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4. Diagrams such as the

E following should be constructed with accuracy, where DE is parallel to BC, and therefore the triangles ABC and ADE similar. AB, BC and AD should then be measured, and DE calculated from the proportion DE BC


x AD,

AB and the accuracy of the construction, measurements and calculation tested by measuring DE with the dividers and scale.

5. The proportionality of the sides of similar triangles may be employed to reduce or enlarge a figure to any scale.


or DE


as ABC

Suppose we wish to obtain a figure the same shape

. , but with linear dimensions half those of ABC .. Take a line 0AA, with OA'=A'A. From O draw a number of lines OA, OB, ... With the parallel rulers obtain B', through A'B' being parallel to AB; also C', through A'C' being parallel to AC;


also D', through A'D' being parallel to AD; and so on. Then, with the judgment of the eye, fill in the contour between A and B similarly to that between A and B; between B' and C similarly to that between B and C; and so on. Any two points in the larger figure should be just twice as far apart as the two corresponding points in the smaller, and this may be used to test the accuracy of the drawing.

Maps may, in this way, be reduced or enlarged, the first drawing being obtained by using translucent paper, or by tracing against a window pane. Of course the drawing of all maps is, in part, a question of the construction of similar figures.



1. Draw any line AB and divide it in the ratio of 7 to 8 by drawing another line ACD, inclined to AB at any angle, such that AC=28 and CD= 32 millimetres, completing construction with parallel rulers. Verify result by measuring segments of AB.

2. Divide a line 4 in. long in the ratio 3.4 to 4.1.

3. There are three lines of lengths 27, 39 and 64 millimetres. Construct geometrically for a fourth proportional to them, and verify result by calculation and measurement.

4. A line is 45 in. in length. Divide it into three parts, such that they shall be to one another as 7:8:9.

5. Draw a line AB an inch long. Draw another line AC of length 50 millimetres, inclined to former at any angle. Divide the inch line into tenths.


6. Divide an inch into twelfths.

7. Draw AB, AC, making an angle of 47° with one another. In either of them take a point P and drop a perpendicular PN on the other. Measure the lengths of the sides of APN, and obtain the numerical values of the following ratios to two decimal places,

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(Most accurate results will be obtained by taking P at some distance from A, and measuring in millimetres.)

8. Take other line, at different dis nce from A, ma similar construction, measure sides of APN, and again find, to two decimal places, the values of the above ratios for 47°.

9. Calling the side opposite 47° the perpendicular, the side opposite the right angle the hypotenuse, and the remaining side the base, whether it be on the upper or lower line, are the above ratios, i.e.,



and hyp. hyp.

base always the same for 47°, or do they depend on where the point P is taken ?

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10. With the explanation in the preceding question, find the values of these same ratios



and hyp. hyp.

base for an angle of 63°, to two decimal places.

11. It is required to find the distance of a point C from an object B on the other side of a chasm. For this purpose a line CA is run at right angles to BC. AC is found to be 278 feet, and the angle to A to be 47°. What is the distance of B from C?


12. In the preceding question, if AC be 344 feet, and the angle at A be 63°, what is the distance of B from C? Find also the length of AB.

13. To find how far a distant object C is from A, a base line AB is measured of 400 ft. and the angles at A and B are found to be 75° and 80°. Then on paper a line DE of length 3 in. is drawn, and angles EDF, DEF are constructed of 75° and 80°, respectively,-and FD is measured in inches and fractions of an inch. What, ap. proximately, is the length of CA?

14. If, in the preceding question, AB be 250 feet, and the angles at A and B be 65° and 77°, respectively, by constructing a similar triangle on paper and measuring the sides, determine approximately the distances AC and BC.

Find ap

15. In triangle ABC, AC=372 feet, A=48°, C=90°. proximately the length of BC, having previously found for 48° the CHAPTER XXI.

perp. ratio

base 16. Draw an irregular quadrilateral, and construct another of same shape and with linear dimensions half those of former. Verify equality of corresponding angles, and ratio of sides and of diagonals.

17. Draw an irregular pentagon, and construct another of same shape and with linear dimensions one-third those of former. Verify equality of corresponding angles, and ratio of sides and of diagonals.

18. Make an outline map of the state of Michigan with linear dimensions half or twice those in map of United States given in your atlas. Verify correctness by finding ratio of distance between pairs of corresponding points.

19. Make a map of the Mississippi and Ohio rivers from Quincy, Ill., and Cincinnati to Memphis, half or twice the size of that given in your atlas. Test correctness by finding ratio of distances between pairs of corresponding points.

20. Construct a triangle with sides 50, 30 and 48 millimetres. Bisect the angle opposite the last side. In what ratio are the segments into which this bisecting line divides this side? Does the same ratio exist elsewhere in the figure ?

Similar Triangles. (Continued).


1. Let ABC and DEF be similar triangles, having the base EF three times the base BC. The other sides of DEF are therefore three times the corresponding sides of ABC. If DK and AG be the perpendiculars to the bases, the triangles ABG and DEK are equiangular, and therefore, since DE is three times AB,


F DK is also three times AG. If rectangles be con


N structed 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 three times CL, which is half of AG.

So that the rectangle EFNP (which is equal to the triangle DEF) is three times as long and three times as high as the rectangle BCLM (which is equal to the triangle ABC). Hence the rectangle EFNP is nine times the rectangle BCLM, and, therefore, the triangle DEF is nine times the triangle ABC. That is, when

side BC: side EF=1:3, then, triangle ABC: triangle DEF=1:3”, the triangles being, of course, similar. .

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