SECTION VI.-QUADRILATERALS. DEFINITIONS. 1. A quadrilateral is a polygon of four sides. 2. A parallelogram is a quadrilateral having its opposite sides parallel. 3. A rectangle is a parallelogram whose angles are all right angles. 4. A square is a rectangle whose sides are all equal. Ques. Why is a square a parallelogram? Why a quadrilateral? 5. A trapezoid is a quadrilateral having two opposite sides parallel. 6. A diagonal of a polygon is a line joining two angles not adjacent. Ques. How many diagonals may a quadrilateral have? A triangle? 7. The area of any plane figure is the amount of surface which it contains. 8. Two plane figures which will not coincide, but which have equal areas, are equivalent. Ques. When are two plane figures equal? (Ax. 10.) THEOREM XX. The opposite sides and angles of a parallelogram are equal to each other. The figure ABCD is a parallelo gram. Draw the diagonal AC. How many pairs of parallels does it cut? Why is the angle DCA equal to A the angle CAB? Why is CAD equal to ACB?. D The triangles ABC and ADC are equal (Theo. XVIII). AD and BC are opposite equal angles; likewise DC and AB. B Therefore, the opposite sides are equal. Ang. ACB ang. CAD. Ang. ACD = Prove ang. D= ang. B. Cor. 1.-A diagonal of a parallelogram divides it into two equal triangles? Cor. 2.-Two parallels are everywhere equally distant. Def.-Any side of a parallelogram may be taken as its base; and a perpendicular let fall from any point in the opposite side on the base, or the base produced, is called the altitude of the parallelogram. THEOREM XXI. If a quadrilateral has two of its sides equal and parallel, it is a parallelogram. B Let the quadrilateral ABCD have the sides AB, DC equal and parallel. Then the figure is a parallelogram. Draw the diagonal AC, and the alternate angles DCA and BAC are equal (?). Then the triangles CDA and CBA are equal (Theo. XII) in all their parts; and, as ang. ACB=ang. CAD, the lines AD and BC are parallel (?). Therefore (Def. 2, Sec. VI), the quadrilateral ABCD is a parallelogram. Ques.-Is a quadrilateral containing two right angles necessarily a parallelogram? Three? THEOREM XXII. The diagonals of a parallelogram bisect each other. Let ABCD be a parallelogram. It is to be proved that its diagonals, AC, BD, bisect each other in E. Α B In the triangles AEB, DEC, since the angle ABD is equal to its alternate angle BDC (Theo. III), and the angle BAC equal to its alternate angle ACD, and the included side AB equal to the included side DC (Theo. XX), it follows that the two triangles are equal (Theo. XVIII); consequently, the side AE is equal to the side EC, and BE to ED (Cor., Theo. XII); that is, the two diagonals bisect each other in E. Therefore, the diagonals, etc. THEOREM XXIII. The area of a rectangle is equal to the product of its base by its altitude. Let ABCD be a rectangle. It is to be proved that its area is equal to the product of its base AB by its altitude AD. D g Let AB be divided into a certain number of equal parts, Ab, bc, etc., taken as the units of A length; also, let AD be divided. e C B into a certain number of the same units, Af, fg., etc. From b, c, etc., draw straight lines parallel to AD, and from f, etc., draw straight lines parallel to AB. Now, it is evident that the whole rectangle is divided into small squares (Cor. 2, Theo. XX), each equal to Abef; which may be taken as the unit of area. Of these equal squares, there are as many in the tier next to number of linear units in AD. Hence, the area of a rectangle, etc. Sch. 1.—This measurement gives the numerical area, as we multiply the length of the base by the length of the altitude. The above form of statement is used for brevity. Sch. 2.-If AB and AD are incommensurable, that is, if no unit can be found into which they can both be divided without leaving a remainder in one of them, the theorem will still hold true; for if the unit be taken smaller and smaller, the remainder can be made less than any assignable quantity. When the linear unit is one inch, the unit of area is a square inch; when the linear unit is one foot, the unit of area is a square foot, etc. Cor. Since the base and altitude of a square are equal (Def. 4, Sec. VI), its area may be found by multiplying one side into itself. THEOREM XXIV. The area of any parallelogram is equal to the area of a rectangle having the same base and altitude. AB, and of the same altitude, namely, the perpendicular distance between the parallels AB, DF. Then will ABFE be equivalent to ABCD. Since AB and DC are opposite sides of a parallelogram, they are equal (Theo. XX); and, for the same reason, AB and EF are equal; therefore, DC is equal to EF (Ax. 1). Taking away each of these in turn from the whole line DF, we have the remainder DE equal to the remainder CF. But DA is equal to CB (Theo. XX), and the included angle ADE is equal to the included angle BCF (Theorem III); therefore, the triangles ADE, BCF are equal (Theo. XII); and hence, if each of them be taken away in turn from the whole figure ABFD, the remainder ABFE will be equivalent to the remainder ABCD. Therefore, the area of any parallelogram, etc. Cor. 1.-The area of any parallelogram is equal to the product of its base by its altitude. Cor. 2. Since any triangle, as ABC, is half of a parallelogram, A BE Cor. 3.-Triangles on the same base or on equal bases, and of equal altitudes, have equal areas. Cor. 4.-Triangles of equal areas, on the same base or on equal bases, have equal altitudes. THEOREM XXV. The area of a trapezoid is equal to half the product of the sum of its parallel sides by its altitude. E G.-4. |