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Fig. 98.

Fig. 97.

Fig. 99.

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Next to this in importance, as a geometrical truth, and nearly connected with it, is the proposition—Similar figures have their homologous measures proportional. In the application of the elementary principles, these two propositions are continually used. 175. If bo + c2 = a-, then b% = a” — co, and co = a” b% ; that is, The square of one side of a rightangled triangle, is equivalent to the square of the hypothenuse, minus the square of the other side. 176. In figure 98, let AB = a, and CB = b. Produce AB to E, making BE = b, then AE will be equal to a + b, and AC will be equal to a - b. Upon AB make the square AD, which will be the square of a or a”; and upon AC construct the square AF, which will be the square of a -b, that is (a —— b)”. If we complete the rectangle AH, it will be the rectangle contained by a + b and a b : and as the part M' is equal to the part M of the figure, instead of the rectangle AH, we may take its equivalent AIKFLB ; but this part of the figure is equal to the square IB with the square KL taken from it, that is, equal to a 2 b%; so that the rectangle AH is equivalent to the difference between the square described upon a and the square described upon b, which gives the equation (a + b) × (a b) = ao bo. The rectangle contained by the sum and difference of two lines, is equivalent to the difference of the squares described upon the two lines. 177. The two rectangles CEGD, DBFG, (fig. 97) of the same height are in the ratio of their bases CD and DB (155); but they are equivalent respectively to the squares of the two sides, CA and AB, to which they are adjacent; consequently—The squares of the two sides of the rightangled triangle, are to each other in the ratio of the adjacent segments of the hypothenuse. . 178. Suppose that the angle A, instead of being a rightangle, is an obtuse angle (fig. 99); produce the side CA till it meetin D, the perpendicular BD, drawn from the opposite vertex; denote the perpendicular by y, and the distance AD by a. The right-angled triangle ABD (172 gives the equation c2 = y2 + 2* ; and the triangle CB gives a” = y2 + (2 + b)* = y2 + 2* + b% + 2 (b X a.) (170). And substituting c” for its value, we have a* = c2 + b% + 2 (b > v). We therefore say—In an obtuseangled triangle, the square of the side opposite the obtuse angle, is equivalent to the sum of the squares of the other two sides, plus twice the rectangle contained by one of those sides and the distance of the verter of the obtuse angle from the foot of the perpendicular drawn from the opposite vertex to this side produced. 179. If the angle A be acute, the perpendicular BD (fig. 100), will fall within the triangle. Designating AD by r, and BD by y, we have (CD)* = (b r)* = b% -- a 2–2 (0 × 2), (171); c3 = y2 + 2* ; a” = y2 + (CD)* = y2 + ba + a 2 – 2 (b > v); and substituting co for its value, we have a2 = c^ + b” 2 (b X r); that is –In an oblique-angled triangle, the square of the side opposite an acute angle, is equivalent to the sum of the squares of the other two sides, minus twice the rectangle contained by one of the sides adjacent to this angle, and the distance of this angle from the foot of the perpendicular drawn to this adjacent side from the opposite vertex. 180. In figure 99, (170) we have (CD)2 = z* + bo + 2 (0 × 2); and subtracting a 2 from each side, we have (CD)*—wo = b% + 2 (b > v). We had also on the same figure, a * = c^ -- b2 + 2 (0 × 2); subtracting co from each side we have a 2 c’ = b% + 2 (b > v); therefore a” — co- (CD)2 — a 2. *~.. In figure 100 (179) we have (CD)2 = x2 + ba 2 (b > v); therefore (CD)2 — 22 = 03 2 (0 × w); but we had in the same article a* = c^ -i- b” 2 (b > v); which gives ao co-bo 2 (b X w.); therefore a” —co = (CD) * a 2. We therefore say—If from the vertex of either of the angles of any triangle, a perpendicu!ar be drawn to the opposite side (produced if necessary) the difference of the squares of the sides containing this angle, will be equivalent to the difference of the squares of the two distances of the foot of this perpendicular from the other vertices of the triangle. S 181. If we would find the side of a square equivalent to two given squares; we have only to make a rightangled triangle, the two sides of which are respectively equal to the sides of the given squares; the hypothenuse will be the side of the square sought (172). If we would find the side of a square equivalent to the sum of several squares, as a” bo + c + d”, &c. (fig. 101); construct a right-angled triangle ABC, whose two Fig.101. sides are a and b, the square of the hypothenuse (m”) will be equivalent to the two squares a” and b*; then make

Fig.102.

Fig.104.

another right-angled triangle with the two sides on and c; the square of the hypothen use (n°) will be equivalent to the sum of the first three squares; that is, = m2 + c2;

therefore = a” + l2 + c2; or n = (do –– U2 + co); ;
proceed in the same way with any number of squares.
1S2. If we would find a square equivalent to the dif-
ference between two given squares, a” and b2 (fig. 102);
draw a line AC, equal to the side of the smaller of the
two squares, b%; at A draw the indefinite line AB, per-
pendicular to AC ; then from C as a centre with a radius
equal to the side of the larger square, draw an arc cutting
the perpendicular in the point B ; BA will be the side of
the square sought; for we have co = d” b%; c =
(a” bo #.
IS3. If from the summit of any triangle, as ACB

