Elements of Geometry: With Practical Applications, for the Use of SchoolsRichardson, Lord & Holbrook, 1829 - 129 sider |
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Resultat 1-5 av 13
Side xvi
... intersection " 6 31 11 66 .38 11 perimter centre O 66 42 " C 40 " C 60 " 41 G = C prism is a circle read A B to C D vertices as centres A E C BED intersection . G = B perimeter centre N prism is a cube * SECTION FIRST -LINES Page.
... intersection " 6 31 11 66 .38 11 perimter centre O 66 42 " C 40 " C 60 " 41 G = C prism is a circle read A B to C D vertices as centres A E C BED intersection . G = B perimeter centre N prism is a cube * SECTION FIRST -LINES Page.
Side 12
... Erect another upon the middle of D C and it must also pass through the centre . But a point which is in two lines at once must be at their intersection Therefore E is the centre sought . If we had 12 ELEMENTS OF GEOMETRY .
... Erect another upon the middle of D C and it must also pass through the centre . But a point which is in two lines at once must be at their intersection Therefore E is the centre sought . If we had 12 ELEMENTS OF GEOMETRY .
Side 17
... intersection G. Hence E F G H. Before leaving this proposition , we will observe that it ex- plains the nature of an instrument , by which parallel lines are drawn with much greater facility than by the pro- cess described in the ...
... intersection G. Hence E F G H. Before leaving this proposition , we will observe that it ex- plains the nature of an instrument , by which parallel lines are drawn with much greater facility than by the pro- cess described in the ...
Side 22
... intersection F. Thus the two triangles coincide throughout . Hence we say a side and its two adjacent angles determine the triangle- . 56. Having a side and its two adjacent angles given , to F32 construct the triangle . Draw A B ( fig ...
... intersection F. Thus the two triangles coincide throughout . Hence we say a side and its two adjacent angles determine the triangle- . 56. Having a side and its two adjacent angles given , to F32 construct the triangle . Draw A B ( fig ...
Side 23
... intersection E , and the two triangles coincide throughout . Hence we say - the hypoth- enuse and a side determine a right triangle― . 60. Having the hypothenuse and another side given , to construct a right triangle- . Draw D F equal ...
... intersection E , and the two triangles coincide throughout . Hence we say - the hypoth- enuse and a side determine a right triangle― . 60. Having the hypothenuse and another side given , to construct a right triangle- . Draw D F equal ...
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Vanlige uttrykk og setninger
A B C D A B fig adjacent angles axis B A C base and altitude base multiplied bisect called centre chord circ circumference coincide convex surface cube cylinder D E F demonstrated diameter divided draw equally distant equivalent found by multiplying frustum geometry given line gles height Hence homologous sides hundredths inches infinite number infinitely small inscribed angles inscribed circle inscribed sphere intersection line A B line drawn linear unit mean proportional method of Exhaustions number of sides parallel sides perimeter perpendicular polyedrons preceding proposition proved pyramid radii radius ratio regular polygon rence right angle right parallelogram right parallelopiped right triangle semicircumference similar triangles solid angles sphere square feet straight line Suppose tangent tion trapezoid triangles A B C triangles are equal triangular prism vertex vertices
Populære avsnitt
Side ii - Co. of the said district, have deposited in this office the title of a book, the right whereof they claim as proprietors, in the words following, to wit : " Tadeuskund, the Last King of the Lenape. An Historical Tale." In conformity to the Act of the Congress of the United States...
Side xiv - Magnitudes which coincide with one another, that is, which exactly fill the same space, are equal to one another.
Side 30 - 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. D c A' D' Hyp. In triangles ABC and A'B'C', ZA = ZA'. To prove AABC = ABxAC. A A'B'C' A'B'xA'C' Proof. Draw the altitudes BD and B'D'.
Side xiv - LET it be granted that a straight line may be drawn from any one point to any other point.
Side 25 - In any proportion, the product of the means is equal to the product of the extremes.
Side 38 - The perimeters of two regular polygons of the same number of sides, are to each other as their homologous sides, and their areas are to each other as the squares of those sides (Prop.
Side 25 - Multiplying or dividing both the numerator and denominator of a fraction by the same number does not change the value of the fraction.
Side xiv - Things which are equal to the same thing are equal to one another. 2. If equals be added to equals, the wholes are equal. 3. If equals be taken from equals, the remainders are equal. 4. If equals be added to unequals, the wholes are unequal. 5. If equals be taken from unequals, the remainders are unequal. 6. Things which are double of the same are equal to one another.
Side 42 - The area of a trapezoid is equal to the product of its altitude, by half the sum of its parallel bases.
Side xiv - If a straight line meets two straight lines, so as to make the two interior angles on the same side of it taken together lesi than two right angles...