 | William Taylor Campbell - Geometry - 1899 - 276 pages
...feet wide instead of 27 feet. How much additional area does it contain f 2. Area of a Parallelogram. The area of a parallelogram is equal to the product of its base and altitude. The base is any one of the sides. The altitude is the perpendicular distance between the base and the... | |
 | Webster Wells - Geometry - 1899 - 424 pages
...the product of 6 and 5, the numbers which express the lengths of the sides. PROP. IV. THEOREM. 309. The area of a parallelogram is equal to the product of its base and altitude. BFC A b D Given O ABCD, with its altitude DF=a, and its base To Prove area ABCD = ax b. Proof. Draw... | |
 | Harvard University - Geometry - 1899 - 39 pages
...Corollary. The area of a rectangle is equal to the product of its base and its altitude. THEOREM IV. The area of a parallelogram is equal to the product of its base and its altitude. THEOREM V. The area of a triangle is equal to half the product of its base and its altitude.... | |
 | George Albert Wentworth - Geometry, Modern - 1899 - 272 pages
...evident by dividing the figure into squares, each BOOK IV. PLANE GEOMETKY". PROPOSITION IV. THEOREM. 400. The area of a parallelogram is equal to the product of its base by its altitude. BE CF CE A b ~ DA J> ' D Let AEFD be a parallelogram, b its base, and a its altitude.... | |
 | George Albert Wentworth - Geometry, Solid - 1899 - 246 pages
...drawn, the tangent is the mean proportional between the whole secant and its external segment. 400. The area of a parallelogram is equal to the product of its base by its altitude. 401. Parallelograms having equal bases and equal altitudes are equivalent. 403.... | |
 | Education - 1899 - 824 pages
...assumed without proof. Among the other error» in proving the second part it wa.s taken for true that the area of a parallelogram is equal to the product of its two adjacent sides. Questions 7. 8, and 9 were correctly demonstrated by nearly all (a considerable... | |
 | William James Milne - Geometry, Modern - 1899 - 258 pages
...=0= A AD F + A BED ; that is, ABCD =0= ABEF. Therefore, etc. QED MILNE'S GEOM. — 12 332. Cor. I. The area of a parallelogram is equal to the product of its base by its altitude. 333. Cor. II. Parallelograms are to each other as the products of their bases... | |
 | George Albert Wentworth - Geometry - 1899 - 498 pages
...by dividing the figure into squares, each T~ BOOK IV. PLANE GEOMETRY. PROPOSITION IV. THEOREM. 400. The area of a parallelogram is equal to the product of its base by its altitude. BE OF B GE A b DAI) D Let AEFD be a parallelogram, b its base, and a its altitude.... | |
 | Alan Sanders - Geometry, Modern - 1901 - 260 pages
...units) in the rectangle is 4 x 3 or 12. The area then is 12 square feet. 588. COROLLARY. The area of any parallelogram is equal to the product of its base and altitude. Let ABCD be any parallelogram f BEC and DE be its altitude. To Prove ABCD = AD x DE. Proof. Draw AF _L to AD, meeting BC prolonged... | |
 | Metal-work - 1901 - 548 pages
...inches; that is, a figure 2 inches square, and its area is 4 square inches. 25. Area of a Parallelogram. The area of a parallelogram is equal to the product of its base and altitude. Let AB CD, Fig. 4, be any parallelogram. Draw A Fand BE perpendicular to the two opposite sides AB and... | |
| |
The area of a parallelogram is equal to the product of its base and altitude. 
