 | Charles Vyse - 1806 - 342 páginas
...7"", by 8 Feet, 6', 7". (I2j Multiply 46 Feet, 6 In. 8', 4'", by 6 Feet, 4 In.8', &". t PROBLEM II. 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 Height or perpendicular Breadth, and the Product will... | |
 | Charles Hutton - 1807 - 464 páginas
...the rectangle is said to contain, which in the present case is 12. PROBLEM I. To find the Area of any Parallelogram ,- "whether it be a Square, a Rectangle, a Rhombus, or a Rhomboid. MULTIPLY the length by the perpendicular breadth, or height, and the product will be the area*. EXAMPLES.... | |
 | Samuel Webber - 1808 - 466 páginas
...fathom 16i feet. or } . . 5$ yards } 1 pole or rod 40 poles 1 furlong SQJJARE MEASURES. PR0BLEM I. Tojmd the area of a parallelogram ; whether it be a square^ a rectangle, a rhombus, or a rhomboid. RULE.* Multiply the length by the breadth, or a perpendicular height, and the product will be the area. *... | |
 | Peter Nicholson - 1809 - 426 páginas
...altitude of the square. Hence we have the following rule for any parallelogram whatever. PROBLEM I. 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. EXAMPLK... | |
 | Charles Hutton - 1811 - 494 páginas
...rectangle is said to contain, which in the present case is 12. PROBLEM I. 3*0 fold the Area of any Parallelogram ; whether it be a Square, a Rectangle, a Rhombus, or a Rhomboid. MULTIPLY the length by the perpendicular breadth, or height, and the product will be the area *. EXAMPLES.... | |
 | John Gummere - 1814 - 398 páginas
...survey and finding its area by dividing it into triangles and trapeziums. • • - • 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 height or perpendicular breadth, and the product will... | |
 | Thomas Keith - 1817 - 306 páginas
...feet 1 inch 10 parts 6 seconds. PROBLEM III. To find the Area of a PJtombus, or of a Rhomboidcs. RULE. Multiply the length by the perpendicular breadth, and the product will be the area *. Perpend. . * DaMMnMMI. Benuse parallelograms upon the same base and between the same parallels arc... | |
 | Charles Hutton - 1822 - 616 páginas
...rectangle is> said to contain which in the present case is 12. PROBLEM f. To find Ike Area, of any Parallelogram ; whether it be a Square, a Rectangle, a Rhombus, or a Rhomboid. MULTIPLY the length by the perpendicular breadth, or height, and the product will be the area.* EXAMPLES... | |
 | John Nicholson - 1825 - 822 páginas
...triangles, are to each other, as the squares of their like sides. MENSURATION OF SUPERFICIES. Prol. 1. To find the area of a parallelogram: whether it be a square, a rectangle, a rhombus, or a rhomboid. Multiply the length by the breadth, or perpendicular height, and the product will be the area. Ex.... | |
 | Peter Nicholson - 1825 - 1046 páginas
...altititude of the square. Hence we have the following rule for any parallelogram whatever. Prob. 1. To find the area of a parallelogram, whether it be a square, a rectangle, a rhombus, or a rhomboïdes. Multiply the length by the perpendicular height, and (he product will be the area. Ex.... | |
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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. 
