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 Chapter 6 : Quadrilaterals Area Area Two of our most important sources of information about mathematics in ancient Egypt are the Ahmes (or Ahmose) papyrus and the Moscow papyrus. The Ahmes papyrus is named for the Egyptian scribe who copied it around 1650 B.C. Ahmes claimed that his material derived from an original manuscript composed during a period of Egyptian history known as the Middle Kingdom, which lasted from about 2000 to 1800 B.C. Some of this knowledge may have originated with Imhotep, architect and physician for Pharaoh Zoser. If so, this information would date back to the time around 2650 B.C., even before the Great Pyramid was built. The problems in the Moscow papyrus are similar to those in the Ahmes papyrus. The Moscow Papyrus, brought to Russia in the mid-1900's, was written about 1890 B.C. Together, the Ahmes and Moscow papyruses, or papyri, contain 112 problems and their solutions for situations dealing with everyday life. Both papyri show that the Egyptians knew that the area of a rectangle could be found by multiplying its base times its length. They were also able to find the length of a rectangle given the area and the width. Problem 50 of the Ahmes Papyrus demonstrates how to find the area of a circle from its diameter. The quantity ¼ was not known, but by assuming that the area of a circle with a diameter of nine units equals the area of a square with a side of eight units, the Egyptians developed a rule for the area of a circle equivalent to the modern formula , where had a value of , or about 3.16, not far from the approximate value 3.14. Problem 51 of the Ahmes Papyrus shows that the area of an isosceles triangle can be found by taking half the base and multiplying this quantity by the height. This is justified by pointing out that an isosceles triangle can be thought of as two right triangles. Shifting one of these triangles and rejoining the right triangles forms a rectangle with half the base of the isosceles triangle and with the same height. Problem 52 similarly handles finding the area of an isosceles trapezoid. By forming a rectangle from the trapezoid so that the height of the rectangle is the same as the height of the trapezoid and the base of the rectangle is one-half the sum of the bases of the trapezoid, the area of a trapezoid was demonstrated to equal the height times one-half the sum of the bases.