76 base of the parallelogram 70 perpendicular from the opposite side Product 5320 square inches, area required. BY THE SLIDING RULE. Set 1 on B to 76 on A, and opposite to 70 on B you will have 5320 on A. If the base had been 83.2, and perpendicular 30, it in like manner would have been by the pen, 83.2 base. 30 perpendicular Product 2496·0 area sought. BY THE SLIDING RULE. When 1 on B is set to 83.2 on A, the number 2496 stands on A against 30 on B. PROBLEM IV. To find the Area of a Right-angled Triangle. RULE. Half the product of the two sides containing the right angle, will be the area. To obtain which, either multiply half the one by the whole of the other; or multiply the one by the other, and divide by 2. Let ABC be a triangle right-angled at B, whereof the sides containing the right angle are respectively 30 inches, and 48 inches; the area is required. Wherefore the area of the given triangle is 720 square inches. BY THE SLIDING RULE. Set 1 on B to 48 on A, and against 15 on B you find 720 on A, for the area of the triangle. PROBLEM V. will To find the Area of any Triangle by a Perpendicular let fall upon the longest Side from the opposite Angle. RULE. Half the product of the base and perpendicular will be the area; to obtain which proceed as in the last problem. Let ABC be a triangle, whereof the base AB measures 44 inches, and a perpendicular CD let fall upon it from the opposite angle measures 20 inches: the area is required. Unity on B being set to 44 on A, we have 440 for the area on A, against 10 on B. As an additional example suppose the base 148.6, and the perpendicular 51; the work will be as under74-3 half the base 51 perpendicular 74.3 51 743 3715 3789.3 product and area. BY THE SLIDING RULE. Set 1 on B to 74.3 on A, and against 51 on B you have 3789.3 on A, for the area. PROBLEM VI. To find the Area of a Triangle without a Perpendicular, having the three Sides given. RULE. Add the three sides together, and from half the sum subtract each side separately, noting the remainders. Then multiply into one continued product the half sum and three remainders. The square root of this product will be the area. Let ABC be a triangle, whereof the side AB measures 28 inches; the side BC, 32 inches; and the base AC, 40 inches: the area is required. And 50 x 22 x 18 x 10-198000, whereof the square root is 445 square inches NEARLY, the area of the given triangle. In like manner if the three sides were respectively 30 inches, 40 inches, and 50 inches, it would be, 30 first side = And 60 x 30 x 20 x 10 360000, whereof the square root is 600 square inches for the area of the triangle. This last example admits of a different mode of solution, and consequently the result may be verified; for it will be found, on constructing the figure, that the sides 30 and 40 contain a right angle. PROBLEM VII To find the Area of a Quadrilateral wherein two unequal Sides are Parallel to one another. RULE. Multiply half the sum of the parallel sides by the perpendicular distance between them, and the product will be the area. |