26. The four triangles will together constitute a circumscribed equilateral triangle.

27. Draw lightly an arc less than 90° in length. After cutting off 60°, darken this portion. No radii

should be drawn.

A

M

FIG. 21.

To construct an angle of 60°, draw an indefinite line AB (Fig. 21). With any convenient radius, as AM, draw an arc MN. With M as a center, and using the same radius, cut the arc at X. This makes MX an arc of 60°. From A draw a line through X.

28. An arc of 30° is constructed by the bisection of an arc of 60°.

To construct an angle of 30°, draw AB (Fig. 21) and the arc MN; and cut off MX, as in 27. With the same, or any other convenient radius, and using M and X as centers, draw two arcs intersecting on the right of MX. A line drawn from A through this intersecting point will make with AB an angle of 30°.

29. To construct an angle equal to 60° +30° (or 30° + 60°), bisect the arc XM (Fig. 21), and from the bisecting point, which is 30° from M, lay off an additional 60°. A line drawn from A through this point will make with AB an angle of 90°.

XF

30. One method of erecting a perpendicular at the end of a line is given in 29. A somewhat similar method consists in prolonging the arc, and marking off at Q and R two divisions of 60° each. Using these two points as centers, draw arcs intersecting at F. From D draw a line through F

The line DF bisects the arc QR, which makes the angle FDE= 60° (PQ) + 30° (QV) = 90°.

D

FIG. 22.

P E

31. The bisection of VP gives an angle of 45°. Do not draw DF

The recommendation so often made as to getting a variety of drawings from the various members of the class, applies to these problems. The erection of a perpendicular at the end of a vertical line should call forth at least four different results. The perpendiculars may be erected at either end, and may run to the right or to the left. In constructing an angle of 450, a greater variety is possible. The first line may be horizontal, vertical, or oblique; the second line may start from either end; and it may extend above or below, to the right or to the left; the lines forming the angles need not be of the same length on all papers, nor need both of them be of the same length on any one paper.

=

221° of 45°; 135° = 90° + 45°; 15° = 1 of 30°; 75° 60° +15°.

=

33. The preceding problem should give a hint as to the method of solving the present one. In 32, a circle was drawn with A as a center

and AX as a radius. This circle cut

A

M

FIG. 23.

XY (Fig. 23) in M. To erect a perpendicular at A, the center of the circle, the Xextremities X and M of the diameter XM were used as centers to draw arcs intersecting at J. A perpendicular was then drawn from A through J.

This problem does not furnish a circle, nor is one necessary. The line XY is given, and the point A at which the perpendicular is to be erected. With A as a center, and a radius equal to AX, cut off AM. This gives us the diameter of the circle employed in the preceding problem.

It will be noticed that a second set of intersecting arcs on the opposite side of XM is not needed, the point A answering instead. This problem amounts to the bisection of an arc of 180°, of which only the chord XM is drawn, A being the center of the circle.

34. The first two divisions of this problem are reviews of parts of 17 and 30. The erection

of a perpendicular at a point .
between the end and the center,
is a variation of the preceding
problem. Let RS (Fig. 24) be
the required line and A the
point. With A as a center, and
any convenient radius, lay off the
points X and M, which will be

FIG. 24.

equidistant from A. Using X and M as centers, draw arcs intersecting at J. Draw the required perpendicular from A through J.

In problem 33 the point M was located at a distance from A, equal to AR; and while greater accuracy is obtained by having the points Mand X as far apart as possible, the present method is suggested to show that the only essential requirement as to their location is to have them equidistant

from A.

35. The base need not be a horizontal line.

36. Proceed as in the construction of a right-angled triangle, base 2 inches, perpendicular 2 inches; but do not B draw the hypotenuse. With B and H as centers (Fig. 25), and a radius of 2 inches (BP or PH), draw arcs intersecting at O. BO and HO will form the remaining sides.

FIG. 25.

-H

To construct the rectangle, PH (Fig. 25) P should be made 3 inches long, BP measuring 2 inches. With B as a center, and a radius of 3 inches, draw an arc. Intersect this by a second are drawn with H as a center, and a radius of 2 inches. From the point of intersection, draw lines to B and H.

Some scholars may prefer to erect a perpendicular at each end of the base, etc.

38. Draw a base line, WV (Fig. 26), 3 inches long. At any convenient point, M, erect a perpendicular, MP, 2 inches long.

With W as a
X on the line
On PO draw

At its extremity P, draw the perpendicular PO. center and a radius of 3 inches, locate the point PO; WX will be the second side of the rhombus. XN 3 inches; and connect NV. This last line should measure 3 inches if the work is properly done.

To construct a 3-inch rhombus containing an angle of 60°, draw WV; at W construct an angle of 60°, and make WX 3 inches. At X or at V, draw a 3-inch line making an angle of 120° with XW or VW, etc.; or, using X and V as centers, and with a radius of 3 inches, draw arcs intersecting at N; draw NX and NV,

39. Proceed as in the first problems of 38; but make WV and XN each 4 inches long.

40. See BPH (Fig. 25); draw BH.

Some pupils, preferring to commence with a horizontal base, calculate the number of degrees in each angle at the base, † (180° — 90°), or 45°. To each end of a base line of any length (Fig. 27), they draw a line making with the

base an angle of 45°.

If one angle is 120°, the angles at the base will be 30° each; if one is 135°, the angles at the base will measure 221° each.

The method shown in Fig. 27 requires the construction of two angles; and is, therefore, not so direct as drawing two equal lines meeting at the given angle.

41. Erect a 3-inch perpendicular at the middle of a 3-inch

30° each, as shown in the figure; draw AX and A Y, forming the required equilateral triangle AXY. Test the work by measuring XY.

42. To construct a scalene triangle of the required dimensions, erect the 3-inch perpendicular at a point not in the center of the 3-inch base, nor at either extremity.

To construct the obtuse-angled triangle, produce the 3-inch base by a dotted line, and erect the 3-inch altitude at some point outside of the base.

X

43. This problem may puzzle the pupils at first. If the triangle were isosceles, XYM, for instance, there would be no difficulty. The solution of the problem consists practically in making an isosceles triangle, although the line XM is not drawn. With X as a center (Fig. 29), and XY as a radius, the point M is located (without drawing the arc shown

Z

M

R

FIG. 29.

in the figure). Using M and Y as centers, and with any convenient radius, arcs are drawn intersecting at R. XR gives the direction of the altitude.

44. The given triangle in this case will be XMZ (Fig. 29). Produce ZM indefinitely towards Y, and with XM as a radius, cut the base at Y; etc. XY is, of course, not drawn; MY should be a light line or a broken line; etc.