II. MENSURATION OF SURFACES. I.-PLANE SURFACES. THE RECTANGLE. PROP. I. To find the area of a rectangle, having given two adjacent sides. Let AB= a feet, AC=b, feet, dividing AB into a equal parts, and AC into b equal parts;* and through the c points of division draw lines D B square foot; and as we have a rows of these rectangles, with b in each row, the whole number of rectangles will be Hence, to find the area, we multiply the two adjacent sides together. From (1) we have Hence, having given the area and one side, to find the adjacent side, we divide the area by the given side. Cor. 1.-Having given the side of a square equal to find S the area. We evidently have S= a. a = a2; a, Cor. 2. Having given the diagonal of a rectangle, and one of the sides, find the area. * We have here considered only the case in which each of the sides contains the unit an exact number of times. For general proof, see page 78. Cor. 3.-Having given the diagonal of a square, to D find its area. Hence we find the area of a square, when the diagonal is given, by taking half the square of the diagonal. THE PARALLELOGRAM. PROP. II. To find the area of a parallelogram, having given its base and perpendicular height. Let AB a, DE=p. Draw through C, CF perpendicular to AB, and meeting it produced in F. ABDC, FEDC are equal, standing on same base and between same parallels; but area of FEDC = CD x DE Cor.-Let AC-b, and perpendicular on it from B=q, we have area = b.q; THE TRIANGLE. PROP. III.-To find the area of a triangle, having given the base and the perpendicular height. Interpreting this, we find the area of the triangle by multiplying half the base by the perpendicular height. Cor. 1.-Let q denote the perpendicular from B on. AC. Then ... BC.AC.q; .. BC: AC :: 9:p. Cor. 2.-A triangle is equivalent to one-half of any parallelogram, having the same base and the same altitude. Cor. 3.-Triangles having equal bases and equal altitudes are equal. Cor. 4.-Triangles having equal altitudes are to each other as their bases; triangles having equal bases are to each other as their altitudes, and two triangles are to each other as the products of their bases by their altitudes. PROP. IV. To find the area of a triangle, when the three sides are given. In the triangle ABC, let BC=a, AC=b, AB =c. It is required to C. find the area. Let denote the per B D pendicular from A on BC. Then, by the last proposition, area of triangle = ap. If, then, we know p in terms of the data, we can find the area of the triangle. a 2 .. area of triangle × — √s(s− a)(s—b)( s − c) 2 = √s(s—a)(s—b)(s—c); ... (1) where s denotes half the sum of the sides of the triangle. Interpreting this formula, we find the area by taking half the sum of the sides; from this subtracting each side separately, multiplying half the sum and the three remainders together, and taking the square root of the product. Remark. It may here be noticed that each of the factors (s-a), (s—b), (s—c) is a positive quantity, for b+ca (Euclid, I. 20); .. a+b+c72a; Cor. 1.-If the triangle is isosceles, having each of the equal sides equal to a, and the remaining side=b, we have, by substituting, in (1), a for c, A similar result would have been obtained for the area, by first finding the perpendicular p and recollecting that the perpendicular in this case bisects the base. Cor. 2.-If the triangle is equilateral, each side being equal to a, we have, by substituting a for b in (2), Interpreting this formula, we find the area of an equilateral triangle by taking one-fourth of the square of the side, and multiplying it by the square root of 3. Conversely, if S denotes the area of an equilateral triangle, we can find the length of its side, 2 √S 4/3 Cor. 3.-Hence we can find the radius of the B inscribed circle. Since the triangle ABC= BOC+AOC+AOB xr=s•r; .. ABC= = |