angle is greater than a right angle, it is an obtuse-angled triangle, as A OC (fig. 1); and if each of the angles is less than a right angle, it is an acute-angled triangle, as A OB (fig. 1). No triangle can have more than one right angle, and the sum of the three angles of every triangle is always equal to two right angles. In every triangle the sum of the lengths of any two sides is greater than the length of the third side.

A figure formed by four right lines is termed a quadrilateral figure, and the straight lines which join the opposite corners are called diagonals, as A D and BC (fig. 29), which are the diagonals of the figure ABDC. An irregular quadrilateral figure, having all the sides of different lengths, is called a trapezium. A parallelogram or rhomboid is a quadrilateral figure, having the opposite sides equal and parallel, as A B DC (fig. 29), and if all the sides are equal it is called a lozenge or rhombus. A quadrilateral figure in which all the angles are right angles is termed a rectangle or oblong, as ABFE (fig. 29), and when all the sides are also equal it is a square (fig. 27). A rectangle whose adjacent sides are equal to two given lines is called the rectangle under those lines, thus ABFE (fig. 29) is the rectangle under A B and A E. A square having its side equal to a given line A B (fig. 27), is called the square upon A B.

A regular polygon is a figure having all its sides equal, and all the angles made by those sides with each other also equal, so that a circle can be drawn through all its angular points.

The area or amount of surface contained in any figure is measured by the number of square units, or squares of a given size, which it covers. The finding of areas of various figures belongs properly to the science of Mensuration.'

Fig. 20.

Two rectilinear figures are said to be similar when the angles in one are severally equal to the angles in the other and the sides about the equal angles are proportional; thus the triangles BAD and CAE (figs. 13 and 14) are similar. 24. To construct an isosceles triangle, of which two unequal sides are given.-Draw a straight line, A B D (fig. 20), and take A B equal to the given base of the triangle, A. D equal to one of the given equal sides. From A and B as centres, and with AD as radius, describe circles cutting each other in the point C; join AC, BC. Then ACB is the isosD celes triangle required, having its two legs A C and BC equal to one

another.

25. To construct an isosceles triangle in which each angle at the base is double that at the vertex.-First, when the length of the equal sides is given.

Fig. 21.

Let A B (fig. 21) be the given length of one of the sides. Divide A B in extreme and mean ratio (19) at the point C. From A as a centre, and with AB for radius, describe a circle; and from B as a centre, with A C for radius, draw an arc cutting the circle in the point D. Join AD, BD; then BAD is the triangle required, each angle at B and D being double the

angle at A.

Secondly, when the length of the base only is given.
Let AC be the given length of the base. Produce A C

to B so that the line A B shall be divided at C in extreme and mean ratio (20). From A as a centre, and with A B as radius, describe a circle; and from B as a centre, with AC for radius, draw an arc cutting the circle in the point D. Join BD and AD; then BAD is the triangle required, having B D equal to A C, the given length of the base, and each angle at B and D double the angle at A.

Fig. 22.

26. To construct an equilateral-triangle upon a given finite right line. Let AB (fig. 22) be the given line. From A and B as centres, and with AB for radius, describe circles intersecting in the point C. Join AC, BC; and ACB is an equilateral triangle, having AC and BC each equal to A B.

The angles are also equal, being each of them two-thirds of a right angle; hence in laying out an equi

lateral triangle on a large scale it is best to construct the two angles CA B and CBA, each equal to two-thirds of a right angle (60°), and the point C will be found.

27. To construct a right-angled-triangle of which two

sides are given.-First, let the

given sides be those which include the right angle.

Draw two lines at right angles to each other (3), as AB, AC (fig. 23); and make AC and AB equal to the given sides

Fig. 23.

respectively. Join BC; then BA C is the triangle required, having the right angle at A. The side BC which subtends or is opposite the right angle, is termed the hypotenuse, and is always greater than either of the other sides.

C

Secondly, let the base and hypotenuse be given. Draw any two lines, as A C and A B, at right angles to one another, and make AC equal to the given base. Take CD, equal to the given hypotenuse, and from C as a centre, and with CD for radius, describe a circle cutting AB in the point B. Join B C, and CAB is the triangle required.

28. To construct a triangle whose three sides are of given lengths.-(In this case the sum of any two of the given lengths must be greater than the third.)

Fig. 24.

Draw a right line, BD (fig. 24), and take A B, equal to one of the given sides, BD equal to another, and D C to the third. From B as a centre, with BD as radius, describe a circle; and from A as a centre, with DC for radius, draw an arc cutting the circle in the point E. Join EA, EB; and BAE is

the triangle required.

29. To construct a triangle of which two sides and the

Fig. 25,

included angle are given.-Let CAB (fig. 25) be the given D angle. Take AB and AC, equal to the given sides respectively, and join B C. Then BAC is the triangle required.

30. To construct a triangle of which two angles and a side are given.-(The sum of the given angles must be less than two right angles.) First, let the base, as A B (fig. 25), be the given side, and CA B, CBA the given angles adjacent to the base. Construct the angles CA B, CB A, respectively

equal to the given angles (7); then the lines B C and A C will meet at C, and the triangle is completed.

Secondly, let the base A B (fig. 25) and the angles CA B, BCA be the given angles, one of which BCA is opposite the base or given side. Construct the angle CAB equal to one given angle, and the angle CAD equal to the other (7). Draw BC parallel to AD (13), meeting AC in the point C; then the angle ACB is equal (12) to the angle CAD, and the triangle BAC is the one required.

Another method is to take any point F on AC and draw FE, making the angle EFA equal to the given angle, and then drawing BC parallel to E F.

Fig. 26.

31. To construct a triangle of which two sides and an angle not included are given.-Draw a right line as AC (fig. 26), and take AB equal to one of the given sides, BC equal to the other. At A make the angle BAD equal to the given angle (7), and from B as a centre, with BC for

radius, describe a circle cutting

AD in the point D. Join BD, and the triangle BAD is the one required, having BA and BD

respectively equal to the given sides, and BAD to the given angle.

32. To construct a square upon a given right line.-Let A B (fig. 27) be the given line. Erect the perpendiculars A C, BD (3), and make AC and BD each equal to AB. Join CD; then

Fig. 27.

the figure A BDC is the square required, having each side equal to A B.