whence, Q = bc sin A cos A. Substituting for sin A and cos A, their values, taken from Lemma, and reducing, we have RULE.-Find half the sum of the three sides, and from it subtract each side separately. Find the continued product of the half sum and the three remainders, and extract its square root; the result will be the area required. It is generally more convenient to employ logarithms; for this purpose, applying logarithms to the last equation, we have [log is + log (†s—a) + log (†s—b) + log (†s—c)]; hence, we have the following RULE.-Find the half sum and the three remainders as before, then find the half sum of their logarithms; the number corresponding to the resulting logarithm will be the area required. Examples. 1. Find the area of a triangle, whose sides are 20, 30, and 40. = We have s 45, is a 25, sb 15, sc = 5. = — = įs By the first rule, Q: = √45 × 25 x 15 x5 = 290.4737, Ans. 2. How many square yards are there in a triangle, whose sides are 30, 40, and 50 feet? Ans. 663. To find the area of a trapezoid. 98. From the principle demonstrated in Book IV., Prop. VII., we may write the following RULE.-Find half the sum of the parallel sides, and multiply it by the altitude; the product will be the area required. Examples. 1. In a trapezoid the parallel sides are 750 and 1225, and the perpendicular distance between them is 1540; what is the area? Ans. 1520750. 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. 13. 3. How many square yards are there in a trapezoid, whose parallel sides are 240 feet, 320 feet, and altitude 66 feet? Ans. 2053 sq. yd. To find the area of any quadrilateral. 99. From what precedes, we deduce the following RULE. Join the vertices of two opposite angles by a diagonal; from each of the other vertices let fall perpendiculars upon this diagonal; multiply the diagonal by half of the sum of the perpendiculars, and the product will be the area required. Examples. 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 sq. ft. b 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? To find the area of any polygon. Ans. 2221. 100. From what precedes, we have the following RULE.-Draw diagonals dividing the proposed polygon into trapezoids and triangles: then find the area of these figures separately, and add them together for the area of the whole polygon. and found AC 36.21, EC 39.11, Bb = 4, Dd 7.26, Aa 4.18 = required the area. Ans. To find the area of a regular polygon. 101. Let AB, denoted by s, represent one side of a regular polygon whose centre is C. Draw CA and CB, and from C draw CD perpendicular to AB. Then will CD be the apothem, and we shall have AD BD. 296.1292. Denote the number of sides of the polygon by n; then will the angle ACB, at the centre, be 360° equal to n (B. V., page 144, D. 2), and the angle ACD, 180° which is half of ACB, will be equal to n In the right-angled triangle ADC, we shall have, formula (3), Art. 37, Trig., CD = Įs tan CAD. But CAD, being the complement of ACD, we have a formula by means of which the apothem may be computed. But the area is equal to the perimeter multiplied by half the apothem (Book V., Prop. VIII.): hence the following RULE. Find the apothem, by the preceding formula; multiply the perimeter by half the apothem; the product will be the area required. The perimeter is equal to 120: hence, denoting the area by Q, Q = 120 x 17.3205 =1039.23, Ans. 2. What is the area of an octagon, one of whose sides is 20? Ans. 1931.37. The areas of some of the most important of the regular polygons have been computed by the preceding method, on the supposition that each side is equal to 1, and the results are given in the following |