MENSURATION OF SURFACES. The area, or content of a surface, is determined by finding how many times contains some other surface which is assumed as the unit of measure. Thus, when we say that a square yard contains 9 square feet, we should understand that one square foot is taken for the unit of measure, and that this unit is contained 9 times in the square yard. The most convenient unit of measure for a surface, is a square whose side is the linear unit in which the linear dimensions of the figure are estimated. Thus, if the linear dimensions are feet, it will be most convenient to express the area in square feet; if the linear dimensions are yards, it will be most convenient to express the area in square yards, &c. We have already seen (Book IV. Prop. IV. Sch.), that the term, rectangle or product of two lines, designates the rectangle constructed on the lines as sides; and that the numerical value of this product expresses the number of times which the rectangle contains its unit of measure. PROBLEM I To find the area of a square, a rectangle, or a parallelogram. Rule.—Multiply the base by the altitude, and the product will be the area (Book IV. Prop. V.). 1. To find the area of a parallelogram, the base being 12.25 and the altitude 8.5. Ans. 104.125. 2. What is the area of a square whose side is 204.3 feet? Ans. 41738.49 sq. ft. 3. What is the content, in square yards, of a rectangle whose base is 66.3 feet, and altitude 33.3 feet? Ans. 245.31. 4. To find the area of a rectangular board, whose length is 12 feet, and breadth 9 inches. Ans. 9. 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. 211 CASE I. product. Or, multiply one of these dimensions by half the other (Book IV. Prop. VI.). a 1. To find the area of a triangle, whose base is 625 and alti. tude 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. 665. 3. To find the number of square yards in a triangle, whose base is 49 and altitude 251 feet. Ans. 68.736). CASE II. When two sides and their included angle are given. RULE.-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 A are given BA, BC, and the included an a gle B. From the vertex A draw AD, perpendicular to the base BC, and represent the B D area of the triangle by Q. Then, R: sin B :: BA : AD (Trig. Th. I.): BAx sin B hence, AD R (Book IV. Prop. VI.); 2 BC x BA X sin B or 2Q= R Taking the logarithms of both numbers, we have log. 2Q=log. BC+log. BA +log. sin B-log. R; which proves the rule as enunciated. 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 2.099715 1.760799 Then, log. 2Q= + log. sin B 57° 25' 9.925626 -log. R -10. log. 2Q 3.786140 and 2Q=6111.4, or Q=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' ? 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. CASE III. When the three sides are known. RULE.-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 remainders, 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 ABC be the given triangle. АС Take CD equal to the side CB, and draw DB; draw AE parallel to DB, meeting CB produced, in E: then DA B H E AB, meeting CA in H, and EA pro A G duced, in I. Lastly, with the cen K tre H and radius HF, describe the circumference of a circle, meeting CA produced in K: this circumference will pass through I, because AI=FB-FD, therefore, HF=H1; and it will also pass through the point G, because FGI is a right angle. Now, since HA=HD, CH is equal to half the sum of the sides CA, CB; that is, CH=CA+{CB; and since HK is equal to IF={AB, it follows that CK={AC+ CB+jAB= S, by representing the sum of the sides by S. Again, HK=HI={IF=AB, or KL=AB. Hence, CLOCK-KL=;8—AB, and AL=DK=CK-CD= S--CB. Now, AGX CG= the area of the triangle ACE, and AG X FG= the area of the triangle ABE; therefore, AG X CF= the area of the triangle ACB. Х Also, by similar triangles, AG : CG :: DF: CF, or AI : CF; therefore, AGX CF= triangle ACB=CG x DF=CGX AI; consequently, AG X CFx CG x AI= square of the area ACB. But CG x CF=CKX CL= S(S-AB), and AG X AI =AK XAL=CS-CA) x (IS-CB); therefore, AG X CFX CG x AI ESGS - AB) x (IS-CA) X (IS–CB), which is equal to the square of the area of the triangle ACB. 1. To find the area of a triangle whose three sides are 20, 30, and 40. 20 45 45 45 half-sum. 40 5 3d rem. 2)90 45 half-sum. 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. a PROBLEM III. To find the area of a trapezoid. RULE.—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 (Book IV. Prop. VII.). 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. 133 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. 2053; PROBLEM IV. To find the area of a quadrilateral. Rule.—Join two of the angles by a diagonal, dividing the quad rilateral into two triangles. Then, from each of the other angles let fall a perpendicular on the diagonal: then multiply А а the diagonal by half the sum of the two perpendiculars, and the product will be the area. 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. 2 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= 39.11, Bb=4, Dd=7.26, Aa=4.18, required the area. Ans. 296.1292. a A. B 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 per pendicular breadths of the figure, and do the same at several intermediate points, at equal distances from each other. 2. Add together the intermediate breadihs 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 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 |