CHAPTER II PROBLEMS OF TRIANGLES, CIRCLE, QUADRILATERALS, ETC. (Part 1)

CHAPTER II PROBLEMS OF TRIANGLES, CIRCLE, QUADRILATERALS, ETC. LINES AND ANGLES Problem 1,—To bisect a given line AB or Arc CD. With A as centre and a radius greater than half the line AB, describe the are EF, With B as centre and same radius, describe GH, intersect- ing EF at 1 and 2. Join 1 and 2, cutting AB at O. Then AO=OB, and the line AB is bisected, and the line 1-2 is perpendicular to AB. Treat the arc CD by the same method as shown, and the line 1-2 is perpendicular to the tangent of the are CD. Problem 2.—To draw a line parallel to a given straight line. From two centres on line AB draw two ares C and D of equal radius. Draw EF resting upon, and just touching the ares Cand D. Then EF is parallel to AB. (Another Method.) To draw a line parallel to a given straight line through a given point. Let A be the given point and BC the given line. With any point 1 as centre and radius 1A describe are A2. With A as centre and same radius describe another are 1D. Cut off 1D equal to A2. Join AD, which will be parallel to BC. 3 LINES AND ANGLES o Problem 3.—To bisect a given angle AOB. (a) With centre O and a convenient radius describe the are CD. “With C as centre describe are E, and with the same radius and D as centre cut are Eat F. Join OF. Then the line OF bisects the angle AOB, and AOF = FOB. (b) If the lines could not be produced so as to meet, then draw two lines at equal distances from and parallel to the two lines, e.g., AB and BC. Problem 4.—At a given point E in a line AB, to make an angle equal to the given angle HIK. With centre I and largest possible radius describe the are K. Then with centre E describe are G, using radius IK. With centre M and distance KL cut are G at N. Draw EN. Then the angle NEM equals the angle HIK. Problem 5.—To construct an angle containing a given number of degrees. The circumference of a circle is supposed to be divided into 360 equal parts called “Degrees.” The radius of a circle may be set off exactly six times round the circumference ; hence, if an are be described, and a portion cut off equal to the radius of the are, an angle containing 60° will be obtained. By applying this principle a number of angles may be constructed: Construction for angles of 60°, 120°, 30°, 15°, 45°, and 75° are shown. From this other angles may be obtained, By the protractor (Fig. 1).—When this is used it is merely necessary to place the protractor on the line with its centre point over the 10 PROBLEMS OF TRIANGLES, CIRCLE, QUADRILATERALS required point in the line where the angle is to be formed, and then prick through the paper at the required degree on the edge of the protractor. Problem 6.—To divide a straight line AB into any number of equal parts: ¢.g., divide AB into 5 equal parts. Draw AC at a convenient angle to AB and of suitable length, Mark off five equal spaces on AC. Join the last, and draw four lines parallel to B5 and cutting AB, thus dividing it into five equal parts. c y to} . y PROB. 5 PROB. 7 ‘ 5 8 ok B TRIANGLES Problem 7.—To construct an equilateral triangle on a given line AB. With centres A and Band radius AB deseribe two ares cutting each other at C. Join AC and CB. Then ACB is the required equilateral triangle. (Euc. I., 1.) Problem 8.—(a) To construct an isosceles triangle when each of the equal sides equals AB, and each of the equal angles equals C. A. A oR S NEES ny |, Fic Draw DG of indefinite length. Make GDE equal to angle C. Cut off DE to length AB. With E as centre and same radius cut DG in G, Join GE. Then DGE is the required isosceles triangle. (6) To construct an isosceles triangle when the lengths of the two sides and base are given. Let AB =length of sides, and CD=base. With radius AB and from QUADRILATERALS 11 centres C and D describe two arcs intersecting at H. Join HC and HD. HCD is the required isosceles triangle. (¢) To construct a right-angled triangle, the hypo- tenuse and base being given. Let AB be the base and C the hypotenuse. At A erect a perpendicular; with B as centre and radius equal to C cut perpendicular in D. Join DB. Problem 9.