if AD be trisected in the points K and M, and through E and F |s be drawn to AD, and through K and Ms be drawn to AB, the figures 1, 2, 3, 4, 5, 6, 7, 8, 9 are all squares equal to one another. K A E F B 1 4 D 2 3 7 P 5 6 M Now, if AB be taken as u.l., AC is the u.a. ; and if AE be taken as u.l., any one of the small squares, as AP, is the u.a. And the segment AB contains AE 3 times, while the square AC contains the square AP in three rows with three in each row, or 32 times. .. if any assumed u.l. be divided into 3 equal parts for a new u.l., the corresponding u.a. is divided into 32 equal parts for a new u.a. And the least consideration will show that this is true for any whole number as well as 3. .. 1. If an assumed u.7. be divided into n equal parts for a new u.l., the corresponding u.a. is divided into n2 equal parts for a new u.a.; n denoting any whole number. Again, if any segment be measured by the u.l. AB, and also by the u.l. AE, the measure of the segment in the latter case is three times that in the former case. And if any area be measured by the u.a. AC, and also by the u.a. AP, the measure of the area in the latter case is 32 times its measure in the former case. And as the same relations are evidently true for any whole number as well as 3, .. 2. If any segment be measured by an assumed u.l. and also by th of the assumed u.l. as a new u.l., the measure of η the segment in the latter case is n times its measure in the former. And if any area be measured by the corresponding u.a.s the measure of the area in the latter case is 2 times its measure in the former case; n being any whole number. This may be stated otherwise as follows: L N 8 9 G H с By reducing an assumed .. to th of its original length, n we increase the measure of any given segment ʼn times, and we increase the measure of any given area n2 times; n being a whole number. 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. 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. I. Let the measures of the adjacent sides with respect to cthe unit adopted be whole numbers. B Let AB contain the assumed u.l. a times, and let AD contain it b times. Then, by dividing AB into a equal D parts and drawing, through each point A of division, lines || to AD, and by dividing AD into ¿ 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 x u.l.s in AD. We express this relation more concisely by writing symbolically AC AB. AD, where AC means "the number of u.a.s in AC," 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 of the new th of the adopted unit, and let AB contain n units, and AD contain q of them. The measure of AC in terms of the new u.a. is pq (152°, 1), and the measure of the AC in terms of the adopted unit is pq (151°, 2) n2 But the measure of AB in terms of the adopted u.l. is 2, and of AD it is ?. (151°, 2) n Illus. Suppose the measures of AB and AD to some unit-length to be 3.472 and 4.631. By taking a u.l. 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 × 4631 or 16078832; and dividing by 10002, the measure of the area with respect to the original u.l. is 16.078832=3.472 × 4.631. 3. Let the adjacent sides be incommensurable. There is now no u.l. that will measure both AB B and AD. EHD If AC is not equal to AB. AD, let it be equal to AB. AE, where AE has a 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 but and □BH=AB. AH, BD=AB. AE; OBH is <BD; AB. AH is < AB. AE, and AB being a common factor (hyp.) AH is < AE; which is not true. it .. If □AC=AB. AE, AE cannot be < AD, and similarly may be shown that AE cannot be > AD; .. AE=AD, or AC AB. AD. q.e.d. 153°. 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°) 2. The area of a triangle is one-half the product of its base and altitude. (141°) 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 E D Similarly, P C OD=OE=OF=r and O'P O'P' O'P"=r'; A :: ▲ABC=▲AOB+^\BOC+▲COA AB. OF+BC. OD+CA.OE =rx perimeter=rs, where s is the half perimeter ; Δ =rs. Ex. 2. ABC=AO'B+AO'C-BO'C where' is the radius of the ex-circle to side a ; ▲=r'(s-a). (153°, 2) O'P. AB+O'P'. AC-O'P". BC 2. r258=(s—a)(s—b) (s— c) (r''q''' + y'll gol +g'g''). 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.? the side is taken as the u.l.? 5. What is the measure of the diagonal of a square when (150°, 5) 6. Find the measure of the area of a circle when the diameter is the u.l. When the circumference is the u.l. 7. If one line-şegment 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 equilateral triangle has an area three-fourths that of the other. 10. The three medians of any triangle divide its area into six equal triangles. II. 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=3BC, CY=}CA, and AZ=}AB. Express the area of the triangle XYZ in terms of that of ABC. 13. Generalize 12 by making BX-BC, etc. n r 14. Show that a=sI- ·(r'"+r'"'). S |