1. To measure a square having equal sides. RULE. Multiply the side of the square into itself, and the product will be the area. Let A be a true square, each side measuring 10 chains; multiply 10 chains, 00 links, by 10 chains, 00 links, thus, 10,000 10,00 10,00 10)00000 Multiply half the base by the whole perpendicular; or the whole base by half the perpendicular; or multiply the whole base and whole perpendicular together, and take half that product for the content. Either of these three ways will do. The longest side of a triangle is usually called the base, except in a right angled triangle, where the longest of the two legs, which include the right angle, is called the base. In the right-angled triangle, A BC, right angled at B; the base A B 10 chains; the perpendicular B C 13 chains, 70 links. Perpendicular 13 70 the base 4. To measure a Trapezium, or figure of four unequal sides, and un equal angles. A D B C RULE. Draw a diagonal line from one of the angles to the opposite angle as A B. and then will the trapezium be divided into two triangles, of which the diagonal is the common base; then letting fall perpendiculars from the other opposite angles, on the diagonal, add those perpendiculars together, and multiply half that sum into the diagonal, or half of the diagonal into the sum of the perpendiculars, and that product will be the area of the Trapezium.In the Trapezium A C B D, the diagonal A B is 4 chains, 58 links, the perpendicular D E, 1 chain, 55 links, and the perpendicular C F, 2 chains, 23 links. Divide the figure into triangles, by drawing diagonals from one angle to another; then measure all the triangles, by either of the rules, already taught, and the sum of their several areas will be the erea of the given figure. In the triangle A FB, the base F A 26,5 rods, and the perpendicular B a 12,5 rods; in the triangle FB E, the base BE 28 rods, and the perpendicular Fd 13 rods; in the triangle EBD, the base B E 28 rods, and the perpendicular Dc 16; in the triangle DC B, the base D C 22 rods, and the perpendicular B b 12 rods. See the work. 14- BE. The number of triangles, in any irregular figure, will be less, by two, and the diagonals less, by three, than the number of the sides of the figure. To measure a circle and its parts. In the annexed circle A D BFG E, the arch line A DBFGE is cal led the periphery, the lengthof which is called the cir cumference: Any line as A B, passing thro' the centre C, cuts the cir. cle into two equal parts, called semicircles, or half circles,and such lines are called diame ters, of the circle. To find the con tent of the whole circle, you must first know the length of the diameter, or the circumference; one of which being known, the other is easily found; for as 7 is to 22: : so is the diameter: to the circumference; and as 22 is to 7: so is the circumference to the diameter. In this annexed figure, the diameter A B is 2 chains or 200 links; which, multiplied by 22, and the product divided by 7, gives 6 ch. 28 lin and something more for the circumference Now, to find the superficial content, multiply half the circumference by half the diameter, the product will be the content; half the circumference is 3 ch. 14 lin.; half the diameter, 1 ch. 00 lin which, multiplied together, the product is 3,1400 square links, or 1 rood, 40 rods, the content of the circle. To find the content by the diameter only. As 14 is to 11 so is the square of the diameter: to the content. The square of the diameter is 40000; which multiplied by 11, makes 440000; which, divided by 14, gives 31428, or one rood. 14 rods, and something more. To find the content of the sector of a circle. Multiply half the compass thereof by the semidiameter of the circle; the product will be the answer. In the foregoing circle I would know the content of that lit tle piece D C B; the arch D B is 78,5 links; the half of it is 39,25 lin.; which, multiplied by the semidiameter, 1 ch. 00 lin. gives 3925 square links, or 6 25 rods To find the content of a segment of a circle, without knowing the di ameter. Let E F G, in the foregoing figure be the segment of a circle, the chord E F is 1 ch 70 lin. or 170 lin.; the perpendicular G H, 50 links; now multiply of the one by the whole of the other, the product will be the content nearly; the two-thirds of 170 is the nearest 113, which, multiplied by 50. produces 5650 square links, or 9 rods. 7. To measure an Ellipsis or Oval. Multiply the two diameters of the oval together; then,multiplying the product by 7854, this last product will be the area of the oval. In the annexed oval, A B C D, the transverse diameter A C, is 34 feet, and the conjugate diameter, B D, is 24, to find the area. 34 24 B 816 ,7854 3264 4080 6528 5712 610,8864 Area 640,8864 feet.. To Plot and find the content of a lot of land from the minutes in the field-book. Bounds. On whom boundest Likewise, if there A rock. be any thing you Stake & stones. wish to remark, as White oak tree.rivers, roads, buil Apple tree. dings, &c. you may 54 corner of a wall. set them down in 33.5 Heap of stones. Ithis column. |