—To construct a right-angled triangle, the hypo- tenuse and an acute angle being given. Let AB be the hypotenuse and C one of the acute angles. Bisect AB in D, With centre D describe a semi-circle on AB. At A construct an angle BAE, equal to C. Join BE. Then BAE is the required triangle, the angle in the semi-circle being a right angle. (£uc. IIL, 3.) Problem 10.—To construct a triangle, the three sides being given. Let AB be the base and C and D the twosides _~ ; respectively. = ° aie With centre A and radius D SN describe an are. With centre B and radius C describe an are intersecting the former are at E. Join EA and EB, Then ABE is the required triangle. (Buc. I. 22.) QUADRILATERALS Problem 11.—To construct a square, the side being given. Let AB be the given side. At B erect a perpendicular BC and make it equal to AB. With centres C and B and radius AB describe ares intersecting at D. Draw CD, AD. iS oO Problem 12.— To construct a square, the diagonal being given. Let AB be the diagonal. 12. PROBLEMS OF TRIANGLES, CIRCLE, QUADRILATERALS Bisect AB. With centre F and radius FA describe circle ADBC. Join AD, DB, BC, and CA. Note.—The foregoing principles apply in the geometrical construc- tion of rectangles, the only difference being in the lengths of the sides. In Problem 12, for a rectangle, the length of one side would be necessary to complete the problem. Problem 13.—To construct a rhombus. (a The side and one angle being given. (6) The side and diagonal being given. (a) Let AB be the given side and C the angle. At A construct an angle equal to C, and make AD equal to AB. With D and B as centres and radius AB describe two arcs intersecting at E. Draw DE and BE, (0) Let AB be the diagonal and C the side. With radius C and centres A and B describe four arcs inter- secting at D and E. Draw AD, BD, BE, and EA. THE CIRCLE Problem 14.—To find the centre of a circle. Draw any two chords AB, BC. Bisect them by lines which intersect at D. This point is the centre of the circle. (Luc. III., Cor. 2.) Problem 15.—To describe a circle passing through three given points not in the same straight line. Let A, B, and C be the three points. Join AB and BC. Bisect both lines as above. From D, the point where these lines meet, describe the circle. Problem 16.—To describe a circle about a triangle. Bisect two sides and proceed as Problem 15. Problem 17.—To draw a tangent to a circle: (a) From a point in the circumference. (6) From a point without. (a) Let A be the given point in circumference of circle. Join this with the centre B and produce to C, making AC equal to AB. With centres B and C describe two ares cutting at D. Join DA, which is a tangent of the circle. THE CIRCLE 13 (® Let X be the point without the circle. Join BX. Bisect and describe semi-circle cutting circumfer- ence of larger circle. Draw XY, the required tangent. : REGULAR POLYGONS Problemy18.—To construct any regular polygon on a given line AB (say a pentagon), 14 PROBLEMS OF TRIANGLES, CIRCLE, QUADRILATERALS Produce AB, and with centre A and radius AB describe a semi-circle. Divide the semi-circle into as many parts as the polygon has sides (by trial with dividers, or by the aid of a protractor). Always join A with 2, giving another side of the polygon, no matter how many sides the polygon required has. Bisect these two sides, also finding the centre of the polygon and of circle containing same. Describe this circle (radius OA). Mark off BE and ED equal to AB. Join BE, ED, and D2, form- ing the required pentagon. Note.—The division of the semi-circle must be accurately done, as upon this and the correct finding of centre of figure depends the accuracy of this construction. Problem 19.