To draw an equilateral triangle whose area shall be that of two given equilateral triangles. Let the given triangles be A and B. Draw a line D E equal in length to one side of the larger triangle, and upon one end thereof, and at a right angle therewith, erect a line equal in length to one side of the less triangle, as D C; then draw the line C E, which will be one leg of the triangle required. The equilateral triangle CEF contains an area equal to the triangles A and B. NOTE. The same process is applicable to rectangular figures. D B To draw a circle whose area shall be that of two given circles. Let the circles A and B be the given circles. Draw a line whose length shall be equal to the diameter of the larger circle, and upon the end thereof erect a perpendicular equal in length to the diameter of the lesser circle, as C D E; draw the line E C, and bisect it in i, and with the radius i C, or i E, and i as the centre, describe the circle ED F, whose area will be that of the two given circles. A To construct a toothed or cog wheel. B Divide the pitch circle, a a, into as many equal parts as there are to be teeth cr cogs; then, with the dividers extended to 14 times one of those parts, and the points as a centre, describe the arcs i v, v i; then, with the same radius, and r and t as centres, describe the arcs vƒ and ƒ v, so continuing until the upper sections of all the cogs are defined. The lower sections are bounded by straight lines tending to the centre of the wheel. See TEETH of WHEELS. NOTE. -The pitch of a wheel is the rectilinear distance from the centre of one cog to the centre of the next contiguous, measured upon the pitch circle; and that portion of the length of a tooth lying between the lines a and b is usually made equal to the pitch; and that portion lying between the lines a and c is usually made equal to the pitch. RULE. H D E H I Rectangle. Rhombus. Rhomboid. To find the area of either the above figures. - Multiply the length by the breadth or perpendicular height, and the product is the area. EXAMPLE. - How many square feet in a floor whose form is a rhomb, or rhombus, each side being 12 feet, and the perpendicular or right-angular distance from the side F G to the side H I being 8 feet? 12 X 896 square feet. ́ ́ Ans. OF TRIANGLES. To find the area of a triangle. RULE. Multiply the base by half the perpendicular height, or the perpendicular height by half the base, and the product is the area. Or, multiply the base by the altitude and divide the product by 2, and the quotient will be the area. EXAMPLE. The base, A C, of the triangle A C B, is 12 feet, and the altitude, D B, is 4 feet and 6 inches; required the area. B A D 4.5 X 12 =27 square feet. Ans. To find the area of a triangle by means of its sides. RULE. Add the three sides together, and from half the sum subtract each side severally; multiply the half sum and the three remainders into each other, and from the product extract the square root, which will be the area sought. The sides of a triangle are 30, 40, and 60 rods. What area has the triangle? 30+40 +60 ÷ 2: = 65 sum of the three sides. 65 30=35, first remainder. Then, 65 X 35 X 25 X 5= √284375 533.26+ square rods. Ans. To find the hypotenuse of a triangle, the base and perpendicular being given. RULE. - Add the square of the base and square of the perpéndicular together, and the square root of the sum is the hypotenuse. EXAMPLE. The distance from the base of a building to the sill of an attic window-perpendicular of the triangle, B C-is 40 feet; what must be the length of a ladder, hypotenuse of the triangle, A C,- placed on a level with the base of the building, and 12 feet therefrom, base of the triangle, A B,- to reach to the sill of said window? 402 + 122 = √1744 = 413 feet. Ans. A B The hypotenuse and one of the sides of a right-angled triangle being given, to find the length of the other side. RULE. Add the hypotenuse and the given leg together, multiply the sum by the difference of their lengths, and the square root of the product is the side sought. EXAMPLE. The sill of a window in a building standing on the edge of a stream is 30 feet above the water, and a line which has been extended therefrom directly across the stream to the opposite shore is found to measure 80 feet; required the width of the stream at that place. 80 3050, and 80 + 30 × 50= √550074.16 feet. Ans. To find the height of the inaccessible object C, or the length of the perpendicular B C. RULE. Upon a plane, at a right angle to the base of the perpendicular, as A B, erect, at any convenient distance from the base, a perpendicular staff D E, and construct the hypotenuse A E in the direction A C; then, as the base A D is to the perpendicular staff D E, so is the base A B to the height B C. A D B EXAMPLE. The perpendicular DE, of the right-angled triangle ADE, being 3 feet, and the base, A D, 5 feet, what is the perpendicular of a similar triangle, A B C, whose base, A B, is 100 feet? 5:3: 100 = 60 feet. Ans. To find the distance from a given point, B, to an inaccessible object, as at A. RULE. Draw the line B in the direction A, and from the point B, at a right angle to B A, draw, to any convenient length, the line B C; form a right angle to B C, from the point C, and convert into a triangle by carrying the hypotenuse, D E, to the line B C, in the direction D A; then, as the distance C E is to the distance C D, so is the distance B E to the distance B A. EXAMPLE. The distance from C to E is 4 rods, the distance from C to D, 15 rods, and the D distance from E to B, 22 rods, whereby required the distance from B to A. RULE. 4 15: 22 = 83 rods. Ans. OF TRAPEZOIDS AND TRAPEZIUMS. To find the area of a trapezoid. Multiply the sum of the two parallel sides by the perpendicular distance between them, and divide the product by 2; the quotient is the area. trapezoid A B A B EXAMPLE. The side A B, of the CD, is 48 feet, the side C D is 72 perpendicular distance between the sides is 40 feet; how many feet area has the figure? 724851203 X 404890322445 ft. 2 in. Ans. To find the area of a trapezium. RULE. -Draw a diagonal through the figure, whereby it is converted into two triangles, and multiply the length of the diagonal by half the sum of the altitudes of the triangles; the product is the area sought. EXAMPLE. The diagonal A D, in the trapezium A B C D, is 54 rods in length, and the altitudes of A the triangles formed by the introduction of the diagonal are 20 rods and 26 rods; required the area of the figure. C 2620223 × 54 = 1242 square rods. Ans. B OF POLYGONS. To find the area of a regular polygon. RULE. - Multiply the length of a side by half the distance from the side to the centre, and that product by the number of sides; the last product will be the area of the figure. EXAMPLE. -The side A B of a regular hexagon is 12 inches, and the distance therefrom to the centre of the figure, d c, is 10 inches; required the area of the hexagon. 10 X 12 X 6 = 360 sq. in. Ad B = 2 sq. feet. Ans. TABLE of angles relative to the construction of Regular Polygons with the aid of the Sector, and of co-efficients to facilitate their construction without it; also, of co-efficients to aid in finding the area of the figure, the side only being given. NOTE." Angle at centre" means the angle of radii, passing from the centre to the circumference, or corners of the figure. "Angle at circumference" means the angle which any two adjoining sides make with each other. Every circle contains its own radius, as a chord line, exactly six times; therefore, To describe a polygon with the aid of the sector. RULE. - Take the chord of 60° on the sector, and describe a circle; then, with the chord, (on the same line of the sector,) of as many degrees as indicated in the TABLE, for the respective polygon-column, "angle at centre space off the circle, and each space will be the side of the polygon required. Thus, for a decagon, take the chord of 60° on the sector for the radius of the circle, and the chord of 36° on the same line of the sectors for a side. |