Ex. 68. Construct a fourth proportional to three lines whose lengths are 8, 9, and 12, respectively. Ex. 69. Construct a fourth proportional to 12, 8, and 6. Ex. 70. Construct a third proportional to lines whose lengths are 8 and 6, respectively. Ex. 71. Construct a mean proportional to lines whose lengths are 9 and 4. Ex. 72. Construct a line whose length is √5. 271. Upon a given side, homologous to a given side of a given polygon, to construct a polygon similar to the given polygon. Given polygon ABCDE, and line A'B'. Required to construct upon side A'B', homologous to AB, a polygon similar to ABCDE. Construction. 1. Divide polygon ABCDE into A by drawing diagonals EB and EC 2. Construct ▲ A'B'E', ABE similar, with ▲ A' = ▲ A, and A'B'E' LABE. (?) 3. In like manner, construct ▲ B'C'E' similar to ▲ BCE, and A CD'E' similar to ▲ CDE. 272. Def. A straight line is said to be divided by a given point in extreme and mean ratio when one of the segments (§ 231) is a mean proportional between the whole line and the other segment. Thus, line AB is divided internally in extreme and mean ratio at Cif and externally in extreme and mean ratio at D if 273. To divide a given straight line in extreme and mean ratio (§ 272). Required to divide it in extreme and mean ratio. Construction. 1. With radius BE AB, draw ○ BFG tangent to AB at B; and line AE cutting circumference at F and G. 2. On AB take AC AF; on BA extended take AD = AG; then, AB is divided at C internally, and at D externally, in extreme and mean ratio. AG AB AG AB Proof. 3. By § 261, or (1) AB AF AB AC Multiplying by 4, and adding m2 to both members, 4x2 + 4 mx + m2 = 4 m2 + m2 = 5 m2. Extracting the square root of both members, 2x+m= ±m √5. Since x cannot be negative, we take the positive sign before the radical sign; then, 2 x = m √5—m; and a (or AC) = m(√5 −1). 2 BOOK IV AREAS OF POLYGONS PROP. I. THEOREM 275. Two rectangles having equal altitudes are to each other their bases. The words "rectangle," "parallelogram,' "triangle," etc., in the opositions of Book IV, mean the amount of surface in the rectangle, rallelogram, triangle, etc. Case I. When the bases are commensurable. Draw rectangles ABCD and EFGH having equal altitudes and the ases AD and EH commensurable. We then have : Given rectangles ABCD and EFGH, with equal altitudes AB and EF, and commensurable bases AD and EH. To Prove ABCD AD EH (1) Proof. 1. Let AK be contained 5 times in AD, and 3 times n EH. 2. Find ratio of AD to EH. 3. Through points of division of AD and EH draw is to hese lines. 4. Find ratio of ABCD to EFGH. 5. Prove equation (1). (§ 111) (?) We then have : equal altitudes Given rectangles ABCD and EFGH, with To Prove ABCD AD EH (2) Proof. 1. Divide AD into any number of equal parts. 2. Let one of these parts be contained exactly in EK, with a remainder KH < one of the parts. 3. Draw line KL 1 EH, meeting FG at L. 5. Let number of subdivisions of AD be indefinitely increased. 7. Prove equation (2) by Theorem of Limits. 276. Since either side of a rectangle may be taken as the base, it follows that Two rectangles having equal bases are to each other as their altitudes. PROP. II. THEOREM 277. Any two rectangles are to each other as the products of their bases by their altitudes. I |