MENSURATION. MENSURATION is that branch of Mathematics which is employed in ascertaining the extension, solidities, and capacities of bodies, capable of being measured. 1. MENSURATION OF SURFACE. To measure or ascertain the quantity of surface in any right-lined figure whose opposite sides are parallel to each other, as a Square, Rectangle, Rhomboid, &c. Rule.-Multiply the length by the breadth; the product is the area or superficial contents. Application of the rule to practical purposes. 1. The side of a square piece of board is 8 inches in length; required the area or superficies. Decimal equivalent to the fraction =1875 (see page 31); and 8.1875 × 8.1875-67 03515625 square inches, the area. 2. The length of the fire-grate under the boiler of a steam engine is 4 feet 7 inches, and its width 3 feet 6 inches: required the area of the fire-grate. 7 in. 5833 and 6 in.=5 (see Table of Equivalents, p. 31): hence 4.5833 x 3.5=16·04155 square feet, the area. 3. Required the number of square yards in a floor whose length is 131⁄2, and breadth 93 feet. 13.5 × 9.75=131.625-9-14625 square yards. Note 1.-The above rule is rendered equally applicable to figures whose sides are not parallel to each other, by taking the mean breadth as that by which the contents are to be estimated. 2. The square root of any given sum equals the side of a square of equal area. 3. Any square whose side is equal to the diagonal of another square, contains double the area of that square. 4. Any sum or area (of which to form a rectangle) divided by the breadth, the quotient equals the length; or divided by the length, the quotient equals the breadth of the rectangle required. TRIANGLES. Any two sides of a right-angled triangle being given, to find the third side. Rule 1.-Add together the squares of the base and perpendicular, and the square root of the sum is the hypotenuse or longest side. Rule 2.-Add together the hypotenuse and any one side, multiply the sum by their difference, and the square root of the product equals the other side. Application to practical purposes. 1. Wanting to prop a building with raking shores, the top ends of which to be 25 feet from the ground, and the bottom ends, 16 feet from the base of the building; what must be their length, independent of any extra length allowed below the surface of the ground? 252+162881-29.6816 feet, or 6816 × 12=8 inches; consequently, 29 feet 8 inches nearly. 2. From the top of a wall 18 feet in height, a line was stretched across a canal for the purpose of ascer taining its breadth; the length of the line, when measured, was found to be 40 feet; required the breadth from the opposite embankment to the base of the wall. 40-18-22, and 40 + 18 × 22=√1276=35.72, or 35 feet 9 inches nearly, the width of the canal. Triangles similar to each other are proportional to each other; hence their utility in ascertaining the heights and distances of inaccessible objects. Thus, suppose the height of an inaccessible object D is required, I find by means of two staffs or otherwise, the height of the perpendicular BC and the length of the C A B G base line A B; also the distance from A to the base of the object G D ; then AB BC:: AG: G D. And suppose A B=6 feet, BC 2 feet, and A G=150; 62: 150: 50 feet, the height of D from G. Again, suppose the inaccessible distance A be required, make the line B A, B C, at right angles, and в C of three or four equal parts of any convenient distance, through one of which and in a line with the object A, determine the triangle C D F, then the portion will be as pro C D B CFCD::BFB A. Let CF 10 yards, c D=53, and B F=30, 10: 53: 30: 159 yards, the distance from B. To find the area of a triangle when the base and perpendicular are given. Rule.-Multiply the base by the perpendicular height, and half the product is the area. 309375 and g=375 (see page 31): 11.09375 x 3.375 hence 2 B D =18.72075 square inches, the area. 2. The base of a triangle is 53 feet 3 inches, and the perpendicular 7 feet 9 inches; required the area or superficies. 53.25 x 7.75 When only the three sides of a triangle can be given, to find the area. Rule. From half the sum of the three sides, subtract each side severally; multiply the half sum and the three remainders together, and the square root of the product is equal to the area required. Required the area of a triangle whose three sides are respectively 50, 40, and 30 feet. 50+40 + 30 2 =60, or half the sum of the three sides. 60-30-30, first difference, 60-40=20, second difference, 60-50-10, third difference; then 30 x 20 x 10 x 60/360000=600, the area required. Triangles are employed to great advantage in determining the area of any rectilineal figure, as the annexed, and by which the measurement is rendered comparatively simple. |