—To construct a regular hexagon on a given line, With centres A and B and radius AB describe two arcs cutting at O. With O as centre and same radius describe a circle and set off AB round it, Join these points and form the required hexagon. ‘The side AB may be obtained by using the 60° set square and the hexagon quickly set up by this means. Problem 20.—To inscribe any regular polygon in a circle. Draw the diameter AB, If the centre of circle be not given, find same first. Divide AB into seven equal parts. With centres A and B deseribe two ares intersecting at C. From C, through the second part on AB, rule CD, cutting off AD, one of the sides of the polygon. Set off AD round the circle, and join the points. Note.—The greatest care is necessary in dividing the line and drawing the line from C exactly through the second division in Point 2 on the line AB. GENERAL NOTES ON POLYGONS If the angle of two adjacent sides is known, and their length, it is an easy matter to construct the regular polygon by means of the protractor. If all the angles of a polygon be joined with the centre, there will be formed as many equal isosceles triangles as the polygon has sides, having their vertical angles at the centre equal (Huc. I., 15, Cor. 2). Hence, if 360° be divided by the number of sides of the regular polygon, the value of the central angles will be obtained. These may be marked round from readings taken from the protractor. CHAPTER HI AREAS, SCALE DRAWING, CURVES, THE ELLIPSE AREAS, GENERAL PRINCIPLES Ir is necessary, before working further problems in this subject, that the following elementary principles should be carefully studied:— (1) The area of a plane figure (a sheet of paper, lead, or metal, or the surface of a body) is the amount of surface enclosed by its boundary or perimeter. It depends upon both the shape and the perimeter of the figure. (2) Parallelograms upon the same base, or on equal bases, and between the same parallels are equal. (Luc. [, 35, 36.) Thus in Fig. 2, ABCD=ABDE, FGHJ=FGKL, and MNOP=QRST, because MN = QR, also MNTS=MNOP. (3) Triangles upon the same base, or on equal bases, and between the same parallels, are equal. Thus in Fig, 8, ABC=ABD and ABC= DEF. (4) If a parallelogram and a triangle be on the same base, or ‘on equal bases, and between the same parallels, the area of the parallelogram shall be double that of the triangle. Thus in Fig. 4, ABCD =twice ABC, and EFGH =twice EFJ. (5) The square on the hypotenuse of a right-angled triangle is equal to the sum of the squares on the other sides. (Euc. I., 47.) Thus in Fig. 5, the square CBDE=the square ABFG, plus the square AHJC. This is a very impor- tant fact for plumbers and engineers, because, coupled with it, all similar figures follow the same law. Thus circles of these diameters, AC and CB, are in area equal to the area of the circle whose diameter is AB. This is applic- able to the comparison of the areas of various sizes of pipes. For example, A and B represent the diameters of two pipes. What will be the diameter of a pipe the area of which will be equal to the two given pipes? Answer. Construct a right-angled 16 16 AREAS, SCALE DRAWING, CURVES, THE ELLIPSE triangle ABC, making AB equal to diameter A, and AC equal to diameter B. Measure BC, which is the required diameter. c D. : oo) a aon Bo: , “ / \ Ae \ , \ Z ; es . ee er. nu, 7 § Fa? coc f c rie eae ; x. x \ LS a ;7 Bo 15 K e ¥ / i Fas Fi0.4 SCALES AND SCALE DRAWING 4 5; Scales are adopted in order to make diagrams either greater or less than the original diagram or object. They are used in making 4 a fe working drawings, plans, ete., showing methods of construction, ete. For example, in the plan of a house, IG. it would be very inconvenient to have a drawing full size; but by drawing every 4 . measurement to a given reduction (i.c., to scale), we have a drawing having the same relative proportions, and from which, by using the scale, we can take