Elements of Geometry: With Practical Applications Designed for BeginnersD. Appleton, 1853 - 320 σελίδες |
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Αποτελέσματα 1 - 5 από τα 100.
Σελίδα 20
... coincide with AC ; then , along the other edge , upon the surface , draw the line CD . Now , reversing the square , apply the first side so as to coincide A V C B with BC ; and then , along the second side 20 ELEMENTS OF GEOMETRY .
... coincide with AC ; then , along the other edge , upon the surface , draw the line CD . Now , reversing the square , apply the first side so as to coincide A V C B with BC ; and then , along the second side 20 ELEMENTS OF GEOMETRY .
Σελίδα 20
... coincide with AC ; then , along the other edge , upon the surface , draw the line CD . Now , reversing the square , apply the first side so as to coincide A F V B C with BC ; and then , along the second side 20 ELEMENTS OF GEOMETRY .
... coincide with AC ; then , along the other edge , upon the surface , draw the line CD . Now , reversing the square , apply the first side so as to coincide A F V B C with BC ; and then , along the second side 20 ELEMENTS OF GEOMETRY .
Σελίδα 23
... draw AF , cut- ting the mirror at G , and G will be the point required . For , com- paring the triangle GFD with GBD , we have DF equal to DB , DG common , and the angle GDF equal to GDB , each being a right- angle ; therefore those ...
... draw AF , cut- ting the mirror at G , and G will be the point required . For , com- paring the triangle GFD with GBD , we have DF equal to DB , DG common , and the angle GDF equal to GDB , each being a right- angle ; therefore those ...
Σελίδα 25
... draw AC , making the an- gle BAC equal to the given angle , and AC equal to the sum of the other two sides ; join BC , and bisect it by the perpendicular DF ; finally , join BF , and ABF will be the triangle required . This is obvious ...
... draw AC , making the an- gle BAC equal to the given angle , and AC equal to the sum of the other two sides ; join BC , and bisect it by the perpendicular DF ; finally , join BF , and ABF will be the triangle required . This is obvious ...
Σελίδα 25
... draw AC , making the an- gle BAC equal to the given angle , and AC equal to the sum of the other two sides ; join BC , and bisect it by the perpendicular DF ; finally , join BF , and ABF will be the triangle required . This is obvious ...
... draw AC , making the an- gle BAC equal to the given angle , and AC equal to the sum of the other two sides ; join BC , and bisect it by the perpendicular DF ; finally , join BF , and ABF will be the triangle required . This is obvious ...
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Elements of Geometry With Practical Applications George R Perkins Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2023 |
Συχνά εμφανιζόμενοι όροι και φράσεις
a+b+c altitude angle ABC angle ACD angle BAC bisect centre chord circ circular sector circumference circumscribed polygon coincide cone consequently convex surface cylinder diagonal diameter distance draw equal and parallel equiangular equilateral triangle equivalent exterior angle ference figure formed given line greater half the arc hypothenuse inscribed circle intersection isosceles join less Let ABC line CD lines drawn magnitude measured by half meet multiplied number of sides opposite angles parallel lines parallelogram parallelopipedon pendicular perimeter perpen perpendicular plane MN point G prism PROBLEM produced Prop proportional PROPOSITION pyramid radii radius rectangle regular polygon respectively equal right-angled triangle Sabc Schol Scholium scribed semicircle semicircumference side AC similar solid angle sphere spherical polygon square straight line suppose tangent THEOREM three sides triangle ABC triangular prism vertex VIII