Elements of Geometry: With Practical Applications, for the Use of Schools |
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Page 5
... remain always in the same plane represented by the surface of the paper , understanding for the present by the word surface - that which has length and breadth without thickness . We shall treat more particularly of surfaces in the next ...
... remain always in the same plane represented by the surface of the paper , understanding for the present by the word surface - that which has length and breadth without thickness . We shall treat more particularly of surfaces in the next ...
Page 15
... remains to prove that F H B⇒H G D. Now FHBA H G because they are vertical ( 22 ) . But we have just proved that A H G = H G D. Consequently F HB = H GD ; for it is an axiom that — two things , each of which is equal to a third , are ...
... remains to prove that F H B⇒H G D. Now FHBA H G because they are vertical ( 22 ) . But we have just proved that A H G = H G D. Consequently F HB = H GD ; for it is an axiom that — two things , each of which is equal to a third , are ...
Page 16
... remain a right angle L M C. Hence E F is perpendicular to C D , which was the first thing to be demonstrated . The second part hardly needs demonstration , but it can be demonstrated as follows . If A B and C D are perpen- dicular to ...
... remain a right angle L M C. Hence E F is perpendicular to C D , which was the first thing to be demonstrated . The second part hardly needs demonstration , but it can be demonstrated as follows . If A B and C D are perpen- dicular to ...
Page 18
... remain after taking from the equal semicircumferences H M I and H O I , the equal arcs H M and H O. Lastly if both ... remains , then , to prove that D G - half of B G A the intercepted arc . Now if D G be taken from B G A , we have ...
... remain after taking from the equal semicircumferences H M I and H O I , the equal arcs H M and H O. Lastly if both ... remains , then , to prove that D G - half of B G A the intercepted arc . Now if D G be taken from B G A , we have ...
Page 20
... remains after taking equal sums from two right angles or 1800. If , having two angles of a triangle given , we wish to find the third , it may be done arithmetically , by adding the degrees in the given angles and then subtract- ing ...
... remains after taking equal sums from two right angles or 1800. If , having two angles of a triangle given , we wish to find the third , it may be done arithmetically , by adding the degrees in the given angles and then subtract- ing ...
Other editions - View all
Elements of Geometry: With Practical Applications, for the Use of Schools Timothy Walker No preview available - 2023 |
Elements of Geometry: With Practical Applications, for the Use of Schools Timothy Walker No preview available - 2019 |
Common terms and phrases
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
Popular passages
Page 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...
Page xiv - Magnitudes which coincide with one another, that is, which exactly fill the same space, are equal to one another.
Page 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'.
Page xiv - LET it be granted that a straight line may be drawn from any one point to any other point.
Page 25 - In any proportion, the product of the means is equal to the product of the extremes.
Page 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.
Page 25 - Multiplying or dividing both the numerator and denominator of a fraction by the same number does not change the value of the fraction.
Page 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.
Page 42 - The area of a trapezoid is equal to the product of its altitude, by half the sum of its parallel bases.
Page 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...