6 ft. 6 in. deep, with lead which costs £1 8s. per cwt., and weighs 8 lbs. to the sq. ft. (37) Find the cost of carpeting a room 18 ft. 9 in. long and 17 ft. 6 in. broad, with carpet 2 ft. wide, at 4s. 9d. per yd. (38) The cost of carpeting a room 21 ft. long with carpet 24 in. wide, and worth 4s. per yd., is £10 10s.; what is the breadth of the room? (39) The perimeter of a square and also of a rectangle, whose breadth is 14 yds., is 392 yds.; what is the difference in size of these two figures? (40) What will be the area of a rectangular field whose diagonal is 415 yds. and its breadth 249 yds. ? (41) Find how many yards of paper 27 in. wide will be required for papering a room 18 ft. long, 12 ft. broad, and 11 ft. high. (42) The flooring of a room 14 ft. 3 in. long and 13 ft. 4 in. broad is composed of planks, each 8 in. wide and 10 ft. long; how many will be required? (43) What is the cost of papering a room 6 yds. 1 ft. 2 in. long, 6 yds. 0 ft. 4 in. broad, and 12 ft. high, with paper of a yard wide, at 44d. per yd.? (44) A joiner requires 3 sq. ft. of wood, and he has only a plank 1 ft. 6 in. wide from which to cut it off; find the length of the piece that he must cut off. (45) The cost of carpeting a room 15 ft. long with carpet 24 in. wide, worth 4s. 6d. per yd., is £7 17s. 6d. the breadth of the room? What is (46) If the breadth of a rectangular field, which is 75 yds. long, were increased by 10 yds., then its area would be 3 roods 28 poles 23 sq. yds.; find its breadth. (47) The area of a rectangular field, whose breadth is 119 yds., is 2 ac. 3 roods 32 poles 2 sq. yds. Find what distance a person could save himself by going from one corner to the opposite corner across the field, instead of keeping to the footpath, which runs along the two adjacent sides of the field. (48) The dimensions of a rectangle are 45 yds. and 28 yds. Will its diagonal be greater or less than the diagonal of a of the same area? square (49) The building of a wall round a square field, at 4s. per yd., costs £112; what would have been the charge if the field had been in the shape of a rectangle, whose length is 196 yds., but still containing the same quantity of land? (50*) How many square yards of painting are there in a room 20 ft. long, 14 ft. 6 in. broad, and 10 ft. 4 in. high; allowing for a fireplace 4 ft. by 4 ft. 4 in., and 2 windows each 6 ft. by 3 ft. 2 in. ? (51**) Find the expense of covering with lead, at a farthing per sq. in., the inside of a cistern, open at the top, of length 10 ft., width 6 ft., and depth 4 ft. (52) If 864 planks, each 134 ft. long, are used in the construction of a platform 54 yds. long and 21 yds. broad; find the width of each plank. (53) The walls of a room, 21 ft. long, 15 ft. 9 in. wide, and 11 ft. 8 in. high, are painted at an expense of £9 12s. 6d. Find the additional expense of painting the ceiling, at the same rate. IV. THE OBLIQUE PARALLELOGRAM, OR THE RHOMBUS AND RHOMBOID. Definitions.-The rhombus (fig. 1) is a four-sided figure which has all its sides equal, but its angles are not right angles. The rhomboid (fig. 2) is a four-sided figure which has its opposite sides equal and parallel, but all its sides are not equal, and its angles are not right angles. Since both the rhombus and rhomboid are parallelograms, though their angles are not right angles, the general term oblique parallelogram will include both figures. In both figures, AB is called the base or length; and DE is the perpendicular height, or more generally called simply the height. RULE. To find the area of a rhombus or rhomboid. Multiply the base (AB) by the perpendicular height (DE). Note 1.-To find either the base or the perpendicular height, when the area and the other dimension are given.— Divide the area by the base, and the quotient is the perpendicular height; or, divide the area by the perpendicular height, and the quotient is the base. Note 2.-The reason for the rule given above for finding the area of an oblique parallelogram may be thus shown: G However oblique a parallelogram ABCD (fig. 3) may be, it has been shown by Euclid (I. 35) that its area is equal to the area of any other parallelogram upon the same base A AB, and between the same D : parallels AB and CG; and therefore it is equal to the area of the rectangle ABEG. But the area of the rectangle ABEG (by Rule, Prob. III.) is=ABX BE. Hence the area of the oblique parallelogram ABCD= ABX BE; that is, base x perpendicular height. = Therefore the area of a parallelogram, whether rectangular or oblique, will be always found by multiplying the base by the perpendicular let fall upon it from the opposite side. (In the case of a rectangular parallelogram, this perpendicular is the breadth of the figure.) It will be found also that the more oblique a parallelogram is, the less does its area become, until, at last, AD and BC form but one straight line, when, consequently, the area becomes 0. A Note 3.-If the length AB, and its adjacent side AD, and also A E, the distance intercepted between the angle at a and the perpendicular let fall from D, are given (see figs. 1 and 2), and it is required to find the area of the parallelogram, we must first find the perpendicular DE by Rule 2, Prob. I. (since DE is the perpendicular of a right-angled triangle EAD). Then, having got the base and the perpendicular height, we can find the area of the oblique parallelogram by the rule given in this chapter. Note 4.-The diagonals of a rhombus intersect each other at right angles. Hence, if the diagonals of a rhombus are given, we may find its area by multiplying together the two diagonals, and dividing the product by 2. Note 5.-If the diagonals of a rhombus are given, and it is required to find the length of each side of the rhombus, we may proceed thus: Since the diagonals bisect each other at right angles, we shall have a rightangled triangle AEB, of which AE is half one diagonal, and BE half the other diagonal. Then AB, the hypothenuse of this right-angled triangle, can be found by Rule 1, Prob. I. A D B Note 6.-The student must always bear in mind that the area of an oblique parallelogram is found not by multiplying together the two adjacent sides AB and AD, as in the case of a rectangular parallelogram, but by multiplying the base AB by DE, the perpendicular let fall upon it from the opposite side (see figs. 1 and 2). Example 1.-Find the area of a rhombus whose length is 6 chains 25 links, and perpendicular height is 4 chains 50 links. Now 6 ch. 25 lks.=625 lks; and 4 ch. 50 lks.=450 lks. Then the area= =625 × 450=281250 links 2.8125 ac. =2 ac. 3 roods 10 poles. sq. Example 2.-The area of a field, in the shape of a rhomboid, is 4 ac. 2 roods 20 poles, and its base is 9 ch. 25 lks. ; find its perpendicular height. Now 4 ac. 2 roods 20 poles=740 poles=462500 sq. links; and 9 ch. 25 lks.=925 links. Then, perpendicular height=462500÷925-500 links =5 chains. [The student must bear in mind that when the word height occurs in the following Examples, it always refers to the perpendicular let fall upon the base from the opposite side.] EXAMPLES. Find the area of an oblique parallelogram, when its base and perpendicular height, are respectively (1) Base 35 in., and height 15 in. (2) Base 40 ft. 6 in., and height 28 ft. 9 in. |