SURVEYING & CIVIL ENGINEERING. SECTION I. The area of any plain figure, is the measure of the space contained within its bounds, without any regard to thick ness. This area is estimated by the number of small squares that may be contained in it. The contents may be square inches, square feet, square yards, or square perches, &c. A four-pole chain, or a two-pole is the instrument used for measuring the boundaries of a survey. A four-pole chain is 66 feet in length, and divided into 100 links. Therefore each link is 7. 92 inches. A two-pole chain is 33 feet, divided into fifty equal parts or links. It has been agreed that the area of any lot of ground, should be as measured by that portion of the horizontal plane which the boundary lines enclose. = 1 Acre 10 square chains=100000 square links. acre 4 roods=160 perches. ➡ Prob. 1. To find the area of any paralleogram; whether it be a square, a rectangle, a rhumbus, or a rhomboid. 1 Rule. Multiply the length by the perpendicular breadth, the product will be the area. (The truth of this rule is evident from (36-1.) Ex. .1. Required the area of a plantation, in the form of a rhombus, whose length is 64 chains and breadth 15 chains. 29 Ans. 6a Or OP. Prob. 2. To find the area of triangle when the base and perpendicular are given. Rule-Multiply the base by the perpendicular and half the product will be the area. (This rule is plain, as the triangle is half the paralleogram upon the same base and altitude 41. 1.) Ex. What is the area of a field, in the form of a triangle, having the base 82 chains and the altitude 20. 12 chains. Ans. 82a 1r 33p. Prob. 3. To find the area of a trapezoid. Rule-Multiply half the sum of the two perpendiculars by the base, and the product will be area. (The truth of this rule is manifest, as the figure may be divided into two right-angled triangles.). Ex. Required the contents of a lot of land in the form of a trapezoid, whose base is 41. 10 perches and ulars 10. 12 and 6. 40 perches. perpendic Ans. 33a 3r 30P. Prob. 4. To find the area of a circle, or an ellipsis. Rule. Multiply the square of the circle's diameter, or the products of the two diameters of the ellipsis, by 7854 for the area. (Proof. The area of a circle whose diameter is 1, is 7854 nearly. And circles are to each other as the squares of their diameters. Eucl. 2. 12, &c. Or thus-Multiply half the circumference by half the diameter, and the product will give the area. (Proof— Every circle may be divided into a regular polygon of an infinite number of sides, and the radius = perpendicular of such polygon; therefore, half the circumference by half the diameter will give the area.) Note. The ratio of the diameter to circumference is as 7 is to 22, or 1 to 3. 1416, &c. Hence it is easy to find either diameter or circumference by having one given, and consequently the area. Ex. .. What is the area of a circle whose diameter is 10 chains? Ans. 78. 54. Ex. 2. What is the area of an ellipsis whose diameters are 70 and 50 feet? Ans. 2748. 9. Prob. 5. To find the area of a plain triangle when the three sides are given. Rule. From half the sum of the three sides subtract each side severally. Multiply the half sum and the three remainders continually together, the square root of the product will be the area. Demonstration.—Let ABC figure 1, be the given triangle. Draw the parallels CP and BX, meeting the two sides CA, AB, and making AX=CX, and AC=AP. Also, draw the perpendicular AN bisecting BX and PC in N and R; and NOS parallel to the side BC, meeting PC produced in S. Again, with the centre O and radius ON, describe the circle meeting MO in Q, which must pass through R, because a right angle, and through S, because NS is parallel to BC, and NO=OS. Now, it is evident that AO is = sum of the two sides AC and AB, and QO, half BC, hence QA is half of the sum of the sides. From this half sum take each side, and the three remainders will be AM, MC, and CQ. By problem 2, CR × RA= CPA, and CR × RN CBP. Consequently, CR. NA is = ABC. But by similar triangles, CR: RA::XN: NA, (E. 16. 6.) CR. AN = RA. NX; therefore, RC2 NA' CR. AN. RA. NX, = CA. AM, MC. CQ. Take the square root, and RC. AN = the square root of half sum by the three diffc. which is the rule for the area of the triangle. = = Ex. What quantity of land is contained in a triangle whose three sides measure 20. 41, 24. 16, and 30. 12 chains. Ans. 24a 2r 9P. Second Limma to find the area of a triangle when the three sides are given. Square the sum of any two sides, take the square of the third side from it; take the diffc. of the same two sides and square it, take this from the square of the third side, multiply these two numbers together and one-fourth the square root will be the area of the triangle. Demonstration.-Let ABC figure 2, be the proposed triangle. Let fall the perpendicular AP, and bisect BC in O. Take AC=b, AB=c, and BC=a. Now, we get b2-c2=2a. PO, and b2-c2 2 a a 2 b2--c2 Square both sides, and PC2=( + PO, add to both CB Take each side from AC, and we get b',-( =PA'. And PA2= =PC; that is, + b2 c22 2 a a+b2-c2, 2 a Factor this and 2 2 a 2 multiply off by c2—(a—h)3 >1×1, gle as required. Prob. 6. To find the area of a triangle when the two sides and the contained angle are given. Rule.-Multiply the product of the two sides by half the natural sine of the included angle, for the area. and you get PAX = >(a+b) c2 × which gives the area of the trian Demonstration.-Let the last figure be the proposed triangle, AC and CB the given sides, and C the contained angle. Make CN=radius, and let fall the perpendicular MN, which is the natural sine of the angle C. Similar triangles, we get CN: NM::CA: AP, and (16. 6.) CN. AP= BC BC NM. CA. Multiply off by, and AP.BC CA, |