Secondly let ▲ A B C have all its angles acute, and let A D and B E be perpendiculars from A and B on B C and AC; then the square on A B is equal to the sum of the squares on AC and CB, less twice either of the rectangles A C, CE, or B C, C D. Make the same construction as before, then since less twice either of the rectangles C Q or CR. Note. Since Theorems LIII., LIV., LV. are mutually-exclusive propositions, their converses are true, namely, Cor. The vertical angle of a triangle is right, acute, or obtuse, according as the square on the base is equal to, less than, or greater than the sum of the squares on the sides. THEOREMS FOR EXERCISE. 104. The square on the perpendicular drawn from the vertex of an acute-angled triangle on the base is equiva lent to the sum of the rectangles contained by the segments of the base and that contained by one of the sides and the projection of the other upon it. 105. The square on the base of an isosceles triangle is equivalent to twice the rectangle contained by either side and the projection of the base upon it. 106. The squares on the two sides of a triangle are together equal to double the square upon the line joining the vertex to the middle point of the base and the square on half the base. 107. The squares on the sum and the difference of two straight lines are together equal to twice the sum of the squares on the lines. 108. The squares on the sides of any parallelogram are together equivalent to the squares on the two diagonals. 109. The squares on the diagonals of any quadrilateral are together equal to twice the squares on the straight lines joining the middle points of the opposite sides. IIO. The squares on the sides of any quadrilateral are together greater than the squares on the two diagonals by four times the square on the line joining the middle points of the two diagonals. III. From B, one of the angles of a triangle A B C, a perpendicular BD is let fall on A C. Show that the difference of the squares of AB, BC is equal to the difference of the squares of A D, D C. 112. Three times the sum of the squares on the sides of a triangle is equal to four times the sum of the squares of the lines joining the middle point of each side with the opposite angle. 113. The squares on the diagonals of a quadrilateral are together equal to double the squares on the sides of the parallelogram formed by joining the middle points of the sides of the quadrilateral. 114. Show that the sum of the squares on the lines joining the angular points of a square to any point within it is double the sum of the squares on the perpendiculars from that point to the sides. 115. A B C D is a quadrilateral; E F G H the middle points of its sides: prove that the quadrilateral E F G H is half A B C D. 116. The sum of the squares described on the sides of a rhombus is equal to the sum of the squares on the diagonals. Easy Exercises. Draw a straight line I inch in length, and find a line whose length is represented by /2 inches, a line whose length is 3 inches, and a line √21 inches. Draw a square on a line 1 inch long, and divide it so as to show how many square inches are contained in it. Draw a triangle with sides of three, four, and five half inches, and find as nearly as possible the numbers of degrees in the angles. The sides of a quadrilateral inscribed in a circle are respectively 8.4, 18.7, 13.3, and 15.6 inches. Find the measure of two of the angles and the diameter of the circle. SECTION III.-Surveying. To find the Area of an Irregular Figure bounded by Straight Lines.1st Method. Every such figure may be divided into triangles (Fig. 143); then the area of the figure will be the sum of the areas of these triangles. 2nd Method.—Draw a diagonal AE (Fig. 144), and from each of the angles let fall perpendiculars upon it. Then A B' x B B' twice area of triangle A B'B; B'C' x (BB' + C C') B = The area of the figure A'A B'B, B'B C C', The sum of these products divided by 2 will give the area. 3rd Method. Instead of a diagonal we may take any line, M N,and draw perpen diculars to it (Fig. 145). the sum of the trapezoids C'C D D', E'EFF', F'FGG' + the A's HGG', KD D'. - the A's HA A', KEE'. The Cross-staff and Theodolite.—Perpendiculars in fieldsurveying are raised by means of a theodolite or of a crossstaff. Fig. 146 represents a cross-staff, consisting of an eightsided 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 is at right angles to one in the direction B B'. To find with a cross-staff a point A, in M N, such that a line, BA, from a given point, B, is perpendicular to M N. 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 one line of sight; and, and, A B the perpendicular to it. FIG. 146 In field-surveying, as the lines are measured, their dimensions are B M FIG. 147. О В В/423 usually entered in a book in three lines. 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. The dimensions of a field of the shape indicated in Fig. 144, and measured in links, for example, may be entered thus:- C C/640 D'D 490 O Left 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' — AC' — A B' — 620 — 230=390. The double areas will be : Hence the whole area equals half the sum of these quantities. |