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therefore to CD), somewhere in the same line as CD. Then, since CD is equal to E F, the points E and F are equally distant from C and D respectively, and it follows that the triangles B C E and AD F are equal. Hence if the triangle B C E were cut off the parallelogram along B C and placed in the position A D F, we should have converted the parallelogram into the rectangle without changing its area. Thus the area of the parallelogram is equal to that of the rectangle. Now the area of the rectangle is the product of the numerical quantity which represents the length of AD into that quantity which represents the length of A B. AB is termed the base of the parallelogram, and A D, the perpendicular distance between its base and the opposite side E F, is termed its height. The area of the parallelogram is then briefly said to be the product of its base into its height.'

Suppose CD and A B were rigid rods capable of sliding along the parallel lines c d and a b. Let us imagine them connected by a rectangular elastic membrane, A B C D; then as the rods were moved along a b and c d the membrane would change its shape. It would, however, always remain a parallelogram with a constant base and height; hence its area would be unchanged. Let the rod A B be held fixed in position, and the rod CD pushed along cd to the position E F. Then any line, GH, in the membrane parallel and equal to A B will be moved parallel to itself into the position IJ, and will not change its length. The distance through which c has moved is CE, and the distance through which G has moved is G I. Since the triangles CBE and G BI have their sides parallel they are similar, and we have the ratio of CE to GI the same as that of BC to BG; or, when the rectangle ABCD is converted into the

parallelogram A B E F, any line parallel to A B remains unchanged in length, and is moved parallel to itself through a distance proportional to its distance from A B. Such a transformation of figure is termed a shear, and we may consider either our rectangle as being sheared into the parallelogram or the latter as being sheared into the former. Thus the area of a parallelogram is equal to that of a rectangle into which it may be

sheared.

The same process which converts the parallelogram ABEF into the rectangle ABCD will convert the triangle AB E, the half of the former, into the triangle

D

F

FIG. 37.

A B C, the half of the latter. Hence we may shear any triangle into a right-angled triangle, and this will not alter its area. Thus the area of any triangle is half the area of the rectangle on the same base, and with height equal to the perpendicular upon the base from the opposite angle. This height is also termed the altitude, or height of the triangle, and we then briefly say: The area of a triangle is half the product of its base into its altitude.

A succession of shears will enable us to reduce any figure bounded by straight lines to a triangle of equal area, and thus to determine the area the figure encloses by finally shearing this triangle into a right-angled

triangle. For example, let ABCDE be a portion of the boundary of the figure. Suppose AC joined; then shear the triangle A B C so that its vertex B falls at B' on D C produced. The area A B' c is equal to the area Hence we may take A B'D E for the boundary of our figure instead of A B C D E; that is, we have reduced the number of sides in our figure by one. By a succession of shears, therefore, we can reduce any figure bounded by straight lines to a triangle, and so find its

A B C.

area.

§ 9. Of Circles and their Areas.

One of the first areas bounded by a curved line which suggests itself is that of a sector of a circle, or the

B

FIG. 38.

portion of a circle intercepted by two radii and the arc of the circumference between their extremities. Before we can consider the area of this sector it will be necessary to deduce some of the chief properties of the complete circle. Let us take a circle of unit radius and suppose straight lines drawn at the extremities of two diameters AB and CD at right angles; then the circle will appear as if drawn inside a square (see fig. 39). The sides of this square will be each 2 and its area 4.

.

Now suppose the figure composed of circle and square first to receive a stretch such that every line

parallel to the diameter A B is extended in the ratio of a 1, and then another stretch such that every line parallel to CD is again extended in the ratio of a 1. Then it is obvious that we shall have stretched the square of the first figure into a second square whose sides will now be equal to 2 a.

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It remains to be shown that we have stretched the first circle into another circle. Let op be any radius and PM, PN perpendiculars on the diameters A B, C D. As a result of the first stretch the equal lengths oм and N P are extended into the equal lengths o' м' and

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as a result of the second stretch MP and oN, which remained unaltered during the first stretch, are con

verted into M' P' and O'N'; so that

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During this second stretch o'M' and N' P' remain unaltered. Thus as the total outcome of the two stretches we find that the triangle O P N has been changed into the triangle o' P'N'. Now these two triangles are of the same shape by what was said on p. 106, for the angles at N and N' are equal, being both right angles, and we have seen that-

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Thus it follows that the third side OP must be to the third side o' p' in the ratio of 1 to a; or, since op is of unit length, o'p' must be equal to the constant quantity a. Further, since the angles PON, P′O'N' are equal, o' p' is parallel to op. Hence the circle of unit radius has been stretched into a circle of radius a. In fact, the two equal stretches in directions at right angles, which we have given to the first figure, have performed just the same operation upon it, as if we had placed it under a magnifying glass which enlarged it uniformly, and to such a degree that every line in it was magnified in the ratio of a to 1.

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It follows from this that the circumference of the second circle must be to that of the first as a is to 1. Or, the circumferences of circles are as their radii. Again, if the arc PQ is stretched into the arc P'q'-that is, if o' p', o'q' are respectively parallel to o P, o Q-then the arc P'Q' is to the arc P Q in the ratio of the radii of the two circles. Since the arcs PQ, P'Q' are equal to other arcs which subtend the same angles at the centres of their respective circles, we state generally that the arcs of two circles which subtend equal angles at their respective centres are in the ratio of the corresponding radii.

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Since the second figure is an uniformly magnified image of the first, every element of area in the first has been magnified at the same uniform rate in the second. Now the square in the first figure contains four units of area, and in the second figure it contains 4 a2 units of area. Hence every element of area in the first figure has been magnified in the second in the ratio of a2 to 1. Thus the area of the circle in the first figure

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