Figure 4-13 shows a method of drawing a right triangle when the hypotenuse and one side are given. The line H is the given hypotenuse; the line S is the given side. Draw AB equal to H. Locate the center of AB (by bisection), and, with the midpoint as a center and a radius equal to one-half of AB, draw the semicircle from A to B as shown. Set a compass or dividers to the length of S, and, with A as a center, strike an arc intersecting the semicircle at C. Draw AC and BC.

Figure 4-13.-Constructing a right triangle with hypotenuse and one side given.

EQUILATERAL TRIANGLE: LENGTH OF SIDE GIVEN

To construct an equilateral triangle when the length of a side is given, you can follow the method previously described for constructing a triangle when the length of each side is given. The sides of an equilateral triangle are equal in length. Each angle in an equilateral triangle measures 60. This fact is applied in the method of constructing an equilateral triangle with given length of side, such as the one shown in figure 4-14. Simply use a 30/60 triangle and a T square or straightedge to erect lines from A and B at 60 to AB.

A circumscribed plane figure is one that encloses another figure, the circumscribed figure being tangent to the extremities of the enclosed figure. An inscribed plane figure is one that is enclosed by a circumscribed figure.

Figure 4-15 shows you how to inscribe an equilateral triangle within a given circumscribed circle. Draw a vertical center line intersecting the given circle at A and B. With B as a center and a radius equal to the radius of the circle, strike arcs intersecting the circle at C and D. Lines connecting A, C, and D form an equilateral triangle.

Figure 4-16.-Equilateral triangle on a given inscribed circle: one method.

Figure 4-17.-Equilateral triangle on a given inscribed circle: another method.

circle. Draw AB parallel to the horizontal center line of the circle and tangent to the circumference.

Then use a 30 triangle to draw AC and BC at 60 to AB and tangent to the circle. Another method of accomplishing this construction is shown in figure 4-17. Draw radii at 30 to the horizontal center line of the circle, intersecting the circumference at C and B. There is a third point of intersection at A, so you now have three radii: OA, OB, and OC. Draw the sides of the triangle at A, B, and C, tangent to the circle and perpendicular to the relevant radius.