27 MENSURATION. Mensuration is the method of calculating the comparative magnitudes of figures, and it is divided into two parts-Mensuration of Superficies or Surfaces, and Mensuration of Solids. The magnitude of a surface is called its area, and is the space inclosed between its boundary lines. The magnitude of a body is called its solid contents, and is expressed in cubic feet, inches, &c. A SQUARE is a quadrilateral figure, which has all its sides equal, and all its angles right angles. A RECTANGLE is a four-sided figure, which has its angles, right angles, and its opposite sides parallel. Rhombus. Rhomboid. , a A RHOMBUS is a parallelogram, whose sides are equal, but whose angles are not right angles. A RHOMBOID is a parallelogram, whose adjacent sides are unequal, and whose angles are not right angles. A TRAPEZOID is a four-sided figure, which has but two of its sides parallel. D A A CIRCLE is a figure bounded by one line, called the circumference, and is such that all lines drawn to the circumference from a certain point within the figure, called the centre, are equal to each other. Any of these lines is called a radius; and a line drawn through the centre, termi B nating both ways in the circumference, is called a diameter. The portion of circle cut off by a diameter is called a semicircle. An ARC of a circle is any portion of the circumference. E A SEGMENT of a circle is a figure contained by an arc and its chord. A VERSED SINE is a line drawn from the middle of a chord perpendicular to the circumference. A SECTOR of a circle is a figure contained by two radii and an arc, as ACBE. PROBLEM I. To find the area of any parallelogram. RULE. Multiply the length by the perpendicular height, and the product will be the area. EXAMPLE. Required the area of a rhomboid whose length AB= 20.5, and perpendicular height a A = 11·75. 205 x 11.75= 240 875, the area. Note. In a square, or rectangle, the perpendicular height is the breadth: therefore, to find the areas of a square and rectangle, multiply the length by the breadth. PROBLEM II. To find the area of a trapezoid. RULE. Add together the two parallel sides, multiply their sum by the breadth or height, and half the product is the area. EXAMPLE. Required the area of a trapezoid whose sides A B and CD are 145 and 10:25, and breadth, a A = 7.25. RULE. Multiply one of its sides as a base by a perpendicular let fall from the opposite angle, and take half the product for the area. Or, from half the sum of the three sides subtract each side separately, and multiply the three remainders so obtained and the half sum together, and the square root of the product will be the area. EXAMPLE 1. Required the area of a triangle ABC, whose base AB= 16.5, and perpendicular D C 10·25. EXAMPLE 2. What is the area of that triangle whose three sides are 8, 12, and 16 respectively? 8+12+16 18 If any two sides of a right-angled triangle be given, the third side may be found by the following rules. 1-To the square of the base add the square of the perpendicular: and the square root of the sum will be the hypothenuse or longest side. 2-Multiply the sum of the hypothenuse, and one side by their difference; and the square root of the product will be the other side. EXAMPLE 1. Given the base A B = 12; required the length of the hypothenuse A C. 16, and perpendicular B C = √16212220, the length of the hypo thenuse A C. C B A EXAMPLE 2. Given the base AB = 16, and hypothenuse A C = 20; required the length of the perpendicular B C. √20 + 16 × 4 = 12, length of the perpendicular B C. Note.-The diagonal line, or hypothenuse in a square, is equal to the square root of And the side of a square is equal to the square root of twice the square of the side. half the square of its diagonal. Thus suppose each side of a square equal 12 feet: 122 × 2 288 16 9705 feet, the diagonal. Or, 14412 feet, the length of each side. Similar triangles, equi-angular to each other, have the sides about their equal angles proportional; thus, in the annexed figure the triangles ABC and CDE are similar, and therefore have the sides about the equal angles proportional: or those which are D E B AC: BC::CE: DE; The utility, then, of the above triangles for practical purposes, as, for instance, ascertaining the heights of buildings, &c., will be seen from the following: Suppose D E to be an eminence, of which it is required to find the height, and EC the length of the shadow cast by the sun; then, in order to find DE, we may erect perpendicularly at C a pole of any known length, as BC, and after measuring the length of its shadow A C, state-as the length of the pole's shadow is to the height of the pole itself, so is the length of the shadow of D E to the height of DE; or, As AC: CB::CE: ED; and supposing A C= 6 feet, BC= 4 feet, and CE = 30 feet, then ED would be 20 feet. A Again, supposing we wished to find the distance between two objects A and B; draw DB of any length at right angles to A B, and in DB take any point C, through which draw A E; also, at D, at right angles to DB, draw D E, making the triangle D E C, and state, As DC: DE:: BCBA. PROBLEM V. To find the area of any regular polygon. RULE. Multiply the sum of its sides by a perpendicular drawn from its centre to one of its sides, and take half the product for the area. Or, multiply the square of the side of a polygon (from three to twelve sides) by the numbers in the fourth column of the table for polygons, opposite the number of sides required, and the product will be the area nearly. E EXAMPLE 1. Required the area of the regular pentagon ABCDE, each side being 7.5, and perpendicular F G = 6.4. B 7.5 x 5 x 6.4 120, the area. EXAMPLE 2. What is the area of a regular hexagon, each side being 8.75 in length? 8.752 × 2.598 = 199-009375, the area. TABLE of multipliers for polygons from three to twelve sides. 1. The breadth of a polygon given, to find the radius of a circle to contain that polygon. RULE. Multiply half the breadth of the polygon by the numbers in the first column opposite to its name, or number of sides, and the product will be the radius of a circle to contain that polygon. And if the polygon have an unequal number of sides, the half breadth is accounted from its centre to one of its sides. 2. The radius of a circle given, to find the length of side. RULE. Multiply the radius of any circle by the numbers in the second column opposite the polygon required, and the product will be the length of side nearly that will divide that circle into the proposed number of sides. And, 3. The length of side given, to find the radius. RULE. Multiply the given length of side by the numbers in the third column opposite the polygon required, and the product will be the radius of a circle to contain that polygon. EXAMPLE 1. Required the radius of a circle to contain an octagon, whose breadth A B 18.5 inches. = = 9.99 Half of 18.5 9.25, and 9.25 × 1.08 or ten inches nearly, the radius of the circle OD. B 9.99 inches, required the 9:99 × 765764235, the length of side. |