 | Thomas Hornby (land surveyor.) - Surveying - 1827 - 318 pages
...Sq. Mile. 64,000,000 102,400 6,400 2,560 640 1 PROBLEM 1. To find the Area of a Parallelogram. RULE. Multiply the length by the perpendicular breadth, and the product will be the area. Note. — The above rule is true, whether the figure be a square, rhombus, or a rhomboid. EXAMPLE.... | |
 | John Gummere - Surveying - 1828 - 404 pages
...of each may be found; and the sum of these areas will be the area of the tract. PROBLEM I. To fold the area of a Parallelogram, whether it be a Square, a Rectangle, a Rhombus, or a Rhomboides. RULE. Multiply the length by the height or perpendicular breadth, and the product will... | |
 | John Bonnycastle - Geometry - 1829 - 256 pages
...observing this rule will generally find his results to agree with those given in the book. , PROBLEM I. To find the area of a parallelogram; whether it be a square, a rectangle, a rhombus, or a rhomboides. RULE.* Multiply the length by the perpendicular height, and the product will be the area.... | |
 | Alexander Jamieson - Industrial arts - 1829 - 654 pages
...contains the altitude of the square. Hence we have the following rule for any parallelogram whatever: — To find the area of a parallelogram, whether it be a square, a rectangle, a rhombus, or a rhomboides. Multiply the length by the perpendicular height, and the product will be the area. MENTHA... | |
 | John Bonnycastle - Measurement - 1833 - 310 pages
...this rule will generally find his results to agree with those given in the book. PROBLEM I. Tojind the area of a parallelogram ; whether it be a square, a rectangle, a rhombus, or a rhomboides. RULE.* Multiply the length by the perpendicular height, and the product will be the area.... | |
 | Zadock Thompson - Arithmetic - 1832 - 186 pages
...inches, square feet, square yards, or square rods, &c. 305. To find the area of a parallelogram (65), whether it be a square, a rectangle, a rhombus, or a rhomboid. RULE. — Multiply the length by the breadth, or perpendicular height, and the product will be the area.... | |
 | Zadock Thompson - Arithmetic - 1832 - 186 pages
...inches, square feet, square yards, or square rods, &c. 305. To find the area of a parallelogram (65), whether it be a square, a rectangle, a rhombus, or a rhomboid. RULE. — Multiply the length by the breadth, or perpendicular height, and the product will be the area.... | |
 | Ireland commissioners of nat. educ - 1834 - 370 pages
...28 feet 9 inches ? Ans. 1164 feet 4 inches 6 parts PROBLEM III. To find the area of a Rhombus. RULE. Multiply the length by the perpendicular breadth, and the product will be the area.* 1 . What is the area of a rhombus, whose side is 16 feet, and perpendicular breadth 10 feet ? Ans.... | |
 | John Gummere - 1836 - 412 pages
...the area of each may be found ; and the sum of these areas will be the area of the tract. PROBLEM I. area of a Parallelogram, whether it be a Square, a Rectangle, a Rhombus, or a Rhomboides. RULE. Multiply the length by the height or perpendicular breadth, and the product will... | |
 | Commissioners of National Education in Ireland - Measurement - 1837 - 284 pages
...feet 4 inches, 6 parts. MENSURATION OF SUPERFICIES. PROBLEM III. To find the area of a Rhombus. RULE. Multiply the length by the perpendicular breadth, and the product will be the area.* 1. What is the area of a rhombus, whose side is 16 feet, and perpendicular breadth 10 feet ? Ans. 16... | |
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PROBLEM I. To find the area of a parallelogram, whether it be a square, a rectangle, a rhombus, or a rhomboides. 
