the diagonal by half the sum of the two perpendiculars, and the product will be the area. 1. What is the area of the quadrilateral ABCD, the diagonal AC being 42, and the perpendiculars Dg, Bb, equal to 18 and 16 feet? Ans. 714. A D 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. 222 PROBLEM V. To find the area of an irregular polygon. RULE.-Draw diagonuls dividing the proposed polygon into trapezoids and triangles. Then find the areas of these figures separately, and add them together for the content of the whole polygon. 1. Let it be required to determine the content of the polygon ABCDE, having five sides. Let us suppose that we have measured the diagonals and perpendiculars, and found AC-36.21, EC= E a A. 39.11, Bb=4, Dd=7.26, Aa=4.18, required the area. B Ans. 296.1292. C PROBLEM VI. To find the area of a long and irregular figure, bounded on one side by a right line. RULE.-1. At the extremities of the right line measure the perpendicular breadths of the figure, and do the same at several intermediate points, at equal distances from each other. 2. Add together the intermediate breadths and ha'f 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 required area, very nearly. Let AEea be an irregular figure, having for its base the right line AE. At the points A, B, C, D, and E, equally distant from each other, erect the perpendiculars Aa, Bb, Cc, Dd, Ee, to the A d C base line AE, and designate them respectively by the letters a, b, c, d, and e. Then, the area of the trapezoid ABba: × AB, the area of the trapezoid BCcb= the area of the trapezoid CDdc= and the area of the trapezoid DEed= a+b hence, their sum, or the area of the whole figure, is equal to a+b b+c c+d d+c + + + 2 2 2 ХАВ, since AB, BC, &c. are equal to each other. But this sum is also equal to which corresponds with the enunciation of the rule. 1. The breadths of an irregular figure at five equidistant places being 8.2. 7.4, 9.2, 10.2, and 8.6, and the length of the base 40, required the area. 8.2 8.6 2(16.8 8.4 mean of the extremes. 4)40 10 one of the equal parts. 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. PROBLEM VII. To find the area of a regular polygon. RULE I.-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 (Book V. Prop. IX.). REMARK I. The following is the manner of determining the perpendicular when only 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 equal side of the polygon, and the angle at the vertex. Hence, the perpendicular can be determined. Also, CAD=90°-ACD=60°; and AD=10. D B : sin CAD .: AD. 60° .10 0.301030 9.937531 1.000000 1.238561 Perimeter 120, and half the perimeter =60. Then, 60 × 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 calcu lated and arranged in the following Now, since the areas of similar polygons are to each other as the squares of their homologous sides (Book IV. Prop. XXVII.), we shall have 12 tabular area :: any side squared area. Or, to find the area of any regular polygon, we have RULE II.-1. Square the side of the polygon. 2. Then multiply that square by the tabular area sel opposite the polygon of the same number of sides, and the product will be the required area. 1. What is the area of a regular hexagon whose side is 20? 202=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. PROBLEM VIII. To find the circumference of a circle when the diameter is given, or the diameter when the circumference is given. RULE.--Multiply the diameter by 3.1416, and the product will be the circumference; or, divide the circumference by 3.1418, and the quotient will be the diameter. It is shown (Book V. Prop. XIV.), 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, 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 circumference is 11652.1904? Ans. 37.09. 4. What is the diameter of a circle whose circumference is 6850? Ans. 2180.41. PROBLEM IX To find the length of an arc of a circle containing any number of degrees. RULE.-Multiply the number of degrees in the given arc by 0.0087266, and the product by the diameter of the circle. 3.1416 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 degree: that is, =0.0087266= = arc of one degree to the diameter 1. This being multiplied by the number of degrees in an arc, the product will be the length of that arc in the circle whose diameter is 1; and this product being then multiplied by the diameter, will give the length of the arc for any diameter whatever. 360 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. 1. To find the length of an arc of 30 degrees, the diameter being 18 feet. Ans. 4.712364. 10', or 121°, the diam 2. To find the length of an arc of 12° eter being 20 feet. 3. What is the length of an arc of 10° cle whose diameter is 68? PROBLEM X. Ans. 2.123472. 15', or 101°, in a cir To find the area of a circle. Ans. 6.082396. RULE I.-Multiply the circumference by half the radius (Book V. Prop. XII.). RULE II-Multiply the square of the radius by 3.1416 (Book V. Prop. XII. Cor. 2). 1. To find the area of a circle whose diameter is 10 and circumference 31.416. Ans. 78.54. |