like the following. up the table : Draw CPQ in the positions named, and fill Length of CP Length of CQ Area of rectangle When CPQ is just ceasing to cut the circle Can you State the property suggested by the last column. explain why the numbers in the last column are not exactly equal ? 29. On squared paper draw two straight lines AB and AC, containing an angle of 35°. Draw in several positions the base of a triangle, of which two sides lie along AB and AC, and whose area is 4 square inches, and so draw the curve that touches the base in all positions. Draw also the curve that touches the base in all positions, when the two sides lie along AB and AC, and the length of the perpendicular from A on the base is 2.3 inches. By laying your ruler against the two curves draw the base of the triangle that has an area of 4 square inches, and has also the perpendicular from A on the base 2.3 inches long. Also state how you could make a triangle having a vertical angle of 35°, area 4 square inches, and perpendicular from vertex on base 2.3 inches, without drawing any other loci than straight lines and circles. 30. About 3 inches away from the lower edge of a page draw a horizontal line MN right across the page, and from about its middle point X erect a perpendicular XF 2 cm. long. On the tracing-paper draw 10 circles having the same centre, the radii being 1, 2, 3, 4, &c., cm. respectively; then place the tracing-paper so that the circumference of the smallest circle passes through F and touches MN, and in this position prick through the centre of the circle. Deal with the second and the remaining circles in a similar way, determining the positions of the centre points on both sides of FX. Draw a freehand curve through the points thus obtained. 31. Draw a straight line to represent a coast, and two points L and M on it to represent lighthouses. Mark a point R to repre sent a rock in the offing. Measure the angle LRM. If a ship S is sailing past, prove that she will certainly be out of danger from the rock if she is steered so that the angle LSM is always less than the angle LRM. Draw the course of the ship, supposing the angle LSM is constantly kept at 125°. State your method. = 32. Once in ancient Greek times the problem of making a cubical altar of double the volume of an existing cubical altar had great importance. Take the edge of the original altar as 80 cm. and solve it (i) in the Greek fashion by the intersection of the graphs y2 160x and x2 = 80y, (ii) by the graph y (iii) by logarithms, and (iv) by the solution of the equation x3 = 2 as follows:-This equation has a root between 1 and 2; by putting x= 1+y find from it another having its roots less by 1. putting y z/10 find an equation in z. This equation has a root between 2 and 3; find from it an equation in u with its roots less by 2, and so on. = By CHAPTER XV THEOREM OF PYTHAGORAS 1. NAME any case you have had of a right-angled triangle in which you knew a relation between the three sides. If C is the centre of a circle, = AB a diameter, and MO the = right-angled. For suppose ABC a triangle in which B is A JE C FIG. 2. a right angle (Fig. 1). With C as centre draw a circle passing through A, and produce BC to form a diameter = DE of this circle; for such a figure it has been proved that AC2 AB2+ BC2. Or, again, with C as centre draw a circle through B (Fig. 2), and produce AC to form a diameter DE of this circle; we know that for this figure AC2 AB2 + BC2. Repeat these two proofs. = Ex. 1. Redraw the triangle ABC on squared paper, and draw the squares on AB, BC, and AC. Count the number of little squares contained in each of these three squares, and compare with the previous result. Ex. 2. With 1 cm. or 2 cm. as unit draw a triangle ABC having LB = 90°, BA 3 units, BC= 4 units. Find the length of AC by measuring, and also by calculation. (This property of the numbers 3, 4, 5 has been known from very early times, and is still used by builders to square the foundations of a house.) = B с F a 2. Draw on squared paper a triangle ABC (Fig. 3) having B = 90°, a = 4, c 3. Draw the squares on the sides of the triangle, and count the number of unit squares included in each. In the case of the square ACDE on b there are fractions to take account of. Can you avoid the discussion of these fractions by considering where D and E lie?-We go from A to E by going 4 units in the direction BA and 3 in the direction BC; let us say 4 units east and 3 units south. From C to D is also 4 units east and 3 south. Prove that -As G D H FIG. 3. E = if D and E are found in this way ACDE is a square.ABC and EFA have LB = LF, BC = FA, BA = FE, so that the As are congruent and AC AE. The congruence gives also LBCA =LFAE, so that LFAE+LBAC= LBCA+LBAC = 90°, leaving LCAE also = 90°. In the same way CD CA and LACD = 90°. || to AE and the figure is a parallelogram. are equal and all the angles right. = the square BFHG of Fig. 3?-BF is Thus CD is and Can you express the square contains 49 unit squares. the square on b by means of it? -We must subtract the 4 triangular corners, each equal to squares. So that b2 is 49-24 or 25 unit squares. AABC or six unit -If BC is a units 3. Generalize the preceding discussion. long and BA c units, the length of BF is a +c, and the area BFHG is (a+c)2 unit squares. And, the area of ABC being ac, we know that b2 = (a+c)2-2 ac. Give a diagram showing the squares on a, c, and a+c, and the rectangle ac.-Fig. 4 shows that (a+c)2-2ac is a2+c2; which we could have found algebraically by multiplying out (a+c) × (a + c). Again, discuss the case of Fig. 5, made by drawing lines to a and c through the corners of the square ACDE.—GF ac a+c FIG. 4. a2 ac is a-c units, GFKĤ is (a–c)2 square units, and b2 = (a − c)2 +2 ac. Fig. 6 is composed of the square on a-c, and two rectangles each ac. If instead we divide it by the dotted B с Α = line, it is composed of the squares on a and c. So that b2 = a2+c2. This also we find by multiplying out (a–c)2. 4. An ancient Indian geometer, to show how the two squares on the sides of a right-angled triangle may be cut up and fitted together so as to form the square on the hypo |