PROBLEM 6 On a straight line of given length to make a parallelogram equal in area to a given triangle and having an angle equal to a given angle. Let ABC be the given ▲, E the given st. line and D the given 2. It is required to make a gm equal in area to ▲ ABC, having one side equal in length to E, and one ▲ equal to D. Construction.-From BC, produced if necessary, cut off BF E Join AF. Through C draw CG | FA meeting BA, or BA produced, at G. Join GF. Bisect BG at H. Through H draw HM || BC. At B make L CBL Through F draw FM || BL. LBFM is the required ||gm. Proof. Join HF. = L D. AS GAF, AFC are on the same base AF and have the same altitude, .. they are equal. (II—5, p. 101.) To each of these equal As add the A ABF, and ▲ GBF = ▲ ABC. ▲ GBF twice ▲ HBF, (II—6, Cor. 2, p. 102.) (II-7, p. 103.) = ||gm LBFM, AREAS OF SQUARES 80. A rectangle is said to be contained by two st. lines when its length is equal to one of the st. lines, and its breadth is equal to the other. The symbol AB2 should be read:-"the square on AB," and not "AB squared." THEOREM 10 The square on the sum of two straight lines equals the sum of the squares on the two straight lines increased by twice the rectangle contained by the straight lines. Hypothesis.-AB, BC are the two st. lines placed in the same st. line so that AC is their sum. To prove that AC2 = AB2 + BC2 + 2. AB.BC. Algebraic Proof Proof.-Let a, b represent the number of units of length in AB, BC respectively. = area of square on AB + area of square on BC + twice the area of the rectangle contained by AB, and BC. THEOREM 11 The square on the difference of two straight lines equals the sum of the squares on the two straight lines diminished by twice the rectangle contained by the straight lines. Hypothesis.-AB, BC are two st. lines, of which AB is the greater, placed in the same st. line, and so that AC is their difference. Proof-Let a, b represent the number of units of length in AB, BC respectively. Area of square on AC = (a - b)2 = x2 + b2 2ab. = the sum of the squares on AB and BC diminished by twice the area of the rectangle contained by AB and Construction.-On AC, AB, BC draw the squares ACED, ABFG, BCKH. Produce DE to meet BF at L. The difference of the squares on two straight lines equals the rectangle of which the length is the sum of the straight lines and the breadth is the difference of the straight lines. A, B are two st. lines, of which A> B. To prove that the square on A diminished by the square on B = the rect. contained by A+B and A - B. Proof.-Let a, b represent the number of units in A and B respectively. The difference of the squares on A and B |