The solution of geometrical exercises, explained and illustrated; with a complete key to the School Euclid1879 |
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Side 14
... point within the circle is at a less distance from the centre than any point in the circumference , while any point ... middle of the line joining two fixed points , it may be shewn by Euc . I. 4 , that any point in this line is equally ...
... point within the circle is at a less distance from the centre than any point in the circumference , while any point ... middle of the line joining two fixed points , it may be shewn by Euc . I. 4 , that any point in this line is equally ...
Side 15
... point ; but if one of them is a circle , their intersection will in general produce two points . Now in problems ... middle point P draw P S PQ at right to AB . Then the required point lies in PQ . 3. Similarly if we bisect BC by the ...
... point ; but if one of them is a circle , their intersection will in general produce two points . Now in problems ... middle point P draw P S PQ at right to AB . Then the required point lies in PQ . 3. Similarly if we bisect BC by the ...
Side 19
... middle points of two sides of a triangle meet in a point which is equi- distant from the three angles of the triangle . 2. Find a point within an acute - angled triangle from which a circle may be described about the tri- angle . 3. Two ...
... middle points of two sides of a triangle meet in a point which is equi- distant from the three angles of the triangle . 2. Find a point within an acute - angled triangle from which a circle may be described about the tri- angle . 3. Two ...
Side 20
... middle point of the base . 17. The sum of the straight lines drawn from any point within a triangle to each of its angles is less than the sum and greater than half the sum of the sides of the triangle . 18. Show that the last exercise ...
... middle point of the base . 17. The sum of the straight lines drawn from any point within a triangle to each of its angles is less than the sum and greater than half the sum of the sides of the triangle . 18. Show that the last exercise ...
Side 22
... point . 36. If the sides of any hexagon be produced to meet the angles formed by their intersection they will be together equal to four right angles . 37. The middle point of the hypotenuse of a right- angled triangle is equally distant ...
... point . 36. If the sides of any hexagon be produced to meet the angles formed by their intersection they will be together equal to four right angles . 37. The middle point of the hypotenuse of a right- angled triangle is equally distant ...
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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.