Elements of Geometry: With, Practical ApplicationsD. Appleton and Company, 1850 - 320 páginas |
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Página 13
... joining figure , it is obvious that the entire space will be divided into four equal portions , each being a right - angle . Therefore the en- tire circumference may , with great propriety , be taken as the measure of 360 ° , or four ...
... joining figure , it is obvious that the entire space will be divided into four equal portions , each being a right - angle . Therefore the en- tire circumference may , with great propriety , be taken as the measure of 360 ° , or four ...
Página 16
... joining the vertices of two angles , not adjacent . ( 24. ) From the above definitions , in connection with the diagrams , it will be readily seen that the triangle has no diagonal , the quadri- lateral has two diagonals , the pentagon ...
... joining the vertices of two angles , not adjacent . ( 24. ) From the above definitions , in connection with the diagrams , it will be readily seen that the triangle has no diagonal , the quadri- lateral has two diagonals , the pentagon ...
Página 23
... Join CD , and bisect it by the perpendicular FG ; then will the point G be the point sought . For , if we join GC , GD , we shall have the triangle GFC equal to GFD , since the side FC is equal to FD , the side FG common , and the angle ...
... Join CD , and bisect it by the perpendicular FG ; then will the point G be the point sought . For , if we join GC , GD , we shall have the triangle GFC equal to GFD , since the side FC is equal to FD , the side FG common , and the angle ...
Página 24
... Join CB , and bisect it by the perpendicular DF ; then will F be the point at which the tree must break . For , joining BF , and comparing the triangle FBD with the triangle FCD , we see that the side DB is equal to DC , the side FD ...
... Join CB , and bisect it by the perpendicular DF ; then will F be the point at which the tree must break . For , joining BF , and comparing the triangle FBD with the triangle FCD , we see that the side DB is equal to DC , the side FD ...
Página 25
... join BC , and bisect it by the perpendicular DF ; finally , join BF , and ABF will be the triangle required . This is obvious from what has already been done . PROPOSITION IV . THEOREM . If two triangles have two angles and the in ...
... join BC , and bisect it by the perpendicular DF ; finally , join BF , and ABF will be the triangle required . This is obvious from what has already been done . PROPOSITION IV . THEOREM . If two triangles have two angles and the in ...
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Términos y frases comunes
a+b+c altitude angle ABC angle BAC angle BCD bisect centre chord circ circular sector circumference circumscribed polygon coincide cone consequently convex surface cylinder denote diagonal diameter dicular distance draw equal and parallel equiangular equilateral triangle equivalent exterior angle figure formed given line greater half the arc hypothenuse inscribed circle intersection isosceles join less Let ABC line AC line CD lines drawn measured by half meet multiplied number of sides parallel planes parallelogram parallelopipedon pendicular perimeter perpen perpendicular plane MN point G prism PROBLEM produced Prop PROPOSITION pyramid radii radius rectangle regular polygon respectively equal right-angled triangle Sabc Schol Scholium scribed semicircle semicircumference side AC similar similar triangles solid angle sphere spherical triangle square straight line suppose tangent THEOREM three sides triangle ABC triangular prism vertex VIII
Pasajes populares
Página 231 - THE sphere is a solid terminated by a curve surface, all the points of which are equally distant from a point within, called the centre.
Página 147 - PROBLEM. To inscribe a circle in a given triangle. Let ABC be the given triangle : it is required to inscribe a circle in the triangle ABC.
Página 17 - A circle is a plane figure contained by one line, which is called the circumference, and is such, that all straight lines drawn from a certain point within the figure to the circumference are equal to one another.
Página 28 - If two sides and the included angle of the one are respectively equal to two sides and the included angle of the other...
Página 233 - The volume of a cylinder is equal to the product of its base by its altitude. Let the volume of the cylinder be denoted by V, its base by B, and its altitude by H.
Página 276 - THEOREM. Two triangles on the same sphere, or on equal spheres, are equal in all their parts, when they have each an equal angle included between equal sides. Suppose the side...
Página 120 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. D c A' D' Hyp. In triangles ABC and A'B'C', ZA = ZA'. To prove AABC = ABxAC. A A'B'C' A'B'xA'C' Proof. Draw the altitudes BD and B'D'.
Página 18 - If equals be added to unequals, the wholes are unequal. V. If equals be taken from unequals, the remainders are unequal. VI. Things which are double of the same, are equal to one another.
Página 232 - A spherical triangle is a portion of the surface of a sphere, bounded by three arcs of great circles.
Página 96 - Similar figures, are those that have all the angles of the one equal to all the angles of the other, each to each, and the sides about the equal angles proportional.