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Show that if two triangles ABC and DEF have B = E
, each ratio being greater than 1, the two triangles are similar. Where does this proof depend on the
condition that the ratios and are greater than 1 ?—In the course of the proof we cut off small equal lengths BW from BA and EY from ED (Fig. 13), and draw WX
parallel to AC, and YZ parallel to DF. The condition that
and are greater than 1, shows that WX > WB and f
YZ YE, and in consequence that the triangles BWX and EYZ are congruent.
30. Now state generally what conditions fix the shape of a triangle:
A triangle is fixed as to shape
(1) by a knowledge of two angles, or
(2) by a knowledge of two ratios of its sides, or (3) by a knowledge of one angle and of the ratio
of the two sides that form this angle, or
(4) by a knowledge of one angle and of the ratio the opposite side bears to another side provided this ratio is greater than unity.
Or, in other words, two triangles are similar (1) if they have two pairs of corresponding angles e qual; or
(2) if they have the ratios of two sides to the third in one triangle equal to the corresponding ratios in the other; or
(3) if they have a pair of corresponding angles equal, and the ratios of the sides forming these angles also equal; or
(4) if they have a pair of corresponding angles equal and the ratio of the side opposite this angle of one triangle to a second side equal to the corresponding ratio in the other triangle, provided these ratios are greater than 1.
35° and a
31. When a triangle is to be made from B value of, for what values of are there two shapes, for what values one, and for what values none?- -We lay down the angle B and put down the point A anywhere on one arm (Fig. 14). We then calculate the value of b,
and with this as radius draw a circle which cuts the other arm BX in C. If is less than 1, this circle cuts BX in only one point, and there is only one shape. If is greater than 1, the circle (if great enough to meet BX at all) cuts it in two points and gives two shapes. Let fall the perpendicular AD or p from A on BX. If the given ratio is equal to 2, the circle cuts BX in only one
point; if it is less, in none; if it is greater than but less
than 1, in two points.
Does this ratio 2 depend on the particular point at which
A was put down on the arm of the angle B?—No, for the triangle ABD is fixed in shape by the angles B and D, namely B = 35° and D 90°. Two triangles fixed by these angles are similar and so have the same value for the ratio P.
Draw a number of such triangles
ABD, measure p and c in each and so find the value
accurately as you can.
-The value is about 0.57.
Tabulate the results as to the possible shapes of a triangle
with an angle B of 35°, and various values of
If ->1, there is one shape,
Ex. 13. Find and tabulate the corresponding results for LB 25°, 30°, 40°, 45°.
32. It is convenient to have a name for this ratio
(Fig. 14) which depends only on the angle B. called the 'sine of the angle B' and written shortly 'sin B'. Thus we have found that sin 35° = 0.57 as nearly as our drawing will give it. A book of mathematical tables gives the result more accurately as the result of calculation; the value of sin 35° taken to 7 decimal places is 0.5735764.
Ex. 14. From the results of Ex. 13 write down the sines of 25°, 30°, 40°, 45°, and beside them show more accurate values from a book of tables.
Ex. 15. Draw four right-angled triangles in which one side is 0.2, 0.4, 0.6, 0.8 times the hypotenuse. From these find the
angles whose sines are 0.2, 04, 0·6, 0·8, and verify your results by reference to the book of tables.
33. Areas of similar figures. Consider again Figs. 1-4, which we now know to be similar. What is the ratio
AB and what is the ratio of the area ABT to the area A'B' A'B'T', that is, how many times does the area ABT contain the area A'B'T'? Pair the four figures in all possible ways, and for each pair give the ratio of the bases and the ratio of the areas.
If two similar triangles have a pair of corresponding sides in the ratio k, find the ratio of their areas.
the similar triangles of Fig. 15 to have
k. Draw the
perpendiculars AM and DN. The areas of the triangles are X AMX BC and x DN× EF. BCkx EF, and AM=kx DN. So that the first area is xkx DNxkxEF; or kxk times x DN× EF. The ratio of the areas is kxk, or with the index notation it is k2. Prove that
AM=kx DN, that is, that
34. If two plans of the playground are made by chain surveying, are they similar? Yes, because each plan is made triangle by triangle, and taking pairs of corresponding triangles in succession we prove their angles equal. Thus the whole figure conforms to our definition of similarity. And if the two plans were made by the measurement of angles only in addition to one base line? The proof that they are similar is much the same.
If the two base lines of the plans, made by either
method, have the ratio k, what is the ratio of the two areas that represent the playground?-Each triangle of one plan is as to area kxk times the corresponding triangle. So that the whole area of the first is kxk times the area of the second. Is the result affected if the playground has a curved boundary ?-By taking triangles enough we can reduce as much as we like the area along the curved boundary with which we cannot deal. So that we can prove the result true to any approximation we like.
35. Volumes of similar solids. If a schoolroom is p cm. long, q cm. broad and r cm. high, what is its volume ?- -It is px qxr cubic centimetres. And if a model of the schoolroom is made, each dimension being k times the original dimension, so that its length, breadth, and height are in centimetres kxp, kxq, and kxr, what is the volume of the model?—It is kxpxkxqxkxr c. cm., or (omitting the signs of multiplication) kpkqkr or kkkpqr, or (with the index notation) k3pqr. What ratio does the
model bear to the room as to volume?—The ratio is k3.
The original room and the model are said to be similar, and in general two solid figures in which corresponding angles are equal and corresponding lines in the same ratio are said to be similar. If this ratio is k, what is the ratio of the volumes? For instance, suppose a model of a mountain to be made, every distance on the model being k times the original distance, and the two being assumed similar. Suppose the mountain cut up into rectangular blocks and the model into corresponding blocks. Then each block of the model is kkk times the corresponding block of the mountain. So that, apart from small pieces left over that do not make rectangular blocks, the ratio is kkk. by cutting out smaller blocks we can make the piece left over as small as we like, so that the ratio of the whole model to the whole mountain is kkk or k3, to as close an approximation as we please.
36. How many conditions fix the shape of a triangle?— Two, as we saw in Art. 30. How many conditions fix the triangle as to size and shape ?-One length is needed in addition, for instance the length of a side. Or considering size and shape together, we saw earlier (chap. V,