Ex. 2. The wall of a building on the bank of a river is 120 feet high, and the breadth of the river 210 feet: what is the length of a line which will reach from the top of the wall to the opposite bank of the river ? Ans. 241.86 ft. ART. 9. To find one of the Legs when the Hypotenuse and the other Leg are known. Rule. Subtract the square of the leg whose length is known, from the square of the hypotenuse, and the square root of their difference will be the answer. Ex. 1. If AC (Prob. 2)=70 feet, and BC=60 feet, what will be the length of the side AB? Ex. 2. The height of a precipice on the bank of a river is 103 feet, and a line of 320 feet in length will just reach from the top of it to the opposite bank; required the breadth of the river. Ans. 302.97 ft. Ex. 3. The hypotenuse of a triangle is 53 yards, and the perpendicular 45 yards; what is the length of the base? Ans. 28 yds. Ex. 4. A ladder 50 feet in length will reach to a window 30 feet from the ground on one side of the street, and by turning it over to the other side, it will reach a window 40 feet from the ground; required the breadth of the street. Ans. 70 ft. ART. 10. If the area of an Equilateral Triangle be given, the sides may be obtained by the following Rule. Divide the area by the decimal,433013, and extract the square root of the quotient: the result will be the length of one side. Ex. 1. The area of an equilateral triangle is one square chain: how many feet long is one side? PROBLEM III. Ans. 100.29 ft. To find the area of a Trapezium. ART. 11. Rule. Divide the trapezium into triangles by drawing diagonals; and the sum of the areas of these triangles will be the area of the trapezium. NOTE. A trapezium is an irregular figure of four unequal sides and angles. And the area=120+72=192 Ex. 2. Required the area of a trapezium whose diagonal is 78, and whose perpendiculars are 22 and 24. Ans. 1794. Ex. 3. What is the area of a trapezium whose diagonal is 42 feet, and the two perpendiculars 18 and 16 feet ? Ans. 714 sq. ft. Ex. 4. What is the area of a trapezium in which the diagonal is 320,75 chains, and the two perpendiculars 69,73 chains and 130,27 chains ? Ans. 3207 A. 2 R. PROBLEM IV. To find the area of a Trapezoid. ART. 12. Rule.-Multiply the sum of the two parallel sides by the perpendicular distance between them, and half the product will be the area. Ex. 1. Required the area of the trapezoid ABCD, having given AB=321.51 feet, DC-214.24 feet, and whose height is 171.16 feet. D C A B OPERATION. We first find the sum of the sides, and then multi- 321.51+214.24=535.75=the ply it by the perpendicu sum of the parallel sides. Then, lar height; after which, 535.75×171.16-91698.97, we divide the product by 2 for the area. And, 91698.97-2-45849.485, Ans. Ex. 2. How many square rods are contained in a field which has two parallel sides, 65 and 38 rods, and whose breadth is 27 rods? Ans. 139.05. Ex. 3. Required the number of square feet in a trapezoid which has two parallel sides, 46 and 38 feet, distant from each other 27 feet. Ans. 1134. Ex. 4. What is the area of a trapezoid whose parallel sides are 10.50 chains and 18.25 chains, and whose perpendicular height is 16.80 chains ? Ans. 24 A. 0 R. 24 P. PROBLEM V. To find the area of a regular Polygon, or any regular figure. Акт. 13. Rule I. - Multiply one of its sides into half its perpendicular distance from the centre, and this product into the number of sides. It is evident, on inspection, that a regular polygon contains as many equal triangles as the figure has sides. A C D Thus, the adjoining hexagon has six triangles, each equal to ABC. Now, the area of ABC is equal to the product of the side AB into of CD, (Art. 6.) The area of the whole, therefore, is equal to this product multiplied into the number of sides. Ex. 1. Required the area of a regular hexagon, each of whose sides, AB, &c., is 45 feet, and the perpendicular, CD, 24 feet. We first multiply one side by of the perpendicular CD, and that product by the number of sides: this gives the area. OPERATION. 45x12x6=3240 ft. Ans. Ex. 2. The side of a regular pentagon is 20 yards, and the perpendicular from the centre, on one of the sides, is 13.76382; required the area. Ans. 688.191 sq. yds. Ex. 3. What is the area of a regular decagon whose sides are each 102 rods, and perpendicular from the centre 60 rods ? Ans. 30600. Ex. 4. Required the area of a regular decagon whose sides are each 87 feet, and perpendicular 28 feet. Ans. 12180 sq. ft. To facilitate the measurement of polygons, the following Table is constructed, showing the Multipliers of the ten regular polygons, when the sides of each is equal to 1: it also shows the length of the Radius of the inscribed circle. Now, since the areas of similar polygons are to each other as the squares of their homologous sides, if the square of the side of a polygon be multiplied by the multiplier of the like figure, the product will be the area sought. And hence we have, 12: tabular area, :: any side squared : area. By this table may be calculated the area of any other polygon of the same number of sides with one of these. Hence, to find the area of any regular polygon, we have the following method, in addition to the rule already given : Rule II.-1. Square the side of the polygon. 2. Multiply the square thus found by the tabular multi |