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diagonal division. The scale to be long enough to measure

15 miles.

EX. 20.-A scale of yards and feet is drawn to a representative fraction of . Construct a comparative scale of yards and metres, long enough for 2 metres, showing divisions of yards and feet, and of 10 centimetres.

SECTION

IV.

CONSTRUCTION OF TRIANGLESQUADRILATERALS-POLYGONS AND ELLIPSES.

THE working of the following problems will present no great difficulty to the student possessing a fair knowledge of Euclid's Elements, as they are simply a direct application of the principles of pure geometry. The questions on triangles and quadrilaterals are inserted because of their frequent occurrence in examination papers, and their general educational value, rather than for their practical use, which is somewhat limited, except in such work as plotting surveys of lands. The construction of the regular polygons and of ellipses is, however, of much importance, as polygonal and elliptical outlines are very common in engineering construction.

PROBLEM IX. (Fig. 14).—To construct a triangle, knowing the perimeter, base, and one base angle.

The perimeter of a figure is the total length of its sides or of its boundary. Thus the perimeter of a square

of 2" side is 8", and of a circle is the length of its circumference.

Let the perimeter be 7", base 3′′, one base angle 45°. Draw the line AB 3′′ long to represent the base and from the end, A, draw an indefinite line at 45° to A B. From this line cut off a part, A C, equal to the perimeter less the base-that is, 7" - 34". Join C to B, and from B draw the line BD, cutting AC in D, so that the angle A DBC= angle DCB. Then DAB is the required triangle.

Fig. 14.

(BA+ AC = given perimeter and DB=DC (Euclid i., 6), .' BA+ AD+DB= given perimeter.)

PROBLEM X. (Fig. 15).-To construct a triangle, knowing the base, altitude, and vertical angle.

The altitude of a triangle is the perpendicular distance from the vertex to the base.

Let the base be 3′′, altitude 2′′, and vertical angle 40°. Draw the line AB 3′′ long to represent the base, bisect it at C, and draw a perpendicular, OD. Draw a line, E F, parallel to the base 2" away, then the vertex of the triangle must be in this line.

Fig. 15.

B

We now use the proof of Euclid iii., 20, which says "the angle at the centre of a circle is double the angle at the circumference on the same base," and we see that we ought to be able to draw a circle having A B for a chord, so that all angles contained in it shall be 40° (Euclid iii., 21), and to do this the angle at the centre must be 80°.

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Therefore, from one end, A, of the base, draw a line making an angle with the base equal to one right angle less the required vertical angle—that is, 90° — 40° 50°. Let this line cut CD in D. Then the angle ADC is 40°, and D is the centre of the required circle. With D as centre draw a circle passing through A and B, and cutting the line EF in E and F. Then either the triangle E AB or FAB is the required triangle.

EXAMPLES.

EX. 1.-Construct a triangle, sides 31", 28", and 2′′.

EX. 2.-Construct an isosceles triangle, base 21′′, vertical angle 40°.

(Find the measure of the base angles, knowing that the three angles equal two right angles.)

EX. 3.-Construct the following right angled triangles-(a) hypotenuse 5", one side 24"; (b) hypotenuse 4", one acute angle 35°.

(The angle in a semicircle is a right angle, therefore draw a semicircle with the hypotenuse as diameter.)

EX. 4.-The distance from the centre of an equilateral triangle to the sides is 11". Construct the triangle.

(Draw a circle of 14" radius, and divide it into three equal parts, the sides of the triangle will be tangents to the circle through these points.)

EX. 5.-Construct a triangle, base 31", angles as 2: 4: 3.

(Draw a semicircle, and divide it into 2 + 4 + 3 = 9 equal parts, then joining the 3rd point from one end to centre, will give one angle, and the 2nd point will give another angle.)

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EX. 6. Construct a triangle BAC

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with base, AB, 28", angle, 50°, and side, BC, 21′′. (S. & A. E., 1892.)

(Two triangles are possible, this being the ambiguous case of Trigonometry.)

EX. 7. Construct a triangle whose sides, ab, bc, ca, are 31′′, 23", and 2′′ respectively. On ac construct a second triangle a dc, whose vertical angle a dc is equal to the angle a b c, and the side a d 1g". (S. & A. E., 1887.)

(The angles upon the same base and in the same segment of a circle are equal.)

EX. 8.-Two points, A B, are 3 miles apart. Find the position of a point, P, so that P A is 1ğ miles, and the angle AP B is 73°. (Scale 1" 1 mile.) (S. & A. E., 1886.)

