9. To find the area of a quadrilateral. Join two of the angles by a diagonal, dividing the quadrilateral into two triangles. Then, from each of the other angles let fall a perpendicular on the diagonal: then multiply the diagonal by half the sum of the two perpendiculars, and the product will be the area. Ex. 1. What is the area of the · quadrilateral ABCD, the diagonal AC being 42, and the perpendic ulars Dg, Bb, equal to 18 and 16 feet? Ans. 714. A B 2. How many square yards of paving are there in the quadrilateral whose diagonal is 65 feet, and the two perpendiculars let fall on it 28 and 33 feet? Ans. 22217 10. To find the area of an irregular polygon. Draw diagonals dividing the proposed polygon into trapezoids and triangles. Then find the areas of these figures sepa rately, and add them together for the contents of the whole polygon. Ex. 1. Let it be required to determine the contents of the polygon ABCDE, having five sides. Let us suppose that we have measured the diagonals and perpendiculars, and found AC = 36.21, D E a d C A B = EC 39.11, Bb = 4, Dd = 7.26, Aa 4.18: required the area. = Ans. 296.1292. 11. To find the area of a long and irregular figure, bounded on one side by a right line. 1. At the extremities of the right line measure the perpendicu lar breadths of the figure; then divide the line into any number of equal parts, and measure the breadth at each point of division. 2. Add together the intermediate breadths and half the sum of the extreme ones: then multiply this sum by one of the equal parts of the base line: the product will be the requir ed area, very nearly. a Let AEea be an irregular figure, having for its base the right line AE. Divide AE into equal parts, and at the points of division A, B, C, D, and E, erect the perpendiculars Aa, to the base line AE, and designate them the letters a, b, c, d, and e. Then, the area of the trapezoid ABba the area of the trapezoid BCcb = the area of the trapezoid CDdc= and the area of the trapezoid DEed = hence, their sum, or the area of the whole figure, is equal to since AB, BC, &c., are equal to each other. But this sum is also equal to e (S + b + c + d + 2 ) × AB, 2 +b+c+d+ which corresponds with the enunciation of the rule. Ex. 1. The breadths of an irregular figure at five equi distant places being 8.2, 7.4, 9.2, 10.2, and 8.6, and the length of the base 40: required the area. 2. The length of an irregular figure being 84, and the breadths at six equidistant places 17.4, 20.6, 14.2, 16.5, 20.1, and 24.4; what is the area? Ans. 1550.64. 12. To find the area of a regular polygon. Multiply half the perimeter of the polygon by the apothem, or perpendicular let fall from the centre on one of the sides, and the product will be the area required (B. V., p. 8). REMARK I.—The following is the manner of determining the perpendicular when one side and the number of sides of the regular polygon are known: First, divide 360 degrees by the number of sides of the polygon, and the quotient will be the angle at the centre; that is, the angle subtended by one of the equal sides. Divide this angle by 2, and half the angle at the centre will then be known. Now, the line drawn from the centre to an angle of the polygon, the perpendicular let fall on one of the equal sides, and half this side, form a right-angled triangle, in which there are known the base, which is half the side of the polygon, and the angle at the vertex. Hence, the perpendicular can be determined. Ex. 1. To find the area of a reg ular hexagon, whose sides are 20 feet Perimeter 120, and half the perimeter Then, = = 60. 60 x 17.3205 = 1039.23, the area. 2. What is the area of an octagon whose side is 20? Ans. 1931.36886. REMARK II.—The area of a regular polygon of any number of sides is easily calculated by the above rule. Let the areas of the regular polygons whose sides are unity or 1, be calculated and arranged in the following Now, since the areas of similar polygons are to each other as the squares of their homologous sides (B. IV., P. 27), we have, 1o : any side squared :: tabular area : area. Hence, to find the area of any regular polygon, 1. Square the side of the polygon. 2. Then multiply that square by the tabular area set opposite the polygon of the same number of sides, and the product will be the required area.. Ex. 1. What is the area of a regular hexagon whose side is 20? 20 400, tabular area = 2.5980762. = Hence, 2.5980762 × 400 = 1039.2304800, as before. 2. To find the area of a pentagon whose side is 25. Ans. 1075.298375. 3. To find the area of a decagon whose side is 20. Ans. 3077.68352. 13. To find the circumference of a circle when the diame ter is given, or the diameter when the circumference is given. Multiply the diameter by 3.1416, and the product will be the circumference; or, divide the circumference by 3.1416, and the quotient will be the diameter. It is shown (B. V., P. 16, s. 1), that the circumference of a circle whose diameter is 1, is 3.1415926, or 3.1416. But, since the circumferences of circles are to each other as their radii or diameters, we have, by calling the diameter of the second circle d, Exc. 1. What is the circumference of a circle whose diameter is 25? Ans. 78.54. 2. If the diameter of the earth is 7921 miles, what is the circumference? Ans. 24884.6136. 3. What is the diameter of a circle whose circumfer ence is 11652.1904? Ans. 3709. 4. What is the diameter of a circle whose circumfer ence is 6850? Ans. 2180.41. 14. To find the length of an arc of a circle containing any number of degrees. Multiply the number of degrees in the given arc by 0.0087266, and the product by the diameter of the circle. Since the circumference of a circle whose diameter is 1, is 3.1416, it follows, that if 3.1416 be divided by 360 degrees, the quotient will be the length of an arc of 1 3.1416 degree: that is, = 0.0087266 = arc of one degree 360 to the diameter 1. This being multiplied by the number of degrees in an arc, the product will be the length of that are in the circle whose diameter is 1; and this product being then multiplied by the diameter, the product is the length of the arc for any diameter whatever. REMARK.-When the arc contains degrees and minutes, reduce the minutes to the decimal of a degree, which is done by dividing them by 60. Ex. 1. To find the length of an arc of 30 degrees, the diameter being 18 feet. Ans. 4.712364. 2. To find the length of an arc of 12° 10' or 12°, the diameter being 20 feet. Ans. 2.123472. 3. What is the length of an arc of 10° 15', or 104°, in a circle whose diameter is 68? Ans. 6.082396 |