The solution of geometrical exercises, explained and illustrated; with a complete key to the School Euclid1879 |
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Side 4
... drawn out will consist of four parts , viz .: ( 1. ) The particular enunciation . [ ( 2. ) The Analysis . ] ( 3. ) The construction . ( 4. ) The demonstration . In the propositions of Euclid the Analysis is omitted , but it is evident ...
... drawn out will consist of four parts , viz .: ( 1. ) The particular enunciation . [ ( 2. ) The Analysis . ] ( 3. ) The construction . ( 4. ) The demonstration . In the propositions of Euclid the Analysis is omitted , but it is evident ...
Side 6
... drawn from a given point to a given line . 1. Particular Enunciation . - Let AB be the given straight line , and P the given point . From P let the perpendicular PQ fall upon the line AB , and let PC be any other line meeting AB in C ...
... drawn from a given point to a given line . 1. Particular Enunciation . - Let AB be the given straight line , and P the given point . From P let the perpendicular PQ fall upon the line AB , and let PC be any other line meeting AB in C ...
Side 10
... draw from a given point a straight line parallel to a given straight line . " By reference to Prop . 31 , page 57 , of the School Euclid , it will be seen that the resolution of this pro- blem depends upon the following , viz . : - - At ...
... draw from a given point a straight line parallel to a given straight line . " By reference to Prop . 31 , page 57 , of the School Euclid , it will be seen that the resolution of this pro- blem depends upon the following , viz . : - - At ...
Side 11
... draw a chord of the circle BAK , which shall be trisected by its intersection with the inner circle CDH . Analysis ... drawn so that AC = CD , then DB GEOMETRICAL EXERCISES . 11.
... draw a chord of the circle BAK , which shall be trisected by its intersection with the inner circle CDH . Analysis ... drawn so that AC = CD , then DB GEOMETRICAL EXERCISES . 11.
Side 12
Charles Mansford. = drawn so that AC = CD , then DB will also be equal to CD . The pro- blem P is there- fore resolved into ( c ) viz .: To draw from a point A in the circumference of the outer circle a secant AD to the inner circle SO ...
Charles Mansford. = drawn so that AC = CD , then DB will also be equal to CD . The pro- blem P is there- fore resolved into ( c ) viz .: To draw from a point A in the circumference of the outer circle a secant AD to the inner circle SO ...
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The Solution of Geometrical Exercises, Explained and Illustrated: With a ... Charles Mansford Uten tilgangsbegrensning - 1875 |
The Solution of Geometrical Exercises, Explained and Illustrated Charles Mansford,Euclides Ingen forhåndsvisning tilgjengelig - 2015 |
Vanlige uttrykk og setninger
ABC is greater ABCD Analysis angular points base BC bisect the base bisectors circle construct diagonal AC Draw the diagonal drawn parallel Drop the perpendicular equal to half equiangular equilateral triangle exterior angles four sides given line given point given straight line given triangle greater than AC half a right Hence hypotenuse intersect isosceles triangle joining the middle last exercise less Let AB Let ABC locus meeting BC middle point opposite angles opposite sides parallel to BC parallelogram pendicular perimeter PQRS produced proof Prop proposition quadrilateral radius rect rectangle contained rhombus right 2³ right 4º right angles right-angled triangle School Euclid Shew shewn equal Similarly solution square straight line drawn trapezium triangle ABC trisected vertex
Populære avsnitt
Side 25 - 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.
Side 29 - Divide a given straight line into two parts such that the square on one of them may be double the square on the other.
Side 20 - In any triangle, if a straight line be drawn from the vertex to the middle of the base, twice the square of this line, together with twice the square of half the base, is equivalent to the sum of the squares of the other two sides of the triangle. Let...
Side 39 - Show that the locus of a point such that the sum of the squares of its distances from two fixed points is constant, is a circle.
Side 24 - ... exterior of two concentric circles, two straight lines be drawn touching the interior and meeting the exterior ; the distance between the points of contact will be half that between the points of intersection.
Side 31 - All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.
Side 22 - ... line is bisected at the middle point of the first. 37. If through any point equidistant from two parallel straight lines, two straight lines be drawn cutting the parallel straight lines, they will intercept equal portions of these parallel straight lines. 38. If the straight line bisecting the exterior angle of a triangle be parallel to the base, shew that the triangle is isosceles. 39. Find a point B in a given straight line CD, such that if AB be drawn to B from a given point A, the angle ABC...
Side 27 - Prove that the square on any straight line drawn from the vertex of an isosceles triangle to the base, is less than the square on a side of the triangle by the rectangle contained by the segments of the base : and conversely.
Side 29 - The straight line drawn from the vertex of a triangle to the middle point of the base is less than half the sum of the remaining sides.
Side 9 - The line which joins the middle points of two sides of a triangle is parallel to the third side, and is equal to half the third side.