Then Whence , by inversion , A B AH AH HB . $ 263 .. A B is divided at H in extreme and mean ratio . Omitte #ABAD : 1 Q. E. F. REMARK . AB is said to be divided at , internally ...
... equivalents in the last proportion . АН : АВ :: НВ : АН . Then Whence , by inversion , A B AH AH HB . $ 263 .. A B is divided at H in extreme and mean ratio . Omitte #ABAD : 1 Q. E. F. REMARK . AB is said to be divided at , internally ...
Σελίδα 177
... Equivalent figures are figures which have equal areas . REM . In comparing the areas of equivalent figures the symbol ( ) is to be read " equal in area . " PROPOSITION III . THEOREM . 319. The area of a COMPARISON AND MEASUREMENT OF ...
... Equivalent figures are figures which have equal areas . REM . In comparing the areas of equivalent figures the symbol ( ) is to be read " equal in area . " PROPOSITION III . THEOREM . 319. The area of a COMPARISON AND MEASUREMENT OF ...
Σελίδα 180
... equivalent . 326. COR . 2. Triangles having equal bases are to each other as their altitudes ; triangles having equal altitudes are to each other as their bases ; any two triangles are to each other as the product of their bases by ...
... equivalent . 326. COR . 2. Triangles having equal bases are to each other as their altitudes ; triangles having equal altitudes are to each other as their bases ; any two triangles are to each other as the product of their bases by ...
Σελίδα 183
... equivalent to the square described on the hypotenuse . C A B 0 Let ABC be a right triangle with its right angle at C. We are to prove Draw COL to A B. Then A C2 + C B2 = A B2 AC2 = AO × A B , $ 289 ( the square on a side of a rt . △ is ...
... equivalent to the square described on the hypotenuse . C A B 0 Let ABC be a right triangle with its right angle at C. We are to prove Draw COL to A B. Then A C2 + C B2 = A B2 AC2 = AO × A B , $ 289 ( the square on a side of a rt . △ is ...
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Συχνά εμφανιζόμενοι όροι και φράσεις
A B C AABC AACB ABCD acute adjacent angles alt.-int altitude apothem arc A B BC² bisect centre chord A B circumference circumscribed coincide COROLLARY describe an arc diagonals diameter divided Draw equal arcs equal distances equal respectively equiangular polygon equilateral equilateral polygon equivalent exterior angles figure given line given point given polygon greater homologous sides hypotenuse Iden isosceles triangle Let A B limit line A B measured by arc middle point number of sides parallelogram perimeter perpendicular plane PROBLEM prove Q. E. D. PROPOSITION quadrilateral radii radius equal ratio rect rectangles regular inscribed regular polygon required to construct rhombus right angles right triangle SCHOLIUM segment sides of equal sides of similar similar polygons subtend tangent THEOREM third side triangle ABC variable vertex vertices