Plane and Solid GeometryGinn, 1899 - 473 sider |
Inni boken
Resultat 1-5 av 31
Side vii
... OBLIQUE LINES PARALLEL LINES TRIANGLES LOCI OF POINTS . QUADRILATERALS POLYGONS IN GENERAL SYMMETRY . METHODS OF PROVING THEOREMS EXERCISES . 7 8 9 12 13 15 24 30 44 47 56 60 64 • 68 BOOK II . THE CIRCLE . DEFINITIONS ARCS , CHORDS vii.
... OBLIQUE LINES PARALLEL LINES TRIANGLES LOCI OF POINTS . QUADRILATERALS POLYGONS IN GENERAL SYMMETRY . METHODS OF PROVING THEOREMS EXERCISES . 7 8 9 12 13 15 24 30 44 47 56 60 64 • 68 BOOK II . THE CIRCLE . DEFINITIONS ARCS , CHORDS vii.
Side 43
... Both these conclusions are contrary to the hypothesis that BC is greater than EF . Since the A is not equal to the D or less than the D , the △ A is greater than the △ D. Q. E. D. LOCI OF POINTS . 156. If it is required to TRIANGLES . 43.
... Both these conclusions are contrary to the hypothesis that BC is greater than EF . Since the A is not equal to the D or less than the D , the △ A is greater than the △ D. Q. E. D. LOCI OF POINTS . 156. If it is required to TRIANGLES . 43.
Side 44
... locus of points that fulfil a given condition , it is necessary to prove 1. Any point in the line satisfies the ... locus ( pronounced lo ́kus ) is a Latin word that signi- fies place . The plural of locus is loci ( pronounced lo ́si ) ...
... locus of points that fulfil a given condition , it is necessary to prove 1. Any point in the line satisfies the ... locus ( pronounced lo ́kus ) is a Latin word that signi- fies place . The plural of locus is loci ( pronounced lo ́si ) ...
Side 45
... locus of points equidistant from the extremities of the line . D B P Let PR be the perpendicular bisector of the line AB , O any point in PR , and C any point not in PR . Draw OA and OB , CA and CB . To prove OA and OB equal , CA and CB ...
... locus of points equidistant from the extremities of the line . D B P Let PR be the perpendicular bisector of the line AB , O any point in PR , and C any point not in PR . Draw OA and OB , CA and CB . To prove OA and OB equal , CA and CB ...
Side 46
... locus of points equidistant from the sides of the angle . F B / P G Let O be any point equidistant from the sides of the angle PAQ . To prove that O is in the bisector of the △ PAQ . Proof . Suppose OF drawn Draw AO . to AP and OG 1 to ...
... locus of points equidistant from the sides of the angle . F B / P G Let O be any point equidistant from the sides of the angle PAQ . To prove that O is in the bisector of the △ PAQ . Proof . Suppose OF drawn Draw AO . to AP and OG 1 to ...
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Vanlige uttrykk og setninger
ABCD AC² adjacent angles altitude angles are equal apothem axis bisector bisects called centre chord circumference circumscribed coincide construct cylinder denote diagonals diameter dihedral angles divided Draw equiangular equidistant equilateral triangle equivalent exterior angle feet Find the area Find the locus frustum given circle given line given point given straight line greater Hence homologous homologous sides hypotenuse inches inscribed intersecting isosceles triangle lateral area lateral edges limit middle point number of sides opposite parallel lines parallelogram parallelopiped perimeter perpendicular plane MN polyhedron prism prismatoid Proof proportional prove Q. E. D. PROPOSITION quadrilateral radii radius ratio rectangle regular polygon respectively rhombus right angle right circular right triangle segments similar slant height sphere spherical square surface tangent tetrahedron THEOREM trapezoid triangle ABC triangle is equal trihedral vertex vertices
Populære avsnitt
Side 66 - If two triangles have two sides of the one equal respectively to two sides of the other, but the included angle of the first greater than the included angle of the second, then the third side of the first is greater than the third side of the second. Given A ABC and A'B'C...
Side 274 - If two planes are perpendicular to each other, a straight line drawn in one of them perpendicular to their intersection is perpendicular to the other plane.
Side 372 - Each side of a spherical triangle is less than the sum of the other two sides. Let ABC be a spherical triangle, AB the longest side.
Side 385 - If two angles of a spherical triangle are unequal, the sides opposite are unequal,, and the greater side is opposite the greater angle...
Side 191 - 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.
Side 360 - A sphere is a solid bounded by a surface all points of which are equally distant from a point within called the centre.
Side 383 - Hence the two last are right angles ; hence the arc drawn from the vertex of an isosceles spherical triangle to the middle of the base, is at right angles to the base, and bisects the vertical angle.
Side 75 - PERIPHERY of a circle is its entire bounding line ; or it is a curved line, all points of which are equally distant from a point within called the centre.
Side 150 - If two triangles have an angle of the one equal to an angle of the other, and the including sides proportional, they are similar. In the triangles ABC and A'B'C', let ZA = Z A', and let AB : A'B' = AC : A'C'. To prove that the A ABC and A'B'C
Side 376 - The sum of the angles of a spherical triangle is greater than two and less than six right angles ; that is, greater than 180° and less than 540°. (gr). If A'B'C' is the polar triangle of ABC...