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" The area of a parallelogram is equal to the product of its base and its height: A = bx h. "
A Treatise on Elementary Geometry: With Appendices Containing a Collection ... - Σελίδα 129
των William Chauvenet - 1872 - 368 σελίδες
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Elements Of Geometry And Trigonometry

Charles Davies - 1874 - 464 σελίδες
...is equal to the area of a rectangle constructed with the lines as sides. D E PROPOSITION V. THEOREM. The area of a parallelogram is equal to the product of its base and altitude. Let AB CD be a parallelogram, AB its base, and BE its altitude : then will the area of ABCD be equal to...

Elements of Geometry and Trigonometry: From the Works of A.M. Legendre

Adrien Marie Legendre - 1874 - 500 σελίδες
...continued product of the number of linear units in each of the three lines. Thus, -when we say that the area of a parallelogram is equal to the product of its base and altitude, we mean that the number of superficial units in the parallelogram is equal to the number of linear...

Manual of Geometry and Conic Sections: With Applications to Trigonometry and ...

William Guy Peck - 1876 - 376 σελίδες
...expresses the area of a rectangle whose adjacent sides are the given lines. PROPOSITION III. THEOREM. The area of a parallelogram is equal to the product of its base and altitude. Let ACDE be a parallelogram; draw AL K_ and CK perpendicular to AC meeting ED, and ED prolonged at L and...

The Model Elementary Arithmetic: Including Oral and Written Exercises

Alfred Kirk, Henry Holmes Belfield - 1876 - 220 σελίδες
...annexing to it, in the / manner shown, an equal / triangle, may form a par/ allelogram. 224=. Since the area of a parallelogram is equal to the product of its base and altitude; and since every parallelogram may be divided into two equal triangles having the same base and altitude...

Annual Statement, Τόμοι 11-20

1876 - 646 σελίδες
...a triangle divides the opposite side into segments which are proportional to the adjacent sides. 3. The area of a parallelogram is equal to the product of its base and altitude. 4. How do you find the area of a trapezoid ? The areas of similar polygons are to each other in what...

The Complete Arithmetic: Combining Oral and Written Exercises in a Natural ...

Albert Newton Raub - 1877 - 348 σελίδες
...are the rules for the measurements of triangles: It was proved in Denominate Numbers (Art. 107) that the area of a parallelogram is equal to the product of its base and altitude, and since a triangle is half a parallelogram, we derive the following RULE. 1. To find the area of...

Complete Arithmetic: Theoretical and Practical

William Guy Peck - 1877 - 430 σελίδες
...its altitude ? Ans. 16 ft. AREA OF A PARALLELOGRAM. 285. It is shown in Geometry (B. 4, P. 3), that the area of a parallelogram is equal to the product of its base and altitude; that is, Area of parallelogram = Base x Altitude. EXAM PLE S. 1. The base of a parallelogram is 14...

Elements of Plane and Solid Geometry

George Albert Wentworth - 1877 - 416 σελίδες
...equal to the unit of measure ; and the area of the figure equals 7X4. PROPOSITION IV. THEOREM. 321. The area of a parallelogram is equal to the product of its base and altitude. BE CFB CE F D AD Let AE FD be a parallelogram, AD its base, and С D its altitude. We are to prove...

Bradbury's Eaton's Practical Arithmetic: Combining Oral and Written Exercises

William Frothingham Bradbury - 1879 - 392 σελίδες
...of a parallelogram is the perpendicular distance from the opposite side to the base ; as К N. 47L The area of a parallelogram is equal to the product of its base and altitude. NOTE. For the Rectangle see Arts. 197-201. 7. What is the area of a parallelogram whose base is 25...

An Elementary Geometry: Plane, Solid and Spherical

William Frothingham Bradbury - 1880 - 260 σελίδες
...will evidently contain 5X4, or 20, of these squares; that is, its area = 5 X 4 = 20. THEOREM XVI. 43. The area of a parallelogram is equal to the product of its base and altitude. lelogram ABCD ; then the area of j 7 ~7 Let DF be the altitude of the paral- EBFC / At A draw the perpendicular...




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