26. What sum will amount to £2469 88. 73d. in 5 years at 2 per cent. simple interest? Let us suppose £100. Amount on £100 for 5 years at 23 per cent. = £115 28. 6d. Hence our Proportion Sum is: (For the whole Paper see 'Preliminary Army Examination Made Easy,' pp. 23-38 and pp. 36, 37.) ELEMENTS OF GEOMETRICAL DRAWING, including the Construction of Scales and the Use of simple Mathematical Instruments. Thursday, 16th February, 1882. 10 A.M.-12.30 P.M. N.B.-The figures should be neatly drawn in clear fine pencil lines, and if time allows, they may be inked in with Indian ink. The solutions must be strictly geometrical, and particular care should be taken to show all the necessary lines of construction. In reading dimensions, a single accent (') signifies feet; a double accent (") inches. Sol. = Solution. (For the figures, see sheet at beginning of book.) Q. 1. On a given map, 325 chains are represented by 33". (1.) Construct a plain scale of chains, showing 400 chains, and divided to show distances of 10 chains. (2.) Construct a comparative scale of yards for the same map, showing 8000 yards, and divided to show distances of 200 yards. Figure your scales properly, show your calculations, and give the representative fraction. Sol.-N.B.-In this case a length of 33" represents a length of 325 chains (on the plan of an estate), and we are required to show a length of 400 chains. Construction :-Draw a line 4.61" long, divide it into four equal parts, each of these will represent 100 chains, divide the first division on the left hand into ten parts, and we thus obtain subdivisions to measure 10 chains as required. For the representative fraction the chains must be brought to inches, 325 chains × 66′ x 12" = 257,400" in 325 chains. For the comparative scale, showing yards, we find that in 325 chains there are 7150 yards, then the equation will be: Construction:-Draw a line equal in length to 4.19", divide it into eight parts, to show primary divisions of 1000 yards, and subdivide the division towards the left hand into five equal parts showing 200 yards, and the scale is completed. Q. 2. Construct a scale of chords at a radius of 4" to read to 5o, and with it construct angles of 35° and 55°. Sol.-In constructing this problem always place it in the position as shown, so that the Examiner may the more readily see your construction. Place the two lines at right angles, and then with a radius of 4" construct the quadrant 0° B 90°; then draw the chord line from 0° to 90°, then from 0° as centre 0° B as radius describe the arc B 60°, cutting the quadrant in D and divide 0° D into six and D-90° into three equal parts, from these points of division, with 0° as centre, describe arcs cutting the chord line 0°-90° in the points 10°-90° as shown, these will give divisions showing 10° and by dividing these divisions into equal parts we are able to measure to 5° as required. To measure off the angles required, draw the indefinite lines A C, E F as shown, from point 0° on the scale stretch your compasses to 60°, and with this length as radius describe arcs CB, FD, along the arc CB measure off the distance from 0°-35°, and on the arc FD the distance 0°-55°; by drawing lines from B to A and from D to E we obtain the angles B A C = 35°, and D E F = 55° respectively. Note the two stars are placed under 0° and 60° respectively (as in my universal scale), to show that this distance must always be taken to obtain any angle and the number of degrees required measured off along the arc of 60°. Q. 3. The base of a triangle is 3", one of the angles at the base 45°, and the sum of the other two sides 4". Draw the triangle. Sol.-Draw the line AB = 3" long, make the angle BAD = 45° and make AD=41" in length, join D B, make the angle b at B equal to the angle a at D, and draw the line B C, then the sum of the two sides A C+BC= 41′′, and the question is solved. Q. 4. Describe a triangle with sides of 3, 4, and 5", and trisect it by lines drawn parallel to its shortest side. Solution to No. 4.-The sides of this triangle are given in such a ratio that the angle B A C = 90° make it so by construction, and construct the triangle with the sides as required, divide the longest line BC into three equal parts, and draw a semi-circle on B C, having that line for its diameter; from the points of division 1, 2, on B C, draw lines at right angles to BC to cut the semi-circle, then from C as a centre, with CP and CQ as radii, describe arcs cutting the line B C in the points D and E, and from these points draw lines parallel to the short side AB, and the problem is finished. Q. 5. The measurements of a four-sided field, ABCD, are as follows:A B 105 yards, B C = 95 yards, C D = 145 yards, A D=80 yards, and the diagonal AC = 165 yards. Draw a plan of the field to a scale of 40 yards to an inch, and write down the length in yards of the diagonal B D. Construct a similar figure of half the area of A B C D, and write down the length in yards of each of its sides. Sol. The first thing is to draw the scale of 40 yards = 1", then draw a line 5" long, and divide the left hand into 10 parts, these will show 4 yards each, and by making a diagonal scale as shown we obtain single yards. For the construction of the figure : Draw the line A B=105 yards, then from B describe an arc with a radius of 95 yards, then from A, with a radius of 165 yards describe an are to cut that struck from B as a centre, and their intersection will determine the position of C; then from C an arc struck with a radius of 145 yards, and from A with a radius of 80 yards, will determine the position of D, and the figure is complete. = To obtain a similar figure half the area of A B C D, draw a line from B to D, produce A B indefinitely, make BO to half of A B, then find a mean proportional to A B, BO, this will be BP, then make B E = BP, then draw the line EF parallel to AD, FG parallel to D C, and the figure BEFG is similar to the figure A B C D and half its area. N.B.