exact full-size measurements. In a scale any given distance may be represented by a measurement decided upon. For example, a scale may consist of 1 in. to the mile—that is, each inch measured on the drawing represents 1 mile ; or, in the case of a scale of } in. to the foot, each } in. on the drawing represents 1 foot full size. In the case of an enlargement, scales such as this may be used (i.e., scale of 1 in. to ygls5 of an inch). This time 1 in. measured on the drawing would only represent yj of an inch on the object or original drawing. The real distance is called the Natural Distance, and the enlarged or reduced length used upon any given scale is the Artificial Distance. Scales to be of use should fulfil the following conditions :—Be SCALES AND SCALE DRAWING 7 divided with great accuracy, and carefully numbered, long enough to measure the principal lines of the drawing. The zero must always be between the unit and its subd: otl109 876643210 4 SCALE OF 1% IN.TOTFT.OR th FULL SIZE, FIG. 6 876543210 10. 15. =—s = SCALE OF #4 IN. TO TFT. OR wth FULL size. FIG. 7 jog 8 5432 1 0 1o| IB scatesim.=sxv0s. FIG. 8 22 c 7 F S i DIAGONAL SCALE; SHOWING TENTHS. FIG. 10 2 £ 4 t i i ia t t t t ] SIMPLE DIAGONAL SCALE. CeeCeeeyt oe ad fetetictalatatehat AE EEC eee rt —s 1 WStatatatesttahat aN . H 5 Sse | 4f2- CECEEreey A) ¢ 7 | © DIAGONAL SCALE, d +—+ J bi fi aaa DIAGONAL SCALE, SHEWING YARDS, FEET AND INCHES. FIG, 12 To construct any scale (say a scale of 1} in. to 1 ft.). Draw two parallel lines AB, CD (Fig. 6). Mark off from A, spaces of 1} in. each. These by scale represent feet. Divide the first space into 12 in., and place figures in the manner shown. ‘This is the most convenient method. If a distance 2 ft. 9 ins. is required, place one leg of dividers on D and open until the other rests on 9 ins. To construct a scale of 8 ft., to 1 in. or ;';, to measure 22 ft. B 18 AREAS, SCALE DRAWING, CURVES, THE ELLIPSE (Fig. 7). Draw lines as before. Make AB 1 in,, and divide into 8 equal parts, and mark off 22-parts, giving the length required, To draw a scale showing 5} yds. to 1 in. tomeasure 20 yds.— If 5} yds. be represented by 1 in., then 2 ins. will represent 11 yds. (Fig. 8). Draw a line and set off AB 2 ins. Divide this distance into 11 équal parts, and add 9 more to make 20 yards, and complete the scale. Paper scales generally in use can be bought very cheaply, and are “very useful. Those used by builders, ete, are 3, 1, & 3, 3, 1, 1, 2, ‘and 3 ins. to the foot, the latter being for details. All scales divided into inches are called Duodecimal Scales. Those divided into tenths or units of ten are Decimal Scales. All the above are called Plain Seales. Diagonal scales.—By means of the diagonal scale very minute distances may be measured with great accuracy. The principle of its construction is as follows :—If the rectangle ABCD (Fig. 9) be divided into 8 equal parts by lines parallel to AB, and the diagonal DB be drawn, then 2 to E=6 parts of AB, and 7 to F=4 of AB. To draw a diagonal scale showing inches and tenths (Fig. 10).—Draw a line AB, and mark off inches on the same. At A erect a perpendicular, and set off 10 equal parts to any convenient unit. Draw lines parallel to AB, and draw diagonal from 0 to 10. Draw a diagonal scale showing inches and hundredths of an inch (Fig. 11),—Draw 11 parallels, and set up verticals. Now divide AO into 10 equal parts, and join first part on AO with 10 at the end of top line. Rule parallels from each of the parts as shown. Note.—Each part on the line AO shows tenths of an inch, and the distance 9A, on the second line from the top, will be 4); of a tenth— that is, y45—of an inch. The distance 1B=1,%, or 1:3. If 1°59 in. be required, place one point of dividers on the point G, where vertical 1 meets horizontal 9; open dividers to point H, where diagonal 5 meets horizontal 9. Draw a scale of ,~. to show yards, feet, and inches (Fig. 12).