PROPOSITION VI. THEOREM. 327. The area of a trapezoid is equal to one-half the sum of the parallel sides multiplied by the altitude. Let ABCH be a trapezoid, and EF the altitude. area of ABC H = } (H C + A B) E F. Draw the diagonal A C. We are to prove = Then the area of the AAHC HC XEF, $324 (the area of a ▲ is equal to one-half of the product of its base by its altitude), 328. COROLLARY. The area of a trapezoid is equal to the product of the line joining the middle points of the non-parallel sides multiplied by the altitude; for the line OP, joining the middle points of the non-parallel sides, is equal to (HC + A B). .. by substituting O P for (HC + A B), we have, 329. SCHOLIUM. The area of an irregular polygon may be found by dividing the polygon into triangles, and by finding the area of each of these triangles separately. But the method generally employed in practice is to draw the longest $142 diagonal, and to let fall perpendiculars upon this diagonal from the other angular points of the polygon. The polygon is thus divided into figures which are right triangles, rectangles, or trapezoids; and the areas of each of these figures may be readily found. PROPOSITION VII. THEOREM. 330. The area of a circumscribed polygon is equal to onehalf the product of the perimeter by the radius of the inscribed circle. Let ABSQ, etc., be a circumscribed polygon, and C the centre of the inscribed circle. Denote the perimeter of the polygon by P, and the radius of the inscribed circle by R. We are to prove the area of the circumscribed polygon = PX R. Draw CA, CB, CS, etc.; also draw CO, CD, etc., I to A B, BS, etc. = The area of the ▲ CAB ЗАВХСО, (the area of a A is equal to one-half the product of its base and altitude). § 324 the area of the sum of all the ACA B, CBS, etc., (for CO, CD, etc., are equal, being radii of the same ). § 187 Substitute for A B+ BS + SQ, etc., P, and for CO, R; then the area of the circumscribed polygon = = PX R. Q. E. D. PROPOSITION VIII. THEOREM. 331. The sum of the squares described on the two sides of a right triangle is equivalent to the square described on the hypotenuse. C Let ABC be a right triangle with its right angle at C. (the square on a side of a rt. ▲ is $ 289 equal to the product of the hypotenuse by the adjacent segment made by the let fall from the vertex of the rt. ≤); By adding, AC2 + B C2 = (A O + B O) A B, =ABX AB, $ 289 = à B2. 332. COROLLARY. The side and diagonal A of a square are incommensurable. Let ABCD be a square, and AC the Divide both sides of the equation by à Â2, Q. E. D. Extract the square root of both sides the equation, Since the square root of 2 is a number which cannot be exactly found, it follows that the diagonal and side of a square are two incommensurable lines. ANOTHER DEMONSTRATION. 333. The square described on the hypotenuse of a right triangle is equivalent to the sum of the squares on the other two sides. Ꮐ Let ABC be a right ▲, having the right angle BA C. On BC, CA, A B construct the squares B E, CH, A F. Through A draw A L to CE. Draw A D and FC. (being on the same base B D, and between the same ls, AL and B D), and square A F is double ▲ FBC, (being on the same base FB, and between the same ||s, FB and G C) ; OBL = square A F In like manner, by joining A E and BK, it may be proved 334. DEF. The Projection of a Point upon a straight line of indefinite length is the foot of the perpendicular let fall from the point upon the line. Thus, the projection of the point C upon the line A B is the point P. The Projection of a Finite Straight Line, as CD (Fig. 1), upon a straight line of indefinite length, as A B, is the part of the line AB intercepted between the perpendiculars CP and D R, let fall from the extremities of the line C D. Thus the projection of the line CD upon the line A B is the line P R. If one extremity of the line CD (Fig. 2) be in the line A B, the projection of the line CD upon the line AB is the part of the line A B between the point D and the foot of the perpendicular C P ; that is, D P. |