ing 5 squares; hence the whole number of squares, which is the entire surface of the rectangle, equals 3 times 5, or 15. Thus, the surface equals the number of linear units in the base multiplied by the number in the altitude. If the base and altitude are not commensurable, the conception by the doctrine of infinities is as follows: The above is true, whatever the size of the unit of measure; hence it is true when the unit becomes indefinitely or infinitely small, as it must be when the sides are incommensurables; therefore the theorem is universally true. SCHOLIUM 2.-The " product of two lines" is often called the "rectangle of the lines," because the product is equal to the area of a rectangle constructed on the lines as sides. Similarly the second power of a number is called the square of the number. PROPOSITION V.-THEOREM. The area of a parallelogram is equal to the product of its base by its altitude. Given.-Let ABCD be any parallelogram, AB its base, and BE its altitude. To Prove.—Then we are to prove that F D. E the area of ABCD = AB × BE. Proof. On the base AB, with the altitude BE, construct the rectangle ABEF. Then, rect. ABEF□ ABCD. Now, the area of rect. ABEF = AB × BE. Th. 1. Th. 4. Hence, the area of ABCD = AB × BE Therefore, etc. COR. 1.-Parallelograms are to each other as the product of their bases and altitudes. COR. 2.-Parallelograms having equal altitudes are to each other as their bases. COR. 3.-Parallelograms having equal bases are to each other as their altitudes. COR. 4.-Parallelograms having equal bases and equal altitudes are equivalent. PROPOSITION VI.-THEOREM. The area of a triangle is equal to one-half of the product of its base by its altitude. Given. Let ABC be a triangle, AB its base, and CD its altitude. To Prove. Then we are to prove that D B the area of ABC= }AB × CD. Proof. From B draw BE parallel to AC; and from C draw CE parallel to AB, completing the parallelogram COR. 1.-Triangles are to each other as the products of their bases and altitudes. COR. 2.-Triangles having equal altitudes are to each other as their bases; triangles having equal bases are to each other as their altitudes. COR. 3.- Triangles having equal bases and equal altitudes are equivalent. COR. 4.-A triangle is equivalent to one-half of any parallelogram having an equal base and an equal altitude. PROPOSITION VII.-THEOREM. The area of a trapezoid is equal to one-half the sum of its parallel sides multiplied by its altitude. Given.-Let ABCD be a trapezoid, AB and CD its parallel sides, and To Prove. Then we are to prove that the = E B area of ABCD (AB+ DC) × DE. Proof. Draw the diagonal AC, dividing the trapezoid into the two triangles ABC and ADC, the altitude of each being DE. Then, Th. 6. area of ABC= AB × DE. 66 66 = Taking the sum of these two equalities, = {(AB + DC) × DE. COR.-The area of a trapezoid is equal to the product of the line joining the middle points of its non-parallel sides by the altitude. Draw FG, joining the middle points of AD and BC. SCHOLIUM.-The area of an irregular polygon can be found by dividing the polygon into triangles and finding the area of each of these triangles, and taking their sum. The method usually preferred is to draw the longest diagonal, and on this diagonal let fall perpendiculars from the other angular points of the polygon. The polygon is thus divided into figures which are right triangles, right trapezoids, or rectangles, and the areas of these figures can be readily found. SQUARES ON LINES. 8. In treating of Squares on Lines we use the notation AB2 to denote a square constructed on the line AB, and not the numerical product of AB × AB. 9. We also use the expression AB × BC to denote a rectangle contained by the two lines AB and BC, and not the numerical product of AB and BC. 10. The projection of a line CD upon a straight line AB is the dis tance on AB included between per pendiculars let fall from the ex B tremities of CD. Thus, EF is the projection of CD on AB. 11. If one extremity of the line CD is in the line AB, the projec tion of the line CD on AB is the A distance between the point C and F the foot of the perpendicular DF; that is, CF. PROPOSITION VIII.-THEOREM. The square described on the sum of any two lines is equivalent to the sum of the squares described on the lines, increased by twice the rectangle of the lines. Given. Let AB and BC be two lines, and AC their sum. To Prove. Then we are to prove that ACAB2+BC2+2AB× BC. Proof.-On AC construct the square ACFD, and on AB construct the H E B square ABHG; prolong BH to E and GH to I; then all the figures are rectangles by construction. = Now, on account of squares, AC AD, and AB AG; hence BC= GD. On account of parallel lines, BC= HI, and GD = HE; hence HE BC; and consequently HF is equal to a square on BC. = Also, since BH= AB, BCIH is a rectangle on AB anɑ BC; and since GH= AB and GD = BC, GHED is a rectangle on AB and BC. Hence the square ACFD consists of the squares on the two lines AB and BC, increased by twice the rectangle of AB and BC; that is, AC2 AB2 + BC2 + 2AB × BC. Therefore, etc. COR.-The square of any line is equivalent to four times the square of half that line. For if AB equals BC, the four figures AH, BI, GE, and HF are all squares. SCHOLIUM. If a and b denote the numbers representing the two lines, we shall have the algebraic formula, (a + b)2 = a2 + 2ab + b2. |