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(Use method of Prob. X.)

EX. 9.-Construct a triangle, perimeter 7", one base angle 42°, altitude 21".

(Adopt method of Prob. IX.)

EX. 10.-Construct an oblong, diagonal 6", short sides 24". (Half the oblong is a right angled triangle, the diagonal being the hypotenuse.)

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EX. 11.-Construct an oblong, diagonal 4", sides as 3 2. (Draw any oblong sides as 3. 2, and then a similar oblong having diagonals 44".)

EX. 12. The line joining one corner of a square to the middle point of the opposite side is 44". Draw the square. (Draw any square and join one corner to centre of opposite side, make this line 44′′ long, and draw a second square parallel to the first.)

EX. 13.-Draw a rhombus, longest diagonal 51⁄2′′, acute angles 50°.

(Diagonal is base of triangle of which each base angle is known.)

EX. 14. Draw a quadrilateral ABCD, A B = 41", angle A B C = 30°, BC = 5′′, angle B C D 95°, angle BA D = 110°.

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EX. 15.-The diagonals of a parallelogram are 21" and 41′′ long, they contain an angle of 61°. Construct the parallelogram.

(Half of each diagonal makes the sides of a triangle of which the vertical angle is known.)

Regular Polygons.-The geometrical construction of regular polygons depends upon Corollary I. of Euclid i., 32, which says that "the interior angles of any straight lined figure together with four right angles are equal to twice as many right angles as the figure has sides."

The most common of the regular polygons used in engineering designs are the pentagon (five-sided), hexagon (six-sided), and octagon (eight-sided).

Pentagon.-Suppose we require to construct a pentagon of 21" side. All the interior angles together with four right angles will equal 2 × 5 × 90 900°, and, therefore, the interior angles will equal 900 360 540°, and each interior angle will be

540

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108°. Hence we could draw two lines 21" long, meeting at an angle of 108°, and they would be two sides of the pentagon, and we could complete the pentagon by drawing other lines at 108° and 21′′ long, until the figure closed. But this is a cumbersome method and would scarcely be accurate. The geometrical construction is as follows:

PROBLEM XI. (Fig. 16).-To construct a regular polygon on a given line.

Let the polygon be a pentagon, and the given line be A B. Produce AB, and with A, as centre, draw a semicircle of radius equal to the given line, A B.

7

6

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Divide this semicircle into the same number of equal parts as the sides in the required figure, in this case 5, and mark the points as shown. Join the centre, A, to the point, 2. Then A 2 is a second side of the pentagon. For A 2 A B, and the angle 2 A B is % of 180° 108°, and hence, by always joining to the second point, counting from the opposite end of A B, we obtain for six divisions, of 180° 120°, the angle of a hexagon; for nine divisions, of 180° 140°, the angle of a nonagon. Now a regular polygon can always be circumscribed by a

Fig. 16.

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circle, and hence, if we draw a circle containing the two sides, 2 A and A B, it will just contain the required polygon.

Draw the circle passing through 2 A B, having its centre at C (see Ex. 10, p. 7). With A B as distance start from B or 2, and step round the circle, marking the points 6 and 7, and complete the figure as shown.

It is difficult for beginners to finish these polygons accurately, the fault generally lies in a bad division of the semicircle, where a very small error makes a large difference in the result. This is the reason why polygons are seldom accurate when the angle 2 A B is set off with a protractor, as a slight error becomes multiplied as the polygon approaches completion.

Hexagon and Octagon.-Since the interior angles of the hexagon and octagon are respectively 120° and 135°, it is unnecessary in constructing

either of these figures to divide the semicircle into either 6 or 8 equal parts. For the exterior angle (2 AD of Fig. 16) is 60° for a hexagon and 45° for an octagon, it can, therefore, be easily found in the first case by taking the radius of the semicircle as distance and marking off from the point D to the point 2, thus making the angle 2 AD 60°; and in the second case by bisecting the right angle, thus obtaining an angle of 45°.

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The two set squares most commonly used are made with angles of 60° and 45°, and this enables a hexagon or an octagon to be very easily and accurately con

Fig. 17a.

Fig 176.

structed, if the set squares are true. The method of using the set squares to draw these figures is shown in Fig. 17a, b.

In Fig. 17a, the hexagon is to be drawn on the line A B, and the set square is shown in position for drawing two other sides. In Fig. 176, the 45° set square is shown in position for obtaining two of the sloping sides of an octagon. It is easy to see how

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