Elements of GeometryHilliard and Metcalf, 1825 - 224 σελίδες |
Αναζήτηση στο βιβλίο
Αποτελέσματα 1 - 5 από τα 100.
Σελίδα 5
... drawn from A to B , which is impossi- ble ( 25 ) . Now let us suppose , if it be possible , that the lines , when produced , separate from each other at a point C , the one becoming CD , and the other CE . At the point C , let CF be drawn ...
... drawn from A to B , which is impossi- ble ( 25 ) . Now let us suppose , if it be possible , that the lines , when produced , separate from each other at a point C , the one becoming CD , and the other CE . At the point C , let CF be drawn ...
Σελίδα 8
... drawn straight lines OB , OC , to the extremities of BC , one of its sides , the sum of these lines will be less than that of AB , AC , the two other sides . Demonstration . Let BO be produced till it meet the side AC in D ; the ...
... drawn straight lines OB , OC , to the extremities of BC , one of its sides , the sum of these lines will be less than that of AB , AC , the two other sides . Demonstration . Let BO be produced till it meet the side AC in D ; the ...
Σελίδα 9
... Draw the straight line AD from the vertex A to the point D the middle of the base BC ; the two triangles ABD , ADC , will have the three sides of the one , equal to the three sides of the other , each to each , namely , AD common to ...
... Draw the straight line AD from the vertex A to the point D the middle of the base BC ; the two triangles ABD , ADC , will have the three sides of the one , equal to the three sides of the other , each to each , namely , AD common to ...
Σελίδα 10
... drawn from the vertex of an isosceles triangle , to the middle of the base , is perpendicular to that base , and ... Draw CD making the angle BCD = B . In the triangle BDC , BD is equal to DC ( 48 ) ; but AD + DC > AC , and AD + DC ...
... drawn from the vertex of an isosceles triangle , to the middle of the base , is perpendicular to that base , and ... Draw CD making the angle BCD = B . In the triangle BDC , BD is equal to DC ( 48 ) ; but AD + DC > AC , and AD + DC ...
Σελίδα 11
... drawn to that line . Demonstration . If it be possible , let there be two AB and AC ; produce one of them AB , so ... draw two perpendiculars to that line ; for , if CD and CE were these two perpendiculars , the an- gle DCB would be a ...
... drawn to that line . Demonstration . If it be possible , let there be two AB and AC ; produce one of them AB , so ... draw two perpendiculars to that line ; for , if CD and CE were these two perpendiculars , the an- gle DCB would be a ...
Περιεχόμενα
1 | |
2 | |
8 | |
14 | |
23 | |
30 | |
46 | |
53 | |
160 | |
168 | |
170 | |
174 | |
180 | |
186 | |
190 | |
192 | |
59 | |
66 | |
81 | |
91 | |
98 | |
99 | |
105 | |
111 | |
117 | |
123 | |
131 | |
136 | |
149 | |
150 | |
13 | |
27 | |
36 | |
8 | |
22 | |
44 | |
77 | |
86 | |
220 | |
Άλλες εκδόσεις - Προβολή όλων
Συχνά εμφανιζόμενοι όροι και φράσεις
a² b³ algebraic Algebraic Quantities Arith arithmetic becomes binomial changing the signs coefficient common divisor consequently contains courier cube root decimal deduce denominator denoted divided dividend division employed entire number enunciation equa evident example exponent expression extract the root figures follows formula fraction given in art given number gives greater greatest common divisor last term letters logarithm manner method multiplicand multiplied negative number of arrangements observed obtain operation perfect square polynomials preceding article proposed equation proposed number quan question quotient radical quantities radical sign reduced remainder represented resolve result rule given second degree second member second term simple quantities square root subtract suppose taken tens third tion tities units unity unknown quantity vulgar fractions whence whole numbers
Δημοφιλή αποσπάσματα
Σελίδα 9 - If two triangles have the three sides of the one equal to the three sides of the other, each to each, the triangles are congruent.
Σελίδα 44 - Divide the first term of the dividend by the first term of the divisor, and write the result as the first term of the quotient. Multiply the whole divisor by the first term of the quotient, and subtract the product from the dividend.
Σελίδα 63 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. A D A' Hyp. In triangles ABC and A'B'C', To prove AABC A A'B'C' A'B' x A'C ' Proof. Draw the altitudes BD and B'D'.
Σελίδα 101 - Which proves that the square of a number composed of tens and units, contains the square of the tens plus twice the product of the tens by the units, plus the square of the units.
Σελίδα 8 - Any side of a triangle is less than the sum of the other two sides...
Σελίδα 122 - ... is negative in the second member, and greater than the square of half the coefficient of the first power of the unknown quantity, this equation can have only imaginary roots.
Σελίδα 180 - CD, &c., taken together, make up the perimeter of the prism's base : hence the sum of these rectangles, or the convex surface of the prism, is equal to the perimeter of its base multiplied by its altitude.
Σελίδα 54 - The sum of the squares on the sides of a parallelogram is equal to the sum of the squares on the diagonals.
Σελίδα 185 - The convex surface of a cone is equal to the circumference of the base multiplied by half the slant height.
Σελίδα 164 - If two triangles have two sides and the inchtded angle of the one respectively equal to two sides and the included angle of the other, the two triangles are equal in all respects.