a In all cases where a u.l. and a. u.a. are considered together, they are supposed to be connected by the relation of 150°, 3 and 4. B A 152°. Theorem.-- The number of unit-areas in a rectangle is the product of the numbers of unit-lengths in two adjacent sides. The proof is divided into three cases. 1. Let the measures of the adjacent sides with respect to ç the unit adopted be whole numbers. Let AB contain the assumed u.l. a times, and let AD contain it b times. Then, by dividing AB into a equal o parts and drawing, through each point of division, lines || to AD, and by dividing AD into b equal parts and drawing, through each point of division, lines || to AB, we divide the whole rectangle into equal squares, of which there are a rows with b squares in each row. the whole number of squares is ab. But each square has the u.l. as its side and is therefore the u.a. u.a.s in AC=u.l.s in AB xu.l.s in AD. We express this relation more concisely by writing symbolically DAC=AB. AD, where DAC means“ the number of u.a.s in DAC," and AB and AD mean respectively “the numbers of u.l.s in these sides.” And in language we say, the area of a rectangle is the product of its adjacent sides ; the proper interpretation of which is easily given. 2. Let the measures of the adjacent sides with respect to the unit adopted be fractional. Then, ::: AB and AD are commensurable, some unit will be an aliquot part of each (150°, 5). Let the new unit be th of the adopted unit, and let AB contain p of the new units, and AD contain a of them. The measure of DAC in terms of the new u.a. is pa а n n or (152°, 1), and the measure of the DAC in terms of the adopted unit is P9 (151°, 2) n2 But the measure of AB in terms of the adopted u.l. is , and of AD it is 9. (151°, 2) pq –P.4, and na n n' DAC=AB. AD. Illus.-Suppose the measures of AB and AD to some unit-length to be 3.472 and 4.631. By taking a ul. 1000 times smaller these measures become the whole numbers 3472 and 4631, and the number of corresponding u.a.s in the rectangle is 3472 x 4631 or 16078832 ; and dividing by 1000%, the measure of the area with respect to the original u.l. is 16.078832= 3.472 X 4.631. 3. Let the adjacent sides be incommensurable. There is now no u.l. that will measure both AB B and AD. If DAC is not equal to AB. AD, let it be equal to AB. AE, where AE has a measure different from AD; and suppose, first, that AE is < AD, so that E lies between A and D. With any u.l. which will measure AB, and which is less than ED, divide AD into parts. One point of division at least must fall between E and D; let it fall at H. Complete the rectangle BH. Then AB and AH are commensurable, and OBH=AB. AH, but OBD=AB. AE; (hyp.) and DBH is < OBD; AB. AH is < AB. AE, and AB being a common factor AH is < AE ; which is not true. .. If DAC=AB. AE, AE cannot be < AD, and similarly it may be shown that AE cannot be > AD; .. AE=AD, or DAC=AB. AD. q.e.d. A EHD 153o. The results of the last article in conjunction with Section I. of this Part give us the following theorems. 1. The area of a parallelogram is the product of its base and altitude. (140) a 2. The area of a triangle is one-half the product of its base and altitude. (141°) B. D 3. The area of a trapezoid is one-half the product of its altitude and the sum of its parallel sides. (145°, Ex.) 4. The area of any regular polygon is one-half the product of its apothem and perimeter. (147°) 5. The area of a circle is one-half the product of its radius and a line-segment equal to its circumference. (149) Ex. 1. Let O, O' be the centres of the in-circle and of the ex-circle to the side BC (131°); and let OD, O'P" be perpendiculars on BC, OE, O'P' perpendiculars on AC, and OF, O'P on AB. Then OD=OE=OF=p and O'P=O'P'=O'P"=p'; :: AABC=AAOB+ABOC + ACOA = }AB. OF+ BC.OD +}CA. OE (153', 2) = r x perimeter=rs, where s is the half perimeter ; A= =O'P. AB+ }O'P'.AC-}O'P". BC ={r'(6+c-a)=r's -a), where q' is the radius of the ex-circle to side a; A=r'ls -a). Similarly, A=r"s -b) =r"(s-c). timer E с P =rs. oto EXERCISES. I 1 I I I. + p" 2. A2=rr'r"q'". 3. What relation holds between the radius of the in-circle and that of an ex-circle when the triangle is equiangular? Note.—When the diameter of a circle is taken as the u.l. the measure of the circumference is the inexpressible numerical quantity symbolized by the letter ?, and which, expressed approximately, is 3.1415926.... 4. What is the area of a square when its diagonal is taken as the u.l. ? 5. What is the measure of the diagonal of a square when the side is taken as the u.l.? (150°, 5) 6. Find the measure of the area of a circle when the di ameter is the ul. When the circumference is the u.l. 7. If one line-segment be twice as long as another, the square on the first has four times the area of the square on the second. (151°, 2) 8. If one line-segment be twice as long as another, the equilateral triangle on the first is four times that on the second. (141°) 9. The equilateral triangle on the altitude of another equilat eral triangle has an area three-fourths that of the other. 10. The three medians of any triangle divide its area into six equal triangles. 11. From the centroid of a triangle draw three lines to the sides so as to divide the triangle into three equal quadrangles. 12. In the triangle ABC X is taken in BC, Y in CA, and Z in AB, so that BX=}BC, CY=}CA, and AZ=}AB. Express the area of the triangle XYZ in terms of that of ABC. 13. Generalize 12 by making BX=-VC, etc. go 14. Show that a=s(I '+g''). n SECTION III. GEOMETRIC INTERPRETATION OF ALGEBRAIC FORMS. 154°: We have a language of symbols by which to express and develop mathematical relations, namely, Algebra. The symbols of Algebra are quantitative and operative, and it is very desirable, while giving a geometric meaning to the symbol of quantity, to so modify the meanings of the symbols of operation as to apply algebraic forms in Geometry. This application shortens and generalizes the statements of geometric relations without interfering with their accuracy. Elementary Algebra being generalized Arithmetic, its quantitative symbols denote numbers and its operative symbols are so defined as to be consistent with the common properties of numbers. Thus, because 2+3=3+2 and 2.3=3.2, we say that a+b=b+a and ab=ba. This is called the commutative law. The first example is of the existence of the law in addition, and the second of its existence in multiplication. The commutative law in addition may be thus expressed :A sum is independent of the order of its addends; and in multiplication-A product is independent of the order of its factors. Again, because 2(3+4)=2.3+2.4, we say that a(6b+c)=ab+ac. This is called the distributive law and may be stated thus :—The product of multiplying a factor by the sum of several terms is equal to the sum of the products arising from multiplying the factor by each of the terms. These two are the only laws which need be here mentioned. And any science which is to employ the forms of Algebra |