Remark. The vertices of all the angles of each triangle constructed in a circle lie in the circumference of the circle. A figure, the vertices of whose angles are thus situated, is said to be inscribed in a circle; and the circle is said to be circumscribed about the figure. A figure, all the sides of which are tangents to a circle, is said to be circumscribed about a circle, and the circle to be inscribed in the figure. 72. To construct an equilateral triangle upon the line AB, (Fig. 64.) Sol. From A as a centre, with a radius equal to AB, describe a circle; and from B, with a radius BA, describe another circle. The circumferences of these circles will intersect at C and D. Draw the straight lines CA, CB, DA, DB. Two equilateral triangles will thus be constructed, which will be equal to each other. 73. To construct an isosceles triangle upon the line AB, (fig. 65.) with equal radii, Draw the straight Sol. From A and B as centres, describe two arcs intersecting at C. lines CA and CB. An isosceles triangle will thus be formed, of which AB is the base. 74. To construct upon the line AB an isosceles triangle, whose equal sides shall be equal to a given line M, (fig. 65.) Sol. From A and B as centres, with a radius equal to M, describe two arcs intersecting at C. Draw CA and CB. CAB is the triangle required. 75. To construct a right-angled isosceles triangle upon the line AB, with a right angle adjacent to the side AB, (fig. 66.) Sol. At either of the points A or B erect a perpendicular to the line AB. In this perpendicular take BCBA. Draw CA, and the required triangle will be constructed. 76. To construct a right-angled isosceles triangle upon a straight line AB, with the right angle opposite to the side AB, (fig. 67.) Sol. Bisect AB. From the division point C as a centre, with a radius equal to CA or CB, describe a semi-circumference. To AB at C erect a perpendicular, which will meet the semi-circumference at its middle point F. Draw FA and FB. FAB is the required triangle. If the proposition were to construct a rightangled triangle in general, we could take any point in the arc, and by connecting it by straight lines with A and B, should construct a right-angled triangle. 77. To construct an obtuse-angled and an acuteangled isosceles triangle, (fig. 67.) Sol. Produce the perpendicular CF beyond the semicircumference, and take in it the point D within, and the point E without the semi-circle. Draw DA, DB, EA, and EB. The triangle DAB is an obtuse-angled isosceles triangle, and the triangle EAB is an acuteangled isosceles triangle. CONSTRUCTION OF QUADRILATERALS. 78. To construct a quadrilateral which shall be equal to a given quadrilateral ABCD, (fig. 68.) Sol. In the given quadrilateral ABCD draw the diagonal AC. It is thereby divided into 2 triangles, CAB and DAC. If we construct 2 triangles equal to these, and place them together in the same relative position, the problem will be solved. Draw a straight line EFAB. From E as a centre, with a radius = AC, and from F as a centre, with a radius = BC, describe 2 arcs cutting one another at G. Draw GE and GF. The triangle GEF = CAB. Again, from E as a centre, with a radius= AD, and from G, with a radius = CD, describe 2 arcs cutting one another at H. Draw HE and HG. Triangle HEG DAC. Consequently the quadrilateral EFGH ABCD. = = 79. To construct a square upon a given straight line AB, (fig. 69.) Sol. 1. At the points A and B erect perpendiculars to the line AB. Take AC and BD each equal to AB. Draw CD and the required square will be constructed. Sol. 2. At the point A erect a perpendicular to the line AB. In this perpendicular take AC = AB. Through the point C draw CD equal and parallel to AB; and through B draw BD equal and parallel to AC. ABCD is the required square. 80. To construct a square in and about a circle, (fig. 70.) Sol. Describe a circle. Draw 2 diameters cutting each other at right angles. Join the ends of these diameters by straight lines, and a square will be constructed within a circle. At the points where the ends of the diameters meet the circumference, draw tangents to the circle. Produce each tangent both ways until they intersect one another; and a square will be constructed about a circle. 81. To construct a square which shall be double a given square, (fig. 72.) Sol. In the given square ABCD, draw the diagonal AC, and construct upon it the square AEFC, which square will be double the square ABCD. For ABCD is composed of 2 equal triangles, ABC and ADC, but AEFC is composed of 4 triangles, each equal to ABC. 82. Remark. In a right triangle the side opposite to the right angle is called the hypothenuse. Now you will observe that the square constructed upon the hypothenuse AC of the isosceles right-angled triangle ABC is equal to the squares constructed upon AB and BC, the sides which include the right angle, taken together. We shall hereafter find this to be true in all right-angled triangles. 83. To construct a square which shall be 4, 9, 16, 25, 36, &c. times as great as a given square, (fig. 71.) Sol. Let a square constructed upon AB be the given square. Produce AB, and take BC= AB, CF — AB, FI=AB, &c. Construct upon AC, AF, AI, &c., the squares ACDE, AFGH, AIKL, &c. ACDE is 4 times greater than the given square; AFGH is 9 times greater; and AIKL is 16 times greater. If the side of a square is 2, its surface will be 4 times as great as that of a square whose side is 1; a square whose side is 3 will have a surface 9 times as great as a square whose side is 1. 84. To construct many rectangular and obliqueangled parallelograms upon a given straight line AB, (fig. 73.) Sol. At the point A erect a perpendicular to AB. Take any points C and D in this perpendicular; through them draw lines parallel to AB. Through B draw a line parallel to AC. Thus we have constructed 2 rectangular parallelograms upon AB. Again, make at A an oblique angle. In the side which together with AB includes this angle take any points E and F, and through these points draw lines parallel to AB. Through the point B draw a line parallel to AE. We have thus constructed 2 obliqueangled parallelograms upon AB. If AC is taken equal to AB, the parallelogram will be a square; if we take AE=AB, the parallelogram will be a rhombus. 85. Two adjacent sides M and N and the included angle O of a parallelogram being given, it is required to construct the parallelogram, (fig. 74.) At A make an Sol. Draw a straight line AB=M. angle BAD=0, and take AD N. centre, with a radius equal to AB, describe an arc, and from the point B as a centre, with a radius equal to AD, describe another arc. From the point C, where these arcs cut each other, draw CD and CB. ABCD is the required parallelogram. 86. The 4 sides and 1 angle of a quadrilateral being given, it is required to construct the quadrilateral, (fig. 75.) |