Plane Geometry: With Problems and ApplicationsAllyn and Bacon, 1910 - 280 sider |
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Side 11
... Altitude Base the base or the base produced . Evidently any side may be taken as the base , and hence a triangle has three different altitudes . 25 . EXERCISES . 1. Is every equilateral triangle also RECTILINEAR FIGURES . 11.
... Altitude Base the base or the base produced . Evidently any side may be taken as the base , and hence a triangle has three different altitudes . 25 . EXERCISES . 1. Is every equilateral triangle also RECTILINEAR FIGURES . 11.
Side 13
... Hence if we make a pattern of a figure , say on tracing paper , and then make a second figure from this pattern , the two figures are congruent to each other . 29. If ABC≈AA'B'C ' , the notation of the triangles may be so arranged that ...
... Hence if we make a pattern of a figure , say on tracing paper , and then make a second figure from this pattern , the two figures are congruent to each other . 29. If ABC≈AA'B'C ' , the notation of the triangles may be so arranged that ...
Side 14
... Hence , side BC will coincide with B'C ' ( § 8 ) . Thus , the two triangles coincide throughout and hence are congruent ( § 27 ) . The process just used is called superposition . It B ' may sometimes be necessary to move a figure out of ...
... Hence , side BC will coincide with B'C ' ( § 8 ) . Thus , the two triangles coincide throughout and hence are congruent ( § 27 ) . The process just used is called superposition . It B ' may sometimes be necessary to move a figure out of ...
Side 16
... hence C must lie on the ray B'C ' . Since the point c lies on both of the rays A'C ' and B'C ' , it must lie at their point of intersection d ' ( § 5 ) . Hence , the triangles coincide and are , therefore , congruent ( § 27 ) . 36 ...
... hence C must lie on the ray B'C ' . Since the point c lies on both of the rays A'C ' and B'C ' , it must lie at their point of intersection d ' ( § 5 ) . Hence , the triangles coincide and are , therefore , congruent ( § 27 ) . 36 ...
Side 19
... Hence , A ABCAA'B'C ' . ( § 28 ) Make an outline of the steps in the above argument , and see that each step is needed in deriving the next . 41. Definition . If one triangle is congruent to another because certain parts of one are ...
... Hence , A ABCAA'B'C ' . ( § 28 ) Make an outline of the steps in the above argument , and see that each step is needed in deriving the next . 41. Definition . If one triangle is congruent to another because certain parts of one are ...
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Plane Geometry: With Problems and Applications Herbert Ellsworth Slaught,Nels Johann Lennes Uten tilgangsbegrensning - 1910 |
Plane Geometry: With Problems and Applications Herbert Ellsworth Slaught,Nels Johann Lennes Uten tilgangsbegrensning - 1910 |
Vanlige uttrykk og setninger
ABCD accompanying design altitude apothem arcs area bounded axes of symmetry axioms base and altitude bisectors bisects central angle chord circumference congruent corresponding sides definite diagonal diameter divided Draw drawn equal angles equal circles equilateral triangle EXERCISES exterior angle feet Find the area Find the locus Find the radius fixed point geometric Give the proof given circle given point given triangle greater Hence hypotenuse inscribed intersect isosceles triangle length line-segment measure meet middle points number of sides Outline of Proof parallel lines parallelogram perimeter perpendicular proof in full Prove quadrilateral radii ratio rectangle regular dodecagon regular hexagon regular octagon regular polygon regular triangle respectively rhombus right angle right triangle secant semicircle Show shown straight angle straight line strip subtend SUGGESTION tangent THEOREM tile trapezoid triangle ABC unit vertex vertices width
Populære avsnitt
Side 215 - If two triangles have two sides of the one equal to two sides of the other...
Side 35 - An exterior angle of a triangle is equal to the sum of the two opposite interior angles.
Side 22 - Any side of a triangle is less than the sum of the other two sides...
Side 113 - Sines that the bisector of an angle of a triangle divides the opposite side into parts proportional to the adjacent sides.
Side 273 - This textbook may be borrowed for two weeks, with the privilege of renewing it once. A fine of five cents a day is incurred by failure to return a book on the date when it is due. The Education Library is open from 9 to 5 daily except Saturday when it closes at 12.30.
Side 52 - The straight line joining the middle points of two sides of a triangle is parallel to the third side and equal to half of it 46 INTERCEPTS BY PARALLEL LINES.
Side 174 - The areas of two regular polygons of the same number of sides are to each other as the squares of their radii, or as the squares of their apothems.
Side 153 - The formula states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the base and altitude.
Side 219 - Find the locus of a point such that the difference of the squares of its distances from two fixed points is a constant.
Side 202 - The area of a rectangle is equal to the product of its base and altitude.