Then the square of 185 is 34,225, and the square of 153 is 23,409; and the difference between them is 10,816. Taking the square root of 10,816, we shall have 104 inches, or 8 ft. 8 in.—the base. Example 3.-The sides AC and AD of the triangle ACD are 55 ft. and 65 ft. respectively, and the perpendicular height AB is 33 ft.; find the base CD. Now, CD=CB+BD; and BDA D2 — AB2= √652-332 =√3025–1089=✓1936=44 ft. ; Example 4.-If the angle at c (see fig. 2) is 30°, and AB is 20 ft., find the length of the base BC. By Note 4.-Double the triangle on the other side of CB, and we have the equiangular triangle ACD, which has therefore all its sides equal. Hence AC AD=2× AB=2 x 20 ft.=40 ft. Therefore BC= √AC2-AB2=√402-202= √1600 — 400 = √1200=34.641 ft. I. AC=√AB2+BC2 Formula II. BC=AC2-AB2=(AC+AB) (AC-AB) EXAMPLES. Find the hypothenuse of the right-angled triangle whose base and perpendicular are, respectively (1) Perp. 36, and base 77. (2) Perp. 28, and base 45. (3) Perp. 19, and base 180. Find the hypothenuse of the right-angled triangle, when the perpendicular and the base are, respectively (4) Perp. 11 ft. 8 in, and base 14 ft. 3 in. (6) Perp. 61-6 yds., and base 66.3 yds. Determine the base of the right-angled triangle, when its hypothenuse and perpendicular are, respectively (7) Hyp. 50 ft., and perp. 30 ft. (8) Hyp. 31 ft. 5 in., and perp. 12 ft. 8 in. (9) Hyp. 945 yds. 1 ft., and perp. 345 yds. 1 ft. (10) Hyp. 22 ft. 5 in., and perp. 21 ft. 8 in. (11) The side of a square is 100 ft.; find its diagonal. (12) The diagonal of a square is 33 ft. 4 in.; what is the length of each side? (13) The diagonal of a rectangle is 46 ft. 5 in., and its length is 44 ft. 4 in.; find its breadth. (14) The length of each side of an equilateral triangle is 120 ft.; find the length of the perpendicular drawn from any angle to its opposite side. (15) The slant height of a cone is 7 ft. 1 in., and the diameter of its base 2 ft. 2 in.; find its perpendicular height. (16) The length of a rectangular room is 27 ft. 8 in., and its breadth is 20 ft. 9 in.; find the distance from one corner of the floor to the opposite corner across the room. (17) A ladder 65 ft. long reaches a window 56 ft. high from the ground; find the distance of the foot of the ladder from the side of the house. (18) A ladder 53 ft. long reaches a window 45 ft. high; how much nearer to the side of the house must it be moved that it may reach a window 48 ft. high? (19) Two columns, whose heights are respectively 17 ft. and 50 ft., stand on the same horizontal plane, with their bases 56 ft. apart; find the distance between the tops of these columns. (20) Find the cost of building a wall round a garden, in the shape of a right-angled triangle, whose hypothenuse is 97 yds. and base 72 yds., at 13s. 6d. per yard. (21) Two persons, travelling at the rate of 6 and 8 miles per hour respectively, proceed in directions at right angles to each other; find the distance between them at the end of 6 hours. (22) Find the length of a beam which shall rest on two walls standing on the same horizontal plane, and which are respectively 4 ft. and 8 ft. 4 in. high, when the distance between them is 13 ft. 9 in. (23) The mast of a ship is partially broken by the wind. The broken part, which is 35 ft. long, though still adhering to the unbroken part, strikes with its top the deck at the distance of 28 ft. from the foot of the mast; find the entire length of the mast. (24) A scaffolding pole 40 ft. long, placed at the distance of 21 ft. from the side of a house, is partially broken by the wind, at the height of 5 ft. from the ground, so that its top strikes against the wall; find at what height from the ground it will strike the wall. (25) There are two upright pillars, in the same horizontal plane, whose heights are respectively 44 ft. and 28 ft. A certain point is taken in that plane between the two pillars, and it is found that the distance of this point from the top of the higher pillar is 125 ft., and from the top of the shorter pillar is 53 ft. Find the distance between the tops of these pillars. (26) A ladder 55 ft. long may be so placed in a street as to reach a window 44 ft. high on one side, and, on being turned round, without changing its position, it will reach another window 33 ft. high on the opposite side of the street; find the breadth of the street. (27) A ladder 37 ft. long is so placed in a street 36 ft. wide that it will reach a window 35 ft. high, and, upon being turned round, without changing its position, it will reach another window upon the opposite side of the street; find the height of the window. (28) The dimensions of a rectangular room are 20 ft. long, 15ft. wide, and 12 ft. high; find the distance from a corner of the floor to the opposite corner of the ceiling across the room. (29) A foot-passenger, instead of keeping to the road, which runs along the two adjacent sides of a rectangular field, which measures 640 yds. long and 480 yds. broad, goes across the field from one corner to the opposite corner of it; find what distance he will save by doing so. (30) The breadth of the gable-end of a house is 56 ft., the perpendicular height of the roof is 15 ft., and the distance of the ridge from one of its eaves is 25 ft.; find the distance of the ridge from the other of the eaves. (31) If AC, the hypothenuse of a right-angled triangle ABC (fig. 1), is 40 ft., and the angle at A is 45°, find the length of the base BC. (32) The angle at c, in a right-angled triangle ABC (fig. 2), is 30°, and the perpendicular AB is 30 ft.; find the hypothenuse AC. (33) BC, the base of a right-angled triangle ABC (fig. 3), is 50 ft., and the angle at c is 60°; find the perpendicular A B. (34) AC, the hypothenuse of a right-angled triangle ABC (fig. 3), is 80 ft., and the angle at c is 60°; find the length of the base BC, and also of the perpendicular AB. (35) A mast, being partially broken by the wind, at the height of 20 ft. from its foot, its top touches the deck at an angle of 30°. Find the entire length of the mast. II. THE SQUARE. Definition. A square is a four-sided A figure, having all its sides equal, and all its angles right angles. BD is the diagonal. The perimeter of a square is the sum B of all its sides, or four times the length of a side. RULES.-(1) To find the area of a square, when a side is given. Square the side given. (2) To find the area of a square, when its diagonal is given. Square the diagonal, and divide by 2. Note 1.-To find the side of a square, when its area is given.—Take the square root of the area. Note 2.-To find the diagonal of a square, when its area is given.—Take the square root of the double of its area. Note 3.-To find the number of tiles or flags required for paving a square or rectangular floor or court.—Divide the area of the floor or court by the area of each tile or flag (always bearing in mind that, before dividing, we must have the areas in the same denomination). Note 4.-To find the area of a B walk or border, surrounding or running round the inside of a square court or plot of land.-First, find the area of the larger square ABCD, including both the court and the walk; and then of the smaller square EFGH. Then, the area of A F E H D the walk or border is the difference between the areas of these two squares. N.B.-If EH, the length of the inner square, is given, then AD, the length of the outer square,=(EH+2 width of walk or border). If AD, the length of the outer square, is given, then EH, the length of the inner square, =(AD-2 width of walk or border). |