. (fig. 103), we draw a straight line, as CE, to the middle

of the base, it will divide the triangle into two triangles,
one of which (if the sides CA and CB are unequal) will be
an obtuse-angled triangle, and the other an acute-angled
triangle. If we draw the perpendicular CD, and designate
the line CB by a, AC by b, AE and EB each by , c, ED by
ar, and CE by m ; the triangle ACE will give the equation
b°= m” -- (#c)? --2 (30 × a ), (178); the other trian-
gle gives the equation a” = m^ -- (4 c)? — 2 (; c X w.),
179); adding these two equations, we have ao –– U2 =
2 m2 + 2 (; c)”; that is—If a line be drawn from the
vertex of any angle of a triangle to the middle of the op-
posite side, twice the square of this line, added to twice
the square of half this opposite side, will be equivalent to
the sum of the squares of the two sides which contain this
angle.
184. If in the parallelogram ACBD (fig. 104) we
draw the two diagonals, they will bisect each other at E;
and according to the last article, the triangle ACB gives
b2+ a” = 2 (AE)2 + 2 (CE)”;
c2 + d2 = 2 (AE)2+2 (DE)”;
adding now the two equations, and considering that CE
equals DE, we have a2+ b” + c2 + d” = 4 (AE)” + 4
(CE)”. But 4 (AE)” is the square of 2AE; that is, the
square of AB; and 4 (CE)” is the square of 2 CE, the
square of CD. The equation, therefore, becomes
a2 + 52 + c2 + do = (AB)2 + (CD)”. That is—The
sum of the squares of the four sides of a parallelogram,

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is equivalent to the sum of the squares of the two diagomals. 185. A regular polygon (fig. 76) is composed of a Fig. 76. certain number of equal isosceles triangles, whose bases make the perimeter of the polygon, and whose common height C a is the apothegm of the polygon. The area of each of these triangles is the product of its base into half the height; therefore–The area of a regwlar polygon is expressed by the product of its perimeter multiplied by half its apothegm ; that is, by half the radius of the inscribed circle. 1S6. As two regular inscribed polygons of the same number of sides are composed each of the same number of equal triangles; and as the triangles in the one polygon are similar to those of the other, these similar trian* * * * * * BOC (BO)2 gles (fig. 105) will give the proportion 7. = % F Fig.105. (ON)2. 6(BOC) (BO)2 (ON)2 (on)? ' Or 6(b o c) T (bo)* T (on)? BOC is the larger polygon, and six times b o c is the smaller polygon ; BO, b 0, are radii of the circumscribed circles; and ON, on, are radii of the inscribed circles. We therefore say—The areas of regular polygons of the same number of sides, are to each other as the squares of the radii of the circumscribed circles, and also of the inscribed circles. 187. It is manifest that as we increase the number of sides in the inscribed regular polygon, we increase its area, which may approach indefinitely near to that of the circle, but can never exceed it. And as we increase the number of sides in the regular circumscribed polygon, we diminish its area, which, though it may approach indefinitely near to that of the circle, can never be less. We have also seen (148) that as the number of sides of the inscribed and circumscribed polygons are increased, their perimeters approach in value to that of the circumference. If we suppose that the number of sides of the circumscribed polygon is so great that the excess of its perimeter over the circumference is less than any assignable magnitude, which excess we denote by d: the area of the polygon will exceed the area of the circle, which we express by A, by an exceedingly small quantity, which

But six times

we denote by m. If we denote the circumference by C. the perimeter of the polygon will be expressed by C+ d . the radius of this circle, which we denote by R, will evidently be the apothegm of this circumscribed polygon, and its area will be expressed by A -i- m = 4 IR × (C+ d).

The perimeter of the inscribed polygon of the same number of sides will be less than the circumference of the circle by an indefinitely small magnitude d'; and the area of this polygon will be less than that of the circle by an indefinitely small magnitude m'. The apothegm of this inscribed polygon will be less than the radius of the circle by an indefinitely small magnitude a . To express the area of this inscribed polygon, we have the equation A m/ = 4 (I& a y X (C– 6"). The area A + m is greater than the area of the circle by an indefinitely small quantity, and the product R × (C+ d) which expresses this area, is greater than , R × C. The area A m/, is less than the area of the circle by an indefinitely small quantity, and the product , (fo a J × (C d') which expresses this area, is less than * R × C. If then, any area greater than the area of the circle requires for its expression a product greater than . It X C; and if any area less than the area of the circle requires for its expression a product less than * R × C.; it manifestly follows that the product #8 × C. cannot be the measure of an area greater or less than that of the circle whose radius is R and circumference C; it must therefore be the measure of an area equal to the area of this circle. Ilence we say—The area of a circle has for its measure half the product of radius multiplied by the circumference. That is—A circle is equivalent to a rectangle whose base is equivalent to the entire circumference, and whose height is equal to half radiots.

188. Having a rectangle equivalent to a given circle, a mean proportional between its two adjacent sides, will be the side of an equivalent square.

To find the side of a square equivalent to a circle of a given diameter, constitutes the hitherto unsolved problem of the quadrature of the circle. This problem, it will be seen, depends upon another, that is, to find the circumference of a circle, the radius or diameter being given, or in other words, to find the definite ratio of the circumference of a circle to its diameter. This is called—The definite rectification of the circumference of a circle.

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