Plane Geometry: With Problems and ApplicationsAllyn and Bacon, 1910 - 280 sider |
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Resultat 1-5 av 55
Side 9
... bisector . is the bisector of a straight angle . 20 . EXERCISES . Obtuse Angle Acute Angle Oblique Line B A B Thus a perpendicular 1. Since we can always place two straight angles so as to make them coincide , what can we say as to ...
... bisector . is the bisector of a straight angle . 20 . EXERCISES . Obtuse Angle Acute Angle Oblique Line B A B Thus a perpendicular 1. Since we can always place two straight angles so as to make them coincide , what can we say as to ...
Side 10
... bisector . 2/1 D 4 B 5. How many rays perpendicular to BD at the point A can be drawn on the same side of BD ? Does the answer to this question depend upon the answers to the questions in Ex . 4 ? How ? 6. Pick out three acute angles ...
... bisector . 2/1 D 4 B 5. How many rays perpendicular to BD at the point A can be drawn on the same side of BD ? Does the answer to this question depend upon the answers to the questions in Ex . 4 ? How ? 6. Pick out three acute angles ...
Side 22
... = ZA'B'C ' . 50. Definition . A line which is perpendicular to a line- segment at its middle point is called the perpendicular bisector of the segment . 51. PROBLEM . To construct the perpendicular bisec- tor of 22 PLANE GEOMETRY .
... = ZA'B'C ' . 50. Definition . A line which is perpendicular to a line- segment at its middle point is called the perpendicular bisector of the segment . 51. PROBLEM . To construct the perpendicular bisec- tor of 22 PLANE GEOMETRY .
Side 23
... bisector of AB . To prove this , show that △ ACD ≈ △ BCD . Hence < 3 = 24 . By what test can it now be shown that Hence △ AOCA BOC ? Z1 = 22 . ( Why ? ) ( Why ? ) Therefore co ( or CD ) is perpendicular to AB ( Why ? ) and also A0 ...
... bisector of AB . To prove this , show that △ ACD ≈ △ BCD . Hence < 3 = 24 . By what test can it now be shown that Hence △ AOCA BOC ? Z1 = 22 . ( Why ? ) ( Why ? ) Therefore co ( or CD ) is perpendicular to AB ( Why ? ) and also A0 ...
Side 24
... bisector of AB , and hence is perpendicular to l from the point P. 55. PROBLEM . To construct a triangle when two sides and the included angle are given . SOLUTION . Let b and c be the given sides , and A the given angle . As in § 47 ...
... bisector of AB , and hence is perpendicular to l from the point P. 55. PROBLEM . To construct a triangle when two sides and the included angle are given . SOLUTION . Let b and c be the given sides , and A the given angle . As in § 47 ...
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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.