same as the unit of length forming its side, with the word square An area of ten square chains is termed an acre, and the fourth of an acre a rood. When we say that the area of a surface is a square yard, we mean that the surface is equivalent to a square each side of which is a yard in length. A surface equivalent to twelve squares with sides I foot long, is said to be 12 feet in area. Rules for finding the Areas of certain Figures. I. The Rectangle.— Let the lengths of the sides be 3 inches and 5 inches respectively. Mark the divisions in inches on these lines, and D C draw parallels through the points. It is evident that there will be as many square inches formed along the base as there are units of length in it, and there will be as many times this number in B the whole rectangle as there are units of length in the other side. Hence the number of units of area in the rectangle is the product of the numbers of units of length in the base and height. FIG. 130. We may abbreviate this rule thus :— The area of a rectangle length breadth, provided we bear in mind that the number of units in each case is intended. Thus the area of a rectangle 3 feet long and 2 feet broad is 3 X 2, or 6 square feet. A rectangle 5 yards long and 3 yards wide contains 5 X 3, or 15 square yards. A rectangle 5 feet long and II inches broad contains 60 X II, or 660 square inches. 2. The Square.-The area of a square is found by multiplying the number of units of length in its side by itself. Thus, if the side be 4 inches long, the square contains 16 square inches. (Fig. 131.) contains 49 square yards. If the side be 7 yards in length, the square If the square contain 144 square inches, the side must be 12 inches long. From the fact that the number of units of area in a square is found by multiplying the number of units of length in the side by itself, multiplying a number by itself is termed squaring the number. The square of the number of units in a line A B is written A B2. 3. Other Rectangular Figures.-From the method of finding the numerical measurement of the area of a rectangle we can deduce that for finding the areas of other rectilineal figures, by means of the following proposition:— Any two parallelograms which have the same base and the same height, or lie between the same parallels, are equal in respect of area. The truth of this may be shown experimentally by taking two such parallelograms on paper and showing that one may be cut into pieces which can be arranged so as to cover the other. D FC E 2 2 B FIG. 132. If the sides of the two parallelograms do not cut one another, as in the figure, we may form ABEF from ABCD by simply moving the part 2 from left to right (Fig. 132). If two sides, B C, A F, do cut one another, then, by drawing through the point of intersection a line parallel to the base, and through the extremities of this line other lines parallel to the sides of the opposite parallelogram, and repeating the process as represented in the figure, it will be seen that the pieces into which the parallelogram on the left is divided may be placed together, 6 5 5 4 4 3 3 2 2 FIG. 133. so as to form the parallelogram on the right (Fig. 133). The same process might be applied to a parallelogram and a rectangle on the same base and between the same parallels. Therefore any parallelogram is equal to a rectangle having the same base and height. Hence we find the area of a parallelogram, A B C D, W N FIG 134. B FIG. 135. E when we find the area of the rectangle, A B E F, having the same base, A B, and the same height, D G. (Fig. 134.) Any triangle, ABC, is half a parallelogram, ABCE; hence the triangle is half a rectangle on the same base, B C, and having the same height, AD (Fig. 135). Thus, applying the abbreviated form above, we have : (a). The Parallelogram.-The area of a parallelogram height. = base X For it is equal to the rectangle with the same base and height. (b). The Triangle.-The area of a triangle For the triangle is half a parallelogram with the same base and height. (c). The Rhombus.-Hence the area of a rhombus: of its two diagonals. (Fig. 129.) = half the product (d). The Trapezoid.—The area of a trapezoid = half the product of the sum of the bases by the height. B H FIG. 136. D = Therefore half of (A DXAH+BC XAH) half of (A D + BC) × AH. It must be remarked that the two dimensions of these areas must be expressed in the same units of length. Example. Find the area of a rectangle the base of which is 3 yards 2 feet and the height I yard 7 inches. 3 yds. 2 ft. I yd. 7 in. Area = 132 in. 5676 sq. in. 4 sq. yds. 3 sq. ft. 60 sq. in. If the area of a square be given, the side can be found by taking the square root. If the area of a rectangle and one dimension be given, the other will be found by dividing the area by the given dimension. Arithmetical Questions 1. Find the area in acres of a square whose side is 956 links. 2. Of a square whose side is 15 chains 67 links. 3. What is the side of a square whose area is 14 acres I rood, 24 perches ? 4. A square field cost £9144, at £640 per acre: what is the length of its side in ya: ds? 5. Find the area of a parallelogram, the length being 12.25 and breadth or height 8.5 ft. 6. Find the area of a rectangular board whose length is 12 ft. and breadth 9 in. 7. Find the area of a piece of land, in the form of a rhombus, its length being 6·20 chains and perpendicular breadth 5:45. 8. Find the number of square yards in a parallelogram whose length is 37 ft. and height 5 ft. 3 in. 9. A door is ft. 3 in. long and 3ft. 6 in. broad: required its area and cost at 2s. 3d. per square ft. 10. A floor 29 ft. 2 in. long cost £20 Is. old. paving, at 6s. 9d. per square yard: find its breadth. 11. How many yards of carpet yards wide will cover a room 14 ft. 3 in. by 9 ft. 4 in. ? 12. How many square feet are there in the triangle whose base is 40 ft. and perpendicular 30 ft. ? 13. Find the number of square yards in a triangle whose base is 49 ft. and height 254 ft. 14. Find the area of a triangle whose base is 18 ft. 4 in. and height II ft. 10 in. 15. From a plank 16 in. in breadth, 6 square ft. are to be sawn off: at what distance from the end must the line be struck? 16. The base of an Egyptian pyramid is a square whose side is 693 ft. how many acres of ground does it cover? 17. The length of a floor is 45 ft. 6 in. and area 94711⁄2 ft.: find the breadth. 18. A diamond-paned window cost £2 9s. 53d. glazing, at Is. 112d. per square foot. The perpendicular diagonals of each pane were 8 in. and 12 in. How many panes were there in the window? 19. Find the area of the gable end of a house, the height of the eaves being 36 ft., of the ridge 45 ft. 4 in., and the breadth of the building 22 ft. 5 in. 20. A B C is a triangular field, and C D its height, A D = 895 links, Ᏼ Ꭰ 1063, CD=994: what is its area? = 21. The parallel sides of a piece of ground measure 856 and 684 links, and their perpendicular distance 985 links: what is the area of the ground? 22. If the parallel sides of a garden be 65 ft. 6 in. and 49 ft. 3 in., and their perpendicular distance 56 ft. 9 in., what would the garden cost at £325 10s. per acre? THEOREMS ON CHAPTER XII. SECTION I. Note. The equality referred to in this chapter is equality in respect of area only. THEOREM XLII. Parallelograms on the same base and between the same parallels, or having the same height, are equal. Let A B C D, A BEF be parallelograms on the same base, A B, and between the same parallels; then the area A B C D the area ABEF. F B E ▲ ECB, 4 DFA = 4 CEB (Theorem XVII.), and therefore also ▲ DA F = 4CBE (Theorem XVIII.) Hence the triangles DA F, CBE are equal. If D A F be taken from the whole figure the remainder is the parallelogram A B E F. If C B E be taken from the whole figure the remainder is the parallelogram A B C D. Therefore the area A B C D the area A B E F. THEOREM XLIII. A parallelogram is double the triangle on the same base and of the same height. Let A B C be a triangle and ABEF a parallelogram on the same base, A B, and of the same height or |