-The length of each side of the smaller figure is shown; care is required in using the scale to get the dimensions correctly. Q. 6. The three sides of a triangle, A B C, are as follows:AB=13", BC= 11", A C=2". Construct the triangle, and in it inscribe a circle. Produce the sides A B and A C indefinitely, and determine the centre of a circle that shall touch B C, and also A B and A C produced. Sol. Draw the triangle with the sides the required length, and by bisecting the angles A and B at the base we obtain the centre for the inscribed circle, we next produce A C towards Q and A B towards P, and by bisecting the angles QCB and PBC we obtain the centre S of the required circle tangential to the lines Q C, CB, BP. Q. 7. Construct a regular pentagon of 2" side, and draw any diagonal. Construct a square equal to the larger portion of the pentagon so divided. Sol.-To construct the pentagon of 2" edge, draw a line A B 2" long, bisect it by a line QP perpendicular to it, make QP = to the length of side (2" in this case) from A and B as centres, describe arcs with a radius equal to A B, draw a line from B through P, and from P mark off PR a distance equal to half A B, from B as a centre with BR as radius describe an arc cutting QP in D, and from D as a centre with a radius equal to A B describe an arc cutting the arcs struck from A and B as centres in the points C and E, join ABCDE and the pentagon is complete. Next draw the diagonal A C, then make A' C' D' E' in the Fig. No. 7 (2) equal to AC DE in Fig. 7, produce the line A' C' both ways, draw a line from D' to A' and a line from E' parallel to D' A' to cut A' C' produced in F', join F' to D', and the triangle F'D' C' is equal in area to the quadrilateral figure A' C' D' E', again the rectangle (parallelogram) F' C'GH is equal to the triangle F'D'C', and C'I is a mean proportional to the two sides of this parallelogram, and therefore the square constructed upon C'I as a side will be equal in area to the rectangle F' G H C', to the triangle F'D' C', or to the quadrilateral figure A' C' D' E', because by construction they are all equal to each other, and the square C'IJK is the solution of the problem, being equal in area to the quadrilateral A' C' D' E'. Solved by PARTON PARRY, 31, Prospect Row, Woolwich. GEOMETRY. (Euclid, Book V.) Thursday, 16th February, 1882. 1.30 P.M.-3.15 P.M. [N.B.-You are not expected to answer more than FOUR of the following.] 1. If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles contained by those sides equal to each other; they shall likewise have their bases or third sides equal, and the two triangles shall be equal, and their other angles shall be equal, each to each, viz. those to which the equal sides are opposite. This is the fourth proposition of the First Book. (See 'Preliminary Army Examination Made Easy,' p. 38.) 2. In the triangle L M N show that the greater side L M has the greater angle L N M opposite to it. This is the eighteenth proposition of the First Book. (See Preliminary Army Examination Made Easy,' p. 38.) 3. Equal triangles on the same base and on the same side of it are between the same parallels. This is the thirty-ninth proposition of the First Book. (See Preliminary Army Examination Made Easy,' p. 39.) 4. Describe a square on the given straight line C D. 5. A B, C D are two parallel straight lines. B and C are joined, and CD through P, the middle point of BC, a straight line is drawn meeting A B in E and CD in F. Show that B E is equal to C F. Because A B is parallel to C D and B C meets them, the angle A B C is equal to the alternate angle BCD. And the angle EPB is equal to the angle CPF, because they are opposite vertical angles. Therefore in the two triangles EPB, FPC there are two angles of the one equal to two angles of the other, each to each, and one side P B equal to one side C P— sides that are adjacent to the equal angles in each. Therefore the other sides are equal, each to each. Hence BE is equal to C F.-Q. E. D. GEOGRAPHY. 1. In the space below draw a map distinctly, with pen or pencil, an outline map of France marking the entire frontier, the chief capes and islands along the coast, the rivers Seine, Loire, Garonne, Rhone, and Saône. Show by dots with the names written near, Paris, Lyons, Marseilles, Rouen, Tours, Rheims, Brest, Toulon, Calais, and Havre. In the map should be placed the following capes: Gris Nez on the Straits of Dover, La Hogue in the English Channel, and Bec du Raz; the Channel Islands, Ushant off the western extremity of Brittany, Belle Isle, Yeu, Ré and Oleron in the Bay of Biscay. The Seine should be drawn rising in the Côte d'Or Mountains and flowing north-west into the English Channel, with Paris and Rouen on its banks and Havre at its mouth; the Loire, rising in the Cevennes Mountains and flowing first north then west into the Bay of Biscay, with Tours on its left bank; the Garonne, rising in the Pyrenees and flowing north-west into the Bay of Biscay; the Rhone, joining the Saône at Lyons and flowing south into the Gulf of Lyons. Marseilles should be marked just north-east of the Gulf of Lyons, and Toulon south-east of Marseilles. Rheims should be placed in the province of Marne in the north, Brest in the north-west corner of Finisterre, and Calais on the Straits of Dover. (See Preliminary Army Examination Made Easy,' p. 46.) 2. Name the four largest islands in the world (excluding Australia) in their order of magnitude. Which are the principal islands included in the term "West Indies," how grouped, and by whom severally owned? Where are the Bahamas and Bermudas respectively situated? Borneo, Madagascar, New Guinea, Sumatra. |