MENSURATION 176. Certain measurements have been in very common use in the development of arithmetical knowledge from the earliest times. 177. The Babylonians and Egyptians used a great variety of geometrical figures in decorating their walls and in tile floors. The sense perception of these geometrical figures led to their actual measurement and finally to abstract geometrical reasoning. 178. The Greeks credited the Egyptians with the invention of geometry and gave as its origin the measurement of plots of land. Herodotus says that the Egyptian king, Sesostris (about 1400 B.C.), divided Egypt into equal rectangular plots of ground, and that the annual overflow of the Nile either washed away portions of the plot or obliterated the boundaries, making new measurements necessary. These measurements gave rise to the study of geometry (from ge, earth, and metron, to measure). 179. Ahmes, in his arithmetical work, calculates the contents of barns and the area of squares, rectangles, isosceles triangles, isosceles trapezoids and circles. There is no clew to his method of calculating volumes. In finding the area of the isosceles triangle he multiplies a side by half of the base, giving the area of a triangle whose sides are 10 and base 4 as 20 instead of 19.6, the result obtained by multiplying the altitude by half of the base. For the area of the isosceles trapezoid he multiplies a side by half the sum of the parallel bases, instead of finding the altitude and multiplying that by half the sum of the two parallel sides. He finds the area of the circle by subtracting from the diameter of its length and squaring the remainder. This leads to the fairly correct value of 3.1604 for π. 180. The Egyptians are also credited with knowing that in special cases the square on the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides. They were careful to locate their temples and other public buildings on north and south, and east and west lines. The north and south line they determined by means of the stars. The east and west line was determined at right angles to the other, probably by stretching around three pegs, driven into the ground, a rope measured into three parts which bore the same relation to each other as the numbers 3, 4 and 5. Since 324252, this gave the three sides of a right triangle. 181. The ancient Babylonians knew something of rudimentary geometrical measurements, especially of the circle. They also obtained a fairly correct value of π. 182. It was the Greeks who made geometry a science and gave rigid demonstrations of geometrical theorems. 183. The importance of certain measurements gives mensuration a prominent place in arithmetic to-day. The rules and formulæ of the present chapter will be developed without the aid of formal demonstration. 184. The Rectangle. If the unit of measure CYXN is 1 sq. in., then the strip CYMD contains 5 x 1 sq. in. 5 sq. in., and the whole area contained in the three strips will be 3 × 5 sq. in., or 15 sq. in. = A B X M Y N C 185. The dimensions of a rectangle are its base and altitude, and the area is equal to the product of the base and altitude. That is, the number of square units in the area is equal to the product of the numbers that represent the base and altitude. If we denote the area by A, the number of units in the base by b, and the numbers of units in the altitude by a, then A=ab and any one of the three quantities, A, b, a, can be determined when the other two are given. Thus, if the area of a rectangle is 54 sq. in. and the altitude is 6 in., the base can be determined from 6b 54, or b = 9 in, = 186. If the dimensions of a rectangle are equal, the figure is a square and the area is equal to the second power of a number denoting the length of its side, or Aa2. For this reason the second power of a number is called its square. A M b B 187. The Parallelogram. Any parallelogram has the same area as a rectangle with the same base and altitude, as can be shown by dividing the parallelogram ABCD into two equal parts, M and N, and placing them as in A'B' C'D', thus forming a rectangle with the same base and altitude as the given parallelogram. Therefore, the area of a parallelogram is equal to the product of its base and altitude, or A= ab. A' D' A α B' M a b 188. The Triangle. Since the line AB divides the parallelogram into two equal triangles with the same base and altitude as the parallelogram, the area of the triangle is equal to one half of the area of the parallelogram. But the area of the parallelogram is equal to the product of its base and B altitude. Therefore, the area of the triangle is one half the product of its base and altitude, or A = 1⁄2 ab. Thus, the area of a triangle with base 6 in. and altitude 5 in. is of 6 × 5 sq. in. 15 sq. in. 189. The Trapezoid. The line AB divides the trapezoid into two triangles whose bases are the upper and lower bases of the trapezoid and whose common altitude is the altitude of the trapezoid. The areas of the triangles are respectively ab1 and ab2, and since the area of the trapezoid equals the sum of the areas of the triangles, therefore the area of the trapezoid is ab1 + 1⁄2 ab2 1a(b1+b2), or the area of a trapezoid is equal to one half of the product of its altitude and the sum of the upper and lower bases, or A= 1⁄2 a(b1 + b2). = Thus, the area of a trapezoid whose bases are 20 ft. and 17 ft. and whose altitude is 6 ft. is of 6 × (20+ 17) sq. ft. = 111 sq. ft. 190. The Right Triangle. The Hindu mathematician Bhaskara (born 1114 A.D.) arranged the figure so that the square on the hypotenuse contained four right triangles, leaving in the middle a small square whose side equals the difference between the sides of the right triangle. In a second figure the small square and the right triangles were arranged in a different way so as to make up the squares on the two sides. Bhaskara's proof consisted simply in drawing the figure and writing the one word "Behold." From these figures it is evident that the area of the square constructed on the hypotenuse will equal the sum of the areas of the squares constructed on the two sides. In general, if the sides of the right triangle are a and b and the hypotenuse is c, a2 + b2 = c2. Thus, the hypotenuse of the right triangle whose sides are 5 and 12 is √25+ 144 = 13. This theorem is known by the name of the Pythagorean theorem, because it is supposed to have been first proved by the Greek mathematician Pythagoras, about 500 B.C. 191. If either side and the hypotenuse of a right triangle are known, the other side can be found from the equation a2+b2 = c2. Thus, if one side is 3 and the hypotenuse is 5, the other side is √25 - 9 = 4. EXERCISE 29 1. The two sides of a right triangle are 6 in. and 8 in. Find the length of the hypotenuse. 2. Find the area of an isosceles triangle, if the equal sides are each 10 ft. and the base is 4 ft. 3. Find the area of an isosceles trapezoid, if the bases are 10 ft. and 18 ft. and the equal sides are 8 ft. 4. What is the area in hectares, etc., of a field in the form of a trapezoid of which the bases are 475m and 580m and the altitude is 1270m? 5. Show that the altitude of an equilateral triangle, each of whose sides is a, is 2√3. α 6. The hypotenuse of a right triangle with equal sides is 10 ft. Find the length of the two equal sides. 7. The diagonal of a square field is 80 rd. How many acres does the field contain? 8. Find correct to centimeters the area of an equilateral triangle each side of which is 1m in length. 192. The Circle. If the circumference (c) and the diameter (d) of a number of circles are carefully meas |