4. To find the area of a rectangular board, whose length is 12 feet, and breadth 9 inches. Ans. 93 sq. ft. 5. To find the number of square yards of painting in a parallelogram, whose base is 37 feet, and altitude 5 feet 3 inches. Ans. 2112. 5. To find the area of a triangle. CASE I. When the base and altitude are given. Multiply the base by the altitude, and take half the product. Or, multiply one of these dimensions by half the other (B. IV., P. 6). Ex. 1. To find the area of a triangle, whose base is 625, and altitude 520 feet. Ans. 162500 sq. ft. 2. To find the number of square yards in a triangle, whose base is 40, and altitude 30 feet. Ans. 663. 3. To find the number of square yards in a triangle, whose base is 49, and altitude 25 feet. Ans. 68.7361. CASE II. 6. When two sides and their included angle are given. Add together the logarithms of the two sides and the logarithmic sine of their included angle; from this sum subtract the logarithm of the radius, which is 10, and the remainder will be the logarithm of double the area of the triangle. Find, from the table, the number answering to this logarithm, and divide it by 2; the quotient will be the required area. Let BAC be a triangle, in which there are given BA, BC, and the included angle B. From the vertex A draw AD perpendicular to the base BC, and represent the area of the triangle by Q. R: sin B :: BA: AD; B D Then (Trig. Th. I.), - hence, by substituting for AD its value, we have, BC X BA X sin B Q = Then, log. 2 Q Taking the logarithms of both members, we have, log. 2 Qlog, BC + log. BA + log. sin Blog R; the formula of the rule as enunciated. = or, 2Q log. 2 Q and 2 Q 6111.4, or Q = Ex. 1. What is the area of a triangle whose sides are, BC= 125.81, BA= 57.65, and the included angle B = 57° 25'? + log. BC + log. BA = = BCX BA X sin B R 125.81 + log. sin B 57° 25′ CASE III. 2.099715 1.760799 9.925626 -10. 3055.7, the required area. 2. What is the area of a triangle whose sides are 30 and 40, and their included angle 28° 57'? 3.786140 Ans. 290.427. 3. What is the number of square yards in a triangle of which the sides are 25 feet and 21.25 feet, and their included angle 45°? Ans. 20.8694. 7. When the three sides are known. 1. Add the three sides together, and take half their sum. 2. From this half-sum subtract each side separately. 3 Multiply together the half-sum and each of the three re mainders, and the product will be the square of the area of the triangle. Then, extract the square root of this product, for the required area. Or, After having obtained the three remainders, add together the logarithm of the half-sum and the logarithms of the respective remainders, and divide their sum by 2: the quotient will be the logarithm of the area. Let ACB be a triangle: and denote the area by Q: then, by the last case, we have, Q = A By substituting in this equation the values of sin A, and cos A, found in Arts. 92 and 93, Plane Trigonometry, we obtain, 45 half-sum. √s (sa) (s—b) (s—c). Ex. 1. To find the area of a triangle whose three sides 20, 30, and 40. 20 30 40 2)90 45 20 25 1st rem. b C 45 30 15 2d rem. B 45 half-sum. 40 5 3d rem. = Then, 45 x 25 x 15 x 5 84375. The square root of which is 290.4737, the required area. 2. How many square yards of plastering are there in a triangle whose sides are 30, 40, and 50 feet? Ans. 663. 8. To find the area of a trapezoid. « Add together the two parallel sides: then multiply their sum by the altitude of the trapezoid, and half the product will be the required area (B. IV., P. 7). Ex. 1. In a trapezoid the parallel sides are 750 and 1225, and the perpendicular distance between them is 1540; what is the area? Ans. 152075. 2. How many square feet are contained in a plank, whose length is 12 feet 6 inches, the breadth at the greater end 15 inches, and at the less end 11 inches? Ans. 131 sq. ft. 3. How many square yards are there in a trapezoid, whose parallel sides are 240 feet, 320 feet, and altitude 66 feet? Ans. 20531. 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 perpendiculars Dg, Bb, equal to 18 and 16 feet? Ans. 714. A 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, EC 39.11, Bb = 4, Dd = 7.26, Aa = = area. E 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. 2221 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 separately, and add them together for the contents of the whole polygon. = a A D D 4.18: required the 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. B C D 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, Bb, Cc, Dd, Ee, to the base line AE, and designate them respectively by the letters a, b, c, d, and e. Then, the area of the trapezoid ABba = α ( a + b + c + d 2 8.2 8.6 2)16.8 a + 8.4 mean of the extremes. 7.4 9.2 10.2 35.2 sum. the area of the trapezoid BCcb the area of the trapezoid CDdc= and the area of the trapezoid DEed = × DE; hence, their sum, or the area of the whole figure, is equal to 150 a + b b+c c + d + + + d + e 2 2 2 2 e b+c 2 c+d 2 d+e 2 с a+b 2 × BC, since AB, BC, &c., are equal to each other. But this sum is also equal to d 35.2 sum. 10 352 area. × AB, × AB, × AB, × CD, which corresponds with the enunciation of the rule. Ex. 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. 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. |