If the side be 7 yds. in length, the square contains 49 square yds. If the square contain 144 square in., the side must be 12 in. long. 153. The area of a parallelogram=basex height. For we have seen it is equal to the rectangle with the same base and height. 154. The area of a triangle basex height 2 2 For the triangle is half a parallelogram with the same base and height. = Hence the area of a rhombus half the product of its two diagonals. From the fact that the number of units of area in a square is found by multiplying the number of units of length in the side by itself, multiplying a number by itself is termed squaring the number. The square of the number of units in a line A B is written A B2. 155. From the preceding proposition, when two sides of a right-angled triangle are given we may find the third. Thus, if A B C be the triangle, and C the right angle, (AB)=(AC)+(BC); (A C)=(AB)2(BC)2; (BC)2=(A B)2—(A `C)2. The following arithmetical facts may be applied at once to geometry. Ist. The difference of the squares of two numbers is equal to the product of their sum and difference. 2nd. The square of the sum of two numbers is equal to the sum of the squares together with twice the product. 3rd. The square of the difference of two numbers is equal to the sum of the squares less twice their product. If we take the numbers as measures of lines, these facts may be written thus: Ist. The difference of the squares on two lines is equal to the rectangle whose sides are respectively the sum and difference of these lines. Thus,- A B C D2 (AB+CD) (AB-C D). = 2nd. If a line be divided into two parts, the square on the whole is equal to the sum of the squares on each part, together with twice the rectangle contained by the parts. 3rd. If a straight line be divided into two parts, the square on one part is equal to the sum of the squares on the whole line and the other part, less twice the rectangle contained by the whole line and this part. To describe a square equivalent to a given parallelogram. C 156. Let A B be the base of the parallelogram, and BH its height. Place these in a line, and with the middle point D of the line A H as centre, and half this line as radius, describe a semicircle. Draw the perpendicular B C to meet the circumference in C. The square on BC will be A equivalent to the parallelogram. H D B For let r be the length of the radius of the circle. Then A D=D H=DC=r; A B=r+ D B; BH=r-D B. If DC be drawn, then BC-DC2-DB2=r2-DB2 =(r+DB)x(r−D B)=AB×B H=the area of the parallelogram. = 157. The area of a trapezoid half the product of the sum of the bases by the height. To find the area of an irregular figure bounded by straight lines. 158. 1st Method. Every such figure may be divided into triangles (fig. 134); then the area of the figure will be the sum of the areas of these triangles. 2nd Method. Draw a diagonal AE (fig. 135), and from each of the angles let fall perpendiculars upon it. Then AB X BB= twice area of triangle A B'B; B'C' x (BB' + CC') = twice area B B' C' C, and so on. B Ꭰ The sum of these products divided by 2 will give N the area. 3rd Method. Instead of a diagonal we may take any line, MN, and draw perpendiculars to it (fig. 136). =the sum of the trapezoids A A'B B', B' BC C', C'C D D', E'EFF', F' FGG' + the triangles HGG', KD D'-the triangles HA A', KEE'. The cross-staff. 159. Perpendiculars in field surveying are raised by means of a theodolite or of a cross-staff. Figure 137 represents this instrument. It Fig. 137. Fig. 138. consists of an eight-sided block about four inches long, and a staff tipped with a spike of iron to enter the ground. The opposite sides of the block are parallel and the alternate sides are at right-angles. Each side is furnished with a slit, or sight, such that a line in the direction of the opposite sights, A A', is at right-angles with one in the direction B B'. 160. From a point A in a straight line, MN, to draw a perpendicular with the cross-staff. Place the foot of the staff at A and fix it vertically. Turn it round until the line of sight in one direction is A N, and in the opposite A M; that is, until the rods marking the points MN can be seen in the opposite directions without moving the staff. look along the line at right-angles to M N, and cause another rod to be planted at B in that direction. The line A B is perpendicular to M N. Now 161. To find a point, A, in M N, such that a line, BA, from a given point, B, is perpendicular to MN. Find as near as possible the point A, by observation, then, by trying various positions, fix the staff so that-1st, the line MN has the direction of M B one line of sight; and 2nd, AB the perpendicular to it. 162. As the lines are measured, their dimensions are usually entered in a book in three lines. N The middle one contains the distances of the stations, perpendiculars or offsets from the starting point. Offsets to the left are entered above, and those to the right below. Fig. 139. The dimensions of a field of the shape indicated in figure 135, and measured in links, for example, may be entered thus :— Left O BB'420 CC'640 D'D 490 O Base line A A B'230 AG'450 AC'620 AF1210 AD'1260 AE 1520 B Right G'G 610 F'F 500 From this we can easily find the bases of the trapezoids and triangles by taking the differences of the distances of the two extremities. Thus B'C' — A C'— A B' 620-230=390. The double areas will be— = ABB' 420 base 230 = B B'C' C 420 C C'D'D 640 D D'E 490 add 640 add 490 base 260 |