 | Trinity College (Hartford, Conn.) - 1888 - 978 pages
...they are eqiml or supplementary. 4. An inscribed angle is measured by one-half its intercepted arc. 5. The area of a parallelogram is equal to the product of its base and altitude. 6. The area of a trapezoid is equal to one-half the product of its nltitiule and the sum of its parallel... | |
 | James Wallace MacDonald - Geometry - 1889 - 80 pages
...altitude are commensurable. CASE II. When they are incommensurable. Proposition II. A Theorem. 237. The area of a parallelogram is equal to the product of its base and altitude. Proposition III. A Theorem. 238. Parallelograms are to each other as the products of their respective... | |
 | James Wallace MacDonald - Geometry - 1894 - 76 pages
...altitude are commensurable. CASE II. When they are incommensurable. Proposition II. A Theorem. 237. The area of a parallelogram is equal to the product of its base and altitude. COROLLARY. Parallelograms having equal bases and altitudes are equivalent. Proposition III. A Theorem.... | |
 | Edward Albert Bowser - Geometry - 1890 - 414 pages
...the second power of its side, being the product of two equal sides. Proposition 4. Theorem. 363. TJie area of a parallelogram is equal to the product of its base and altitude. Hyp. Let ABCD be a en, AB its FD base, and BE its altitude. j / To prove CD ABCD = AB X BE. / Proof.... | |
 | Seth Thayer Stewart - Geometry, Modern - 1891 - 428 pages
...sides. PROP. XIII. The area of a rectangle is equal to the product of its base and altitude. PROP. XIV. The area of a parallelogram is equal to the product of its base and altitude. * PROP. XV. The area of a trapezoid is equal to the product of its altitude and half the sum of its... | |
 | Edward Albert Bowser - Geometry - 1891 - 424 pages
...to the second power of its side, being the product of two equal sides. Proposition 4. Theorem. 363. The area of a parallelogram is equal to the product of its base and altitude. base, and BE its altitude. j / Hyp. Let ABCD be a CD, AB its f D To prove CD ABCD = AB X BE. 1 / Proof.... | |
 | Seth Thayer Stewart - Geometry - 1891 - 428 pages
...that : The area of a rectangle is equal to the product of its base and altitude. PROPOSITION XIV. 353. Theorem : The area of a parallelogram is equal to the product of the base and altitude. Statement : Let S be any parallelogram. The area of S is equal to the product... | |
 | William Chauvenet - 1893 - 340 pages
...PROPOSITION IV. The area of a rectangle is equal to the product of its base and altitude. PROPOSITION V. The area of a parallelogram is equal to the product of its base and altitude. PROPOSITION VI. The area of a triangle is equal to half the product of its base and altitude. Corollary... | |
 | William Shaffer Hall - Measurement - 1893 - 88 pages
...9.5781 P. § 23. To find the area of a parallelogram when the base and altitude are given. [14] EULE: The area of a parallelogram is equal to the product of its base and altitude. For Proof: See Ww. Geom., 321. §. 24. To find the area of a trapezoid. [15] RULE: The area is equal... | |
 | William C. Bartol - Geometry, Solid - 1893 - 112 pages
...sides proportional. 371. A Parallelogram is a quadrilateral which has its opposite sides parallel. 372. The area of a parallelogram is equal to the product of its base and altitude. 373. The area of a circle equals the square of the radius multiplied by n, ie A = n K 2 . 374. In the... | |
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The area of a parallelogram is equal to the product of its base and its height: A = bx h. 
