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F=B E and A D=B C from the nature of parallelograms. Now if from the whole figure A B C F, we take A FD, there will remain A B C D. Again if from the whole figure we take B E C, there will remain A B E F. But it is an axiom that if equals be taken from the same thing, equivalents will remain. Therefore A B C D=A BEF, and the area of A B C D=A BxB E, this being the measure of A B E F (100).

102. ---The area of any triangle is equal to half the product of its base by its altitude- By the altitude of a triangle we mean a perpendicular let fall from one of the vertices to the opposite side, produced if necessary; and by

the base the side upon which the perpendicular falls. F67 Thus in the triangle A CD (fig. 67) C E is the altitude

and A D the base. Then we say that the area of A C D-half of A DXC E. Why?-Because the triangle A C D=half the parallelogram A B C D of the same base and altitude (84). But the area of A B C D=A DXC E (101)

Therefore the area of A CD=half of A DXC E.

103. --The area of a trapezoid is equal to the product of its altitude by half the sum of its parallel sides, By the altitude of a trapezoid we mean the perpendicular let fall

from one of the parallel sides to the other. Thus in the F68 trapezoid A B C D (fig. 68), C E is the altitude, B C and

A D being the parallel sides. Then we say that the area of ABCD=C Exhalf of (A D+BC). Why?-Because the diagonal. A C divides the trapezoid into two triangles having the same altitude as the trapezoid, namely CË. Now the area of A C D=C Exhalf of A D (102). Also the area of A B C, taking A for the vertex and BC for the base is equal to C Exhalf of B C, since C E=A F (38). Therefore, since the trapezoid is equal to the sum of the triangles, its area must be C Exhalf of A D+C Exhalf of B C or C Exhalf of (A B+B C).

104. –The area of a regular polygon is equal to the pro

duct of its perimeter by half the radius of the inscribed cirF64 cle- Let A B C D E F (fig. 64) be the polygon, and

NP the radius of the inscribed circle. From the centre O draw lines to all the vertices. These will all be equal, being radii of the circumscribed circle. Then the polygon will be divided into as many equal triangles as it has sides Moreover these triangles have for their common altitude

the radius of the inscribed circle, and the sum of their bases is the perimeter of the polygon. Therefore, adding their measures (102), we have for the area of the polygon its perimeter multiplied by half the radius of the inscribed circle.

105. The area of a circle is equal to the product of its circumference by half the radius. This follows directly from the preceding. For the circumference of the circle is the perimeter of a regular polygon of an infinite number of sides, and the radius of the inscribed circle is its own radius.

106. - The area of a sector is equal to its arc multiplied by half the radius. Let C AMB (fig. 69) be the sec-F69 tor. Suppose the arc A M B to be made up of infinitely small straight lines, and radii drawn from C to each of the points. Then the sector would be divided into triangles, the sum of whose bases would be the arc A M B and wbose common altitude would be the radius A C. Therefore the area of the sector, being the sum of the areas of these triangles, is A M B xhalf of C A. If now it were proposed to find the area of the segment formed by the arc À M B and the chord A B, we have only to subtract the area of the triangle C A B from the area of the sector C A MB.

107. To find the area of an irregular polygon—. This may be done in two ways. First, by drawing diagonals as in ABCDE (fig. 58), the polygon is divided into trian-F58 gles which may be measured separately, and the sum of their areas will be the area. of the polygon. Secondlyevery polygon may be converted into an equivalent triangle- Let A B C D E (fig. 70) be the polygon to be F70 measured. Draw the diagonal C E, and through D draw DF parallel to C E to meet A E produced. Then draw CF, and the triangle C E F will be equivalent to C E D. Why ? — Because they have the same base C E and their altitudes are equal, since their vertices F and D are in a line parallel to CE. Consequently, having the same measure, the triangles are equivalent." Then by leaving out C D E and taking C E F in its stead, we have the quadrilateral A B C F equivalent to the pentagon A B C D E. In the same mander we may leave out the triangle A B C and take an equivalent one A G C in its stead. Then the triangle GCÉrthe quadrilateral A B C F=the

pentagon A B C D E. The same process would apply to any number of sides.

Then by finding the area of the triangle, we have the area of the polygon.

108. The square described upon the hypothenuse of a right triangle is equivalent to the sum of the squares described upon

the other two sides—. This is the celebrated proposition, with the discovery of which Pythagoras is said to have been so delighted, that he sacrificed a hundred oxen

to the Muses. We are to prove that the square BCG F71 F (fig. 71) is equivalent to the sum of the squares A BH

L and A C I K; or more briefly that B c2=B A2+a c2, From A, the vertex of the right angle, let fall the perpendicular A D and produce it to E. The square B G is thus divided into two right parallelograms B E and C E. If then we prove that B E=the square A H, and C E=the square A I, the proposition will be demonstrated. Draw the diagonals A F and HC. Thus we have two triangles A B F and H BC. These are equal. Why? --Because the angle A B F=H B C, since each is equal to a right angle plus the angle A B C. Also A B=B H and BF=B C, from the definition of a square. Therefore (53) the triangle A B F=H B C. But H B C=half the square A H since it has half its measure (102), namely H B.xhalf of A B. Again A B F=half of B E, since it has half its measure, namely B Fxhalf of B D. Now if two halves are equivalent, their wholes must be equivalent; that is, the square A H=the oblong B E. In precisely the same manner we might prove that the square A I=the oblong C E. But the two oblongs B E and C E make the square B G. Therefore B G=A H+A I or B C2=A B?+

