An elementary course of mathematics, Volum 2 |
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Resultat 1-5 av 36
Side 5
... shown that RM , RN , RO , & c . are all equal ; and there- fore a circle described with the radius RK will circumscribe the poly- gon IKLMNOPQ . And since the equilateral polygons ABCDEFGH , IKLMNOPQ are inscribed in circles , their ...
... shown that RM , RN , RO , & c . are all equal ; and there- fore a circle described with the radius RK will circumscribe the poly- gon IKLMNOPQ . And since the equilateral polygons ABCDEFGH , IKLMNOPQ are inscribed in circles , their ...
Side 7
... shown that the excess of Ć above I is less than eight times MR , with much greater reason is the excess of C above I less than DE . Wherefore in a given circle , & c .: which was to be proved . Cor . 1. Because the circumference of the ...
... shown that the excess of Ć above I is less than eight times MR , with much greater reason is the excess of C above I less than DE . Wherefore in a given circle , & c .: which was to be proved . Cor . 1. Because the circumference of the ...
Side 8
... shown that no polygon can be inscribed in the circle but that its perimeter will be less than the circumference of the circle by a line greater than Z , which is likewise absurd . Therefore the straight line S and the circumference are ...
... shown that no polygon can be inscribed in the circle but that its perimeter will be less than the circumference of the circle by a line greater than Z , which is likewise absurd . Therefore the straight line S and the circumference are ...
Side 9
... shown that P is to its excess above Q , as the square of AC to the square of AF , and the square of AC is greater than P , the square of AF is greater than the excess of P above Q ; with much greater reason is the square of AE , that is ...
... shown that P is to its excess above Q , as the square of AC to the square of AF , and the square of AC is greater than P , the square of AF is greater than the excess of P above Q ; with much greater reason is the square of AE , that is ...
Side 10
... shown to be less than the area of any polygon described about the circle ; therefore the rectangle LA , AS is equal to the area of the circle ABC ( Prop . 8. Cor . 2 ) ; that is , the area of the circle ABC is equal to the rectangle ...
... shown to be less than the area of any polygon described about the circle ; therefore the rectangle LA , AS is equal to the area of the circle ABC ( Prop . 8. Cor . 2 ) ; that is , the area of the circle ABC is equal to the rectangle ...
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ABCD allel altitude angle formed angle of inclination auxiliary plane circle described circumference circumscribed coincide cone consequently construction Descriptive Geometry determined diameter dicular dihedral angle contained distance ellipse equal and similar equal bases equilateral polygon faces ASB figure given angle given plane given point given straight line greater hemisphere horizontal plane horizontal projection horizontal trace inscribed isometric line joining line of level line parallel meets the plane parallel planes parallel to xy parallelepiped parallelogram pendicular perimeter perpen perpendicular to xy plane angles plane MN plane passing plane Prop planes BM planes of projection point of intersection prism Prob PROBLEM projecting plane pyramid rectangle right angles right-angled triangle scale of slope series of cylinders sides solid angle space straight line drawn THEOR third face trihedral vertical plane vertical projection vertical trace Wherefore
Populære avsnitt
Side 5 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz.
Side 18 - FD. Join AC, BD, AD, and let AD meet the plane KL in the point X; and join EX, XF. Because the two parallel planes KL, MN are cut by the plane EBDX, the common sections EX, BD are parallel (Prop.
Side 13 - For the same reason, CD is likewise at right angles to the plane HGK. Therefore AB, CD are each of them at right angles to the plane HGK.
Side 4 - BC above it : and since the straight line AB is in the plane, it can be produced in that plane : let it be produced to D ; and let any plane pass through the straight line AD, and be turned about it until it pass through the point C; and because the points B, C, are in this plane, the straight line* BC is in it: »7Def.1.
Side 9 - Note. (3. 11.) line; let this be BF: therefore the three straight lines AB, BC, BF are all in one plane, viz. that which passes through AB, BC : and because AB stands at right angles to each of the straight lines BD, BE, it is also at right angles (4. 1 1.) to the plane passing through them; and therefore makes right angles (3.
Side 16 - BGH are together equal* to two right angles: and BGH is a right angle; therefore also GBA is a right angle, and GB perpendicular to BA. For the same reason GB is perpendicular to BC. Since therefore the straight line GB stands at right angles to the two straight lines BA, BC, that cut one another in B, GB is perpendicular...
Side 9 - If three straight lines meet all in one point, and a straight line stand at right angles to each of them in that point ; these three straight lines are in one and the same plane. Let the straight line AB stand at right angles to each of the straight lines BC, BD, BE, in B, the point where they meet ; BC, BD, BE are in one and the same plane. If not, let...
Side 1 - A plane is perpendicular to a plane, when the straight lines drawn in one of the planes perpendicular to the common section of the two planes, are perpendicular to the other plane. 5. The inclination of a straight line to a plane...
Side 28 - Cor. 1.) therefore all the angles of the triangles are equal to all the angles of the polygon together with four right angles : (i. ax. 1.) but all the angles at the bases of the triangles are greater than all the angles of the polygon, as has been proved ; wherefore the remaining angles of the triangles, viz. those of the vertex, which contain the solid angle at A, are less than four right angles.
Side 5 - If a straight line stand at right angles to each of two straight lines in the point of their intersection, it will also be at right angles to the plane in which these lines are.