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2. Again, let ABC and DEF be similar triangles, having the base EF one and three-quarter times the base BC. That is, the base BC is to the base EF as 4 is to 7, since 1:13-4:7. Since the angles are similar, the other sides of DEF are 13 times the corresponding sides of ABC. If AG and DK be the perpendiculars to the bases, the triangles ABG and DEK are equiangular, and, therefore, since DE is 12 times AB, DK is also 1 times AG.
If rectangles be constructed on the bases equal to the triangles, the heights of these rectangles are half the heights of the triangles (Ch. VIII., 5). Hence FN, which is half of DK, is 13 times CL, which is half of AG.
So that the rectangle
EFNP (which is equal to the triangle DEF) is 13 times as long and 12 times as high as the rectangle BCLM (which is equal to the triangle ABC). That is, of such parts as EF contains 7, BC contains 4; and of such parts as FN contains 7, CL contains 4. Hence of such small areas as the rectangle EFNP contains 72=49, the rectangle BCLM contains 42=16. And therefore the triangle ABC is to the triangle DEF as 16 is to 49. That is, when
then, triangle ABC: triangle DEF 16:49=42:72 or 1:(12)2.
3. Make figures as in § 1 and § 2 for the following problems:
Two similar triangles, ABC and DEF, have their corresponding sides BC and EF, 1 and 2 inches in length respectively; show that their areas are as 1 to 4, i.e., as 1 to 22.
Two similar triangles, ABC and DEF, have their corresponding sides BC and EF, 1 and 1 inches in length respectively; show that their areas are as 4 to 9, i.e., as 1 to (1)2.
Two similar triangles, ABC and DEF, have their corresponding sides BC and EF, 30 and 50 millimetres in length respectively; show that their areas are as 9 to 25, i.e., as (30)2 to (50)2.
(For the three preceding constructions, the method of article 4, which follows, should also be employed.) The result of our observations in such cases as the preceding may be stated thus:
Similar triangles are to one another as the squares of corresponding sides.
Note: In the preceding examples it will be observed that the lengths of the corresponding sides are supposed commensurable, i.e., a unit of length can be found that is contained in both an exact number of times. All lines are not commensurable, though the preceding statement in italics is true of all similar triangles, whether the corresponding sides be commensurable or not.
4. The following is possibly a more striking way of presenting the preceding proposition:
Let any side, say the base, of a triangle be divided into as many parts as it contains units of length. Through the points of division draw lines parallel to
1.3+5=32 Small Triangles
the sides, and, through the points of intersection of these lines, draw lines parallel to the base. The triangle is thus divided into a number of triangles equal to one another in all respects, and all similar to the original triangle. It will be observed that, considering these triangles in rows, the rows contain 1, 3, 5, 7, . . triangles, respectively. Hence if the base be 2 units in length, the large triangle contains 1+3=22 small triangles; if 3 units in length, 1+3+5=32 small triangles; if 4 units in length, 1+3+5+7=42 small triangles; and so on. Thus if there be two similar triangles, the base of one containing 3 units of length, and the base of the other 4 units of length, the number of small triangles in one will be 32, and in the other 42, all such triangles being equal to one another. Hence the areas of the triangles are as 32 to 42, i.e., as the squares of the bases.
1. Construct two angles, the sides of one being 36, 48 and 50, and the sides of the other 54, 72, 75 millimetres. On the base of each construct a rectangle equal to it; and divide up the rectangles so as to show that the triangles are as (36)2 to (54)2.
2. Divide the triangles of the preceding question into smaller triangles, all equal to one another. Hence show that the original triangles are as (48)2 to (72)2.
3. Draw two straight lines which are to one another as these triangles.
4. Divide a line 31⁄2 in. in length into two segments, such that, when equilateral triangles are described on the segments, one triangle shall be four times the other.
Construct the equilateral triangles, and divide the greater into four triangles, each equal to the smaller.
5. Construct two triangles on bases of 45 and 75 millimetres, with angles adjacent to each base 70° and 50°. Divide the triangles into smaller ones, all equal to one another, showing that the areas of the triangles are as (45)2 to (75)2.
6. Draw a line AB of length 1 in., and produce it to C so that AB may be to BC as the areas of the two triangles in the preceding question.
7. Describe an irregular pentagon, and, after the manner of § 5, Ch. XX., construct another pentagon with linear dimensions half those of former. Divide each pentagon into three triangles by lines drawn from corresponding angles.
How are the sides and angles of corresponding triangles related? Test with bevel and dividers.
How many times is a triangle in the first pentagon greater than the corresponding triangle in the second?
How many times is one pentagon greater than the other?
8. ABC is any triangle, and in AB a point D is taken such that AD is one-quarter of AB. DE is drawn parallel to BC. What fractional part is ADE of the whole triangle? What ratio does ADE bear to the rest of ABC?
9. Construct an equilateral triangle with sides of 1 in., and construct another with area twice the former.
10. Construct a right-angled triangle with sides 30, 40 and 50 millimetres. On the sides describe equilateral triangles. Divide the triangles into smaller ones, so that the smaller ones may all be equal to one another. What relation do you discover between the area of the triangle on the hypotenuse and the areas of the two other triangles ?
11. In the preceding question, instead of equilateral triangles, construct triangles with angles adjacent to the sides of 50° and 80°, so that the three triangles are similar. Again compare areas of smaller triangles with area of greatest.
12. Any line being taken as unity, construct for other lines which shall represent √2 and 2.
Hence draw lines parallel to the base of any triangle so as to form with sides, or sides produced, triangles half and twice the original.
13. The areas of the following states being, Texas, 265780; New York, 49170; Illinois, 56650; California, 158360; Kansas, 82080; Massachusetts, 8315; South Carolina, 30570 square miles; and the square roots of these numbers being 515, 222, 238, 398, 286, 91, 175, or approximately as 52, 22, 24, 40, 29, 9, 18; construct seven equilateral triangles, all with the same vertex, whose areas shall represent proportionately the areas of these states.
14. Draw also seven parallel lines, near one another, and all terminated at one end by the same straight line to which they are perpendicular, so that these lines may approximately represent the areas of these states.
15. Given the following populations,—Pennsylvania, 6302115; Ohio, 4157545; Missouri, 3106665; Indiana, 2516462; Vermont, 343641; construct five squares, with one angle in common, which shall represent proportionately and approximately the populations of these states.
16. Draw also five parallel lines, as in 14, which shall represent approximately the populations of these states.