A c?, which was to be demonstrated. From this equation we have A B2=B C2-A C2 and a c2=B C2 --A B2. Also by extracting the square root, we have B C=(À B2+A c2)Ě, A B=(B 02-A c2)t, and A C=(B C2- -A B2), by which we are enabled in all cases

-to find the third side of a right triangle when the other two are giv

109. – To make a square equal to the sum or the difference F72 of two giren squares—. Suppose A (fig. 72) is the side of

one of the given squares and B that of the other. To make a square equal to their sum, take E D=A, at E erect a perpendicular E F, and take E G=B. Then G D will be the side of the square required. For by the

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preceding proposition G D2=GE?+E D2. Secondly, to make à square equal to the difference of the squares upon A and B, supposing A the greater, make a right angle Ğ E H and take E G=B. Then with G asa centre and a radius equal to A, describe an arc cutting the other side in a point H. E H will be the side of the square required. For (108) E H2=H G2-E G2.

110. - To make a parallelogram equivalent to a given square, and having the sum of its base and altitude equal to a given line. Let C (fig. 73) be the given square and A F73 B the given line. On A B as a diameter describe a semicircle. At A erect a perpendicular A D and make it equal to the side of the given square. Through D draw D E parallel to A B. From E let fall a perpendicular E F. Then A F will be the base and F B the altitude required, for they satisfy both conditions. First A FXF BEF E2=C, that is the parallelogram is equivalent to the square. For (80) we have the proportion A F:FE::FE:FB, whence (63) A FRF B=F E2. Secondly A F+F B=A B, that is the sum of the base and altitude is equal to the given line.

111. - To make a parallelogram equivalent to a given square, and having the difference of its base and altitude equal to a given line— Let the side of the given square be equal to AD (fig. 74) and let the given line be A B. Place F 74 these so as to make a right angle at A and describe a circle upon A B as a diameter. Through D and O draw D F. Then D F will be the base and Ď E the altitude required, for they satisfy both conditions. First D FXD E=D A?, that is the parallelogram is equivalent to the square. For (81) we have the proportion D F:DA::D A:DE, whence (63) D FXD E=D A2. Secondly D F-D E=E F=A B, that is the difference of the base and altitude is equal to the given line.

112. To make a square which shall be to a given square in any given ratio Let G H (fig. 75) be the side of the F 75 given square, and suppose it is required to make a square which shall be to to the given square in the ratio of 3 to 7. Make D C=3 and B D=7. On B C as a diameter describe a semicircle. At D erect the perpendicular D A. Through A and B draw A E=G H. Through E draw E F parallel to B C. Through A and C draw A C to meet E F. Then A F will be the side of the square re

quired. We are to prove that A F2: A E2::3:7. Now (70) we bave A F:A E:: A C:A

B, whence (67) A F2: : A E? :: A C2: A B2. It will be sufficient then to show

that a c2: A B2::3:7. For this purpose we recur to P71 fig. 71. Since A B2=B E (108) and A C2=C E, we

have A B2: A 02::BE:C E. Bút B E=B FXB D and
C E=B FxC D. Hence A B2: A C2::B FXB D:B
FxC D. Leaving out the common factor B F we have
A B2: A 02::BD:C D, that is the squares of the sides of

a right triangle are to each other as the adjacent segments of F75 the hypothenuse“. Therefore (fig. 75) A C2: A B2::3:7.

Consequently A F2: A E2::3:7, which was to be demonstrated. The process would be the same for any other numbers instead of 3 and 7.

113. To find the approximate ratio of the circumference of a circle to its radius or diameter—. This problem, on account of its vast practical importance, has received a variety of solutions. We propose the following as simpler than any we have met with. We have already shown (8) how to find the ratio of two straight lines; but it is obvious from the definition of a curve (10), that we cannot, in the same manner, find the exact ratio between a curve and a straight line, since we cannot find an infinitely small

We can, bowever, approximate to perfect accuracy, just in proportion to the smalldess of the

common measure which we make use of. All this will be F 76 evident from what follows. Let B C (fig. 76) be the side

of an inscribed hexagon. Then B C=the radius. A B (92). Now if we take B C as a common measure of the circumference and radius, it is contained 6 times in the circumference and 1 time in the radius. Accordingly our first approximate ratio is that of 6 to 1. This cannot be very near the truth, because the chord B C differs very perceptibly from the arc BDC. We shall therefore seek a smaller common measure. For this purpose we draw A D perpendicular to the middle of the chord BC, and it will bisect the arc B C in D (28). B D will then be the chord of 1 of the circumference, and its value is found as follows. Calling A B=1 we bave B E=}. Then (108): A E=(À B2—B E2)=(1-1)=0.866. Knowing A E we have D E=AD-AE=1-0.866=0.134. Then

common measure.

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