 | American School (Lansing, Ill.) - Algebra - 1902 - 80 pages
...the rectangle AEFG is equal to the product of its base and altitude. From this we can state : 194. The area of a parallelogram is equal to the product of the base and altitude. Example. The altitude of a parallelogram is 9 inches and the base is 12 inches.... | |
 | William Estabrook Chancellor - Arithmetic - 1902 - 168 pages
...triangles and trapezoids. Since the diagonal of a parallelogram divides it into two equal triangles, and the area of a parallelogram is equal to the product of the numbers expressing the base and altitude, the area of a triangle is one-half the product of the numbers... | |
 | American School (Chicago, Ill.) - Engineering - 1903 - 426 pages
...the rectangle AEFG is equal to the product of its base and altitude. From this we can state : 194. The area of a parallelogram is equal to the product of the base and altitude. Example. The altitude of a parallelogram is 9 inches and the base is 12 inches.... | |
 | Yale University. Sheffield Scientific School - 1905 - 1074 pages
...in this examination.] 1. Derive values of all trigonometric functions of 225° and 12o°. 2. Prove that the area of a parallelogram is equal to the product of two sides and the sine of the included angle. 3. Simplify: — (a) tan f-ir+xJ + tan ( •ir— x);... | |
 | George William Myers, William Rockwell Wickes, Ernst Rudolph Breslich, Harris Franklin MacNeish, Ernest August Wreidt - Mathematics - 1906 - 208 pages
...altitude; ie, A=$ ba. Find A, if (1) 6 = 12 ft., and a = 16 ft.; (2) 6=8.2 rd., and a = 7.78 rods. 4. The area of a parallelogram is equal to the product of the base by the altitude; ie, A =ba. Find A, if (1) 6 = 28, and a = 19; (2) 6 = 16.3, and a = 14.6 5. Find... | |
 | Joseph Claudel - Mathematics - 1906 - 758 pages
...their bases coincide, their vertices fall on the same line C'C parallel to their common base AB. 721. The area of a parallelogram is equal to the product of the base and the altitude (641). Having B = 5 feet and // = 3, we have: g-BXff-5X3-15sq. ft. It is seen... | |
 | John Henry Walsh - Arithmetic - 1908 - 314 pages
...are the dimensions in each case of the rectangle thus formed ? -D— r The number of sqiiare units in the area of a parallelogram is equal to the product of the number of units in the base by the number in the altitude. NOTE. The dimensions must be expressed in... | |
 | Ernst Rudolph Breslich - Mathematics - 1909 - 402 pages
...indicated: (x*+2x+ 1) (x+ 1) ; (a'+2a6+62) (a2+2a6+62) 490. Area of parallelogram and triangle. 1. Show that the area of a parallelogram is equal to the product of the base and altitude. 2. Show that the area of a triangle is equal to one-half of the product of the base... | |
 | Edward Rutledge Robbins - Logarithms - 1909 - 184 pages
...21.43. 11. a = 5.3881, 12. a = 9.0705, с = 9.0774, 6 = 8.4046, £ = 73° 14' 40". с =7.9555. 13. Prove that the area of a parallelogram is equal to the product of two adjoining sides times the sine of their included angle. 14. Prove that the area of any quadrilateral... | |
 | Irving Elgar Miller - Thought and thinking - 1909 - 352 pages
...that a triangle is equivalent to one half of a parallelogram with the same base and altitude. I know that the area of a parallelogram is equal to the product of its base and altitude. Then the area of a triangle is found by taking one half the product of its base... | |
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I label the two new points e and /." FIG. 2 With the help of this figure he then proceeds to the usual proof of the theorem that the area of a parallelogram is equal to the product of the base by the altitude, establishing the equality of certain lines... 
