25°. Given the hypothenuse of a right angled triangle = 35 rods, and the side of the inscribed triangle. square = 12 rods, to determine the Ans. base = 28, or = 21; perpendicular = 21, or = 28. SECTION FOURTH. Comparison of Plane Figures. PROPOSITION I. Rectangles are to each other as the products of their di- (143) mensions. We will take the general case at once, and suppose the containing sides, a, a', of the rectangle A are incommensurable with b, b', those of B. Let b be increased by z, and b' by x', so as to make b+x commensurable with a, and b'+x' with a'. Suppose, for instance, that while a is divided into m parts, that b+x contains n of the same parts, or that (2) A B Fig. 31. and that, while a' is divided into m' equal parts, which may be different from those of a, b'+x' contains n' of the same parts, or that = Through the points of division draw lines parallel to the sides of the rectangles and denote the part added to B by X. Then will the partial rectangles be all equal (95); and the number constructed on the base a being to m, and the number of tiers corresponding to the divisions in a', n', we have mm' for the total number of partial rectangles in A; so the rectangle B+X contains nn' of the same. Hence (24) but = since a, a'; b+x, b' + x', are as their measures m, m' ; n, n' ; whence or B+X_(b+x) (b' + x') __ bb' + bx' + b'x+xx' A = B X bb' bx' + b'x + xx' aa' + + aa' απ' RECTANGLE AND TRAPEZOID. 91 For it is obvious that X depends for its value upon x, x', decreasing as these decrease so as to become nothing if these were to become = 0, and that x, x', being less than the parts into which a, a' are divided, may be made less than any assignable quantity, by sufficiently increasing the points of division. Therefore may be made less than any assignable quantity. Cor. 1. The rectangle is measured by the product of its (144) dimensions. For suppose A to become a measuring square, or that while A = 1 is taken as the unit of surface, a = a' = 1 becomes the linear unit, we have Fig.312. Cor. 2. The right angled triangle is measured by half (145) the product of its base into its altitude. Thus T = +bh. འ་ Fig. 313. PROPOSITION II. The Trapezoid is measured by half the sum of its par- (146) allel sides multiplied into their perpendicular distance. Let a, b, be the parallel sides and from their extremities drop the perpendiculars h, h, on a, b, produced, forming the rectangle X+Trp+Y, Trp α Fig. 32. composed of the trapezoid Trp and the right angled triangles X, Y, having the bases x, y. We have (144), (145), X+ Trp+Y= (x+a) • h=2(x+a)•h Cor. 1. The parallelogram is measured by the product (147) of its base into its altitude. For, when b = a, the trapezoid becomes a trapezium or parallelogram, Z P Fig. 322. Cor. 2. The triangle is measured by half the product of (148) its base and altitude. For becomes Trp = (a + b) • h, Fig. 323. Cor. 3. Parallelograms are to each other, and triangles (149) are to each other, as the product of their bases and altitudes. Cor. 4. Parallelograms of the same or equal altitudes are (150) to each other as their bases. Cor. 5. Triangles of the same or equal altitudes are to (151) each other as their bases. Fig. 324. Cor. 6. Parallelograms of the same or equal bases are to (152) each other as their altitudes. Cor. 7. Triangles of equal bases are to each other as (153) their altitudes. Cor. 8. Parallelograms of equal bases and altitudes, are (154) equivalent. W Fig. 325. Cor. 9. Triangles of equal bases and altitudes, are equiv- (155) alent. X Fig. 326. Cor. 10. Parallelograms, or triangles, between the same (156) parallels, and on the same or equal bases, are equivalent. Cor. 11. Conversely, if parallelograms or triangles, stand- (157) ing on equal bases in the same straight line, and having their vertices turned in the same direction, be equivalent, the line of vertices will be parallel to the line of bases. For if the triangles, having the bases a, a, and Vthe perpendiculars p, p', be equivalent, there results B tap = tap', A Fig. 327. whence the line of vertices VV is parallel to the line of bases BB'. Cor. 12. The parallelogram is double the triangle of the (158) same base and altitude. [Compare (147) with (148).] a Fig. 328. PARALLELOGRAM AND TRIANGLE. 93 PROPOSITION III. Two triangles, having an angle of the one equal to an (159) angle of the other, are to each other as the products of the sides about the equal angles. Let the equal angles of the triangles A, B, be made vertical, and join the extremities of the sides a, b, forming the triangle X; then (151) B Χ Fig. 33. Cor. 1. Equiangular parallelograms are to each other (159) as the products of their dimensions (158). . Cor. 2. Similar triangles are to each other as the squares (160) of their homologous sides. For, then we have b' b b = Cor. 3. If figures, resolvable into the same number of (161) similar triangles, be constructed upon the three sides of a right angled triangle, that on the hypothenuse will be equal to the sum of those described upon the other two sides. For we have Fig. 333. Scholium. It is obvious that (132) is but a particular case of this more general theorem. EXERCISES. 1o. Demonstrate (132) by turning two squares into one. Fig. 34. 2o. On the oblique sides of any triangle describe parallelo |