For the same reason, determining the quantity of surface in a figure is called squaring it; that is, determining the square or number of squares to which it is equal. Thus, if the figure to be measured be the D square ABCD, in the adjoining figure, and the little square E, whose side is one inch, be the measuring unit fixed upon, then as often as the square E is contained in the whole figure, so many square inches it is said to contain, which, in the present case, A we find to be 16. Again: supposing the 8 C 回 E square figure to contain 4 square miles of land, then 4 miles of land would be 4 times 4, or 16 square miles, making a difference of 12 square miles, (fig. 8.) The fundamental problem in the mensuration of superficies, is the very simple one of determining the area of a right parallelogram, as has been shown above. The contents of other figures may readily be obtained by finding parallelograms which are equal to them. C 9 C B If the parallelogram be divided into D small parallelograms, by drawing lines as shown in the subjoined figure, it is e obvious, on inspection, that the number of the little parallelograms must always be equal to the product of the length and breadth. If we count the parallelograms in the upper row of this figure, we find the number to be 4; in two rows, twice 4, or 8; in three rows, three times 4, or 12. Hence, to obtain the number of small parallelograms, ce, contained in the large parallelogram, ABCD, we have only to multiply the number of such small parallelograms contained in the side AB, into the number of such contained in the side BC, which, in the present case, we find to be, 4×3=12. ART. 3. The superficial unit is generally called by the. same name as the linear unit, which forms the sides of the square. If the side be an inch, it is called a linear inch; the side of a square foot, a linear foot; the side of a square rod, a linear rod; and so of any other fixed quantity. It should be remarked, however, that there are some superficial measures which have no corresponding denominations of length with which to compare them. For instance, the acre is not a square which has a line of the same name for its side. Hence, the utility of the following Tables, which contain the linear measure, in common use, with their corresponding square measures. By reducing the denominations of square measure, it will readily be seen that 1 square mile=640 acres=102400 rods=27878400 ft. 1 acre 10 chains-160 rods=43560 feet. From what has been already shown, we learn that multiplying the length of any square or parallelogram by its breadth, will give its square measure, or as it is sometimes called, its square contents, as will be seen in the demonstration of the following problems. PROBLEM I. To find the area of a Four-sided Figure, whether it be a parallelogram, square, rhombus, or rhomboid. ART. 4. Rule.-Multiply the length by the breadth or perpendicular height, and the product will be the area. It is manifest that the number D of square inches in the parallelogram ABCD is equal to the number of linear inches in the length AB,taken as many times as there are inches in the breadth BC. To obtain, then, the number of squares in the large parallelɔ- A C C B gram, we have only to multiply the number of squares in one of the small parallelograms (ABcd,) by the number of such parallelograms contained in the whole figure. Now, the number of square inches in one of the small parallelograms is 9, and the number of linear inches in the breadth BC is 7. Therefore the product of the length into the breadth is 63 square inches. It is therefore said concisely, that the area of a parallelogram is equal to the length multiplied into the breadth. EXAMPLE 1. How many square feet are there in a floor 23 feet long and 18 feet broad? 188.00 Or thus, 231×18=423, Ans. 235 Ans. 423.00 Ex. 2. What is the area of a parallelogram whose length is 12 feet 3 inches, and whose breadth is 8 feet 6 inches? Ans. 104.125. Ex. 3. How many square feet are there in the four sides of a room which is 22 feet long, 17 feet broad, and 11 feet high? 22 feet-length of one side.. 242=square feet in one side. 484 square feet in two sides. Again, 17 feet-length of the short side. 11 feet-height of the short side. 187 square feet in the short side. 2 374 square feet in both short sides. Then, 484+374-858 sq. ft. Ans. NOTE.-The very convenient Arithmetical rule of Duodecimals should be thoroughly learned, (and it may be in a few hours of close application,) by every one who would make progress in Mathematical science. Ex. 4. How many feet of glass are contained in a window 4 feet 11 inches high and 3 feet 5 inches broad? Ans. 16 ft. 9' 7". NOTE. When the dimensions are given in feet and inches, the most convenient way of performing the operation is by the Arithmetical rule of Duodecimals, in which each inferior denomination is one-twelfth of the next higher. Ex. 5. How many square yards of painting are there in a rhomboid, whose length is 37 feet, and height 5 feet 3 inches ? Ex. 6. Herodotus estimated the largest and most remarkable of the Egyptian Pyramids to be 800 feet square at the base. Now, how long a road, 4 rods wide, would occupy as much land as the base of the pyramid ? Ans. 1 mile, 6 fur. 27757, rods. ART. 5. We have already seen that the area of any parallelogram is obtained by multiplying the length into the breadth. Now, if the area and one side of any parallelogram be given, the other side may be found by dividing the area by the given side. So, also, if the area of a square be given, either side may be found by extracting the square root of the given area. This is merely reversing the rule in Art. 4, where a given side is squared. Ex. 1. What is the width of a street 12 rods long, which contains 40 square rods? 40÷12-31 rods, or 55 feet, Ans. Ex. 2. What is the side of a piece of land containing 1681 square rods? Ans. 41. Ex. 3. How much carpeting 1 yard wide will cover a floor 18 feet long and 16 feet wide? 18x16-297 square feet. 9 sq. ft. 1 sq. yd. and 297÷9-33 yds. |