and a window 60 feet high on the other side: what is the breadth of the street?

Answer: 180 feet, nearly.

(4) Of two ships from the same port, one has sailed 50 leagues due east, and the other 84 leagues due north: what is their distance from each other?

Answer: 97 leagues, nearly.

(5) Find the circumference of a circle whose radius is 6.3662 yards.

Answer: 40 yards, nearly.

(6) If the diameter of the earth be 7912 miles, find the length of a French metre, which is one ten-millionth part of a fourth part of its circumference. Answer: 39.37206 inches, nearly.

tions to the known numerical value of the circumference of a circle whose diameter is 1, and point out which is the nearest.

(8) Prove that

501 + 80 √√10

is a close approxi

240

mation to the semicircumference of a circle whose radius is represented by 1.

THE PRACTICE OF SUPERFICIAL MEASURE.

(1) Parallelogram. The area is equal to the product of the base and the perpendicular altitude.

(2) Triangle. The area is equal to half the product of the base and the perpendicular altitude.

(3) Triangle. From half the sum of the three sides, subtract each side separately: multiply together the half-sum and the three remainders, and the square root of the product will be equal to the area.

(4) Trapezium. The area is equal to half the product of either diagonal, and the sum of the perpendiculars let fall upon it, from the opposite angles.

(5) Circle. The area is equal to the square of the radius, multipled by 3.14159, nearly.

(6) Sector. The area is equal to half the product of the radius and the subtending arc.

(7) Ellipse. The area is equal to the product of the semi-axes, multiplied by 3.14159, nearly.

(8) Hence, the areas of similar plane figures are as the squares of their homologous lineal dimensions.

Ex. 1. Find the area of a triangle whose sides are 18, 24 and 30 poles.

Here, we have according to the directions above, half the sum of the three sides = (18 + 24+ 30) = 36:

whence, the area = 36 × 18 × 12 × 6 = √√46656 = 216 square poles.

Ex. 2. If the radius of a circle be 2 feet, find the side of the square whose area shall be equal to it. The area of the circle 12.56636 square feet, nearly: whence, by Article (194), the side of the required square = √12.56636 = 3.545 feet, nearly.

4 x 3.14159

Examples for Practice.

=

(1) If the sides of a triangle be 16.6, 18.32 and 28.6: find its area.

Answer: 143, nearly.

(2) If the diagonal of a trapezium be 498 yards, and the perpendiculars let fall upon it from the opposite angles be 10.8 and 18.8 yards: what is its area? Answer: 7370.4 yards.

(3) Each side of a hexagon is 24 feet, and the perpendicular upon each side from a certain point within it is 12√3 feet: find its area.

Answer: 864√3 feet.

(4) Find the sides of the squares whose areas are 4970.25 square inches, and 885 square feet.

Answers: 70.5 inches, and 29 feet.

(5) How much must be cut off from a rectangular surface 2 feet broad, to make a square yard?

Answer: 4 feet.

(6) If two acres of land be laid out in the form of a circle, what is its radius?

Answer: 55 yards, nearly.

(7) Find the radius of a circle, whose area is equal to that of a square whose side is 5.317 yards. Answer: 3 yards, nearly.

(8) The semiaxes of an ellipse are 25 and 49: find the radius of a circle of equal area.

Answer: 35.

(9) The base of a triangle is 14.1 yards, and its area is 64.86 yards: find its perpendicular height.

Answer: 9.2 yards.

(10) The side of an equilateral triangle is 6: find its area.

Answer: 15.588, nearly.

(11) The two equal sides of an isosceles triangle are 12 feet, and the base is 8 feet; required its area.

the

Answer: 45.2548 feet, nearly.

(12) Compare the area of a circle with the area of square inscribed in it.

Answer: 3.14159: 2, nearly.

(13) What is the relation between the area of a square, and that of the circle inscribed in it?

Answer: 4 3.14159, nearly.

4:

(14) Required the area of the sector of a circle, whose arc and radius are each 2.57 inches.

Answer: 3.30245 inches.

(15) The radii of two concentric circles are 10 and 12 yards: find the space included between them. Answer: 138.22996 yards, nearly,

(16) The areas of squares, circles, similar parallelograms and triangles, are as the squares of their homologous lineal dimensions.

THE PRACTICE OF SOLID MEASURE.

(1) Parallelopiped. The content is equal to the area of the base multiplied by the perpendicular height.

(2) Prism and Cylinder. The content is equal to the area of the base multiplied by the perpendicular height.

(3) Pyramid and Cone. The content is equal to the area of the base multiplied by one third of the perpendicular height.

(4) Sphere or Globe. The content is equal to the cube of the radius multiplied by 4.18879, nearly.

(5) Hence, the contents of similar solid bodies are as the cubes of their homologous lineal dimensions.

Ex. 1. Required the depth of a parallelopiped 29 long, and 44 broad, so that its content shall be equal to that of a cube whose edge is 89.

Here, the area of the base of the parallelopiped

Ex. 2. The content of a cylinder is equal to the sum of the contents of a cone and hemisphere, having the same base and altitude.

Taking 1 to represent the radius of the hemisphere, we shall have immediately from the directions contained in this page:

the content of the hemisphere = 2.09439, nearly:

the content of the cone = 1.04719, nearly: the content of the cylinder 3.14159, nearly:

=

whence, we find the sum of the two former

= 3.14158 nearly, which is the last, very nearly: and this would have been an exact equality were it not for the circumstance of each of the contents being only an approximation to its true value.

Examples for Practice.

(1) Each side of a square prism is 34 inches, and its height is 12 feet: how many solid feet does it

contain?

Answer: 99 ft. 1172 in.

(2) A rectangular cistern whose length is 93 feet, and breadth 6 feet, contains 294 cubic feet: find its depth.

Answer: 524 feet.

(3) What length of a cylindrical stone roller 18 inches in diameter, must be taken to make 14.137155 solid feet?

Answer: 2 feet.

(4) The sides of the base of a triangular pyramid are 3, 4 and 5 feet, and its altitude is 6 feet: find its solid content.

Answer: 12 feet.

(5) The solid content of a sphere is two thirds of that of its circumscribed cylinder.

(6) A right cone, hemisphere and cylinder of the same base and altitude, are as the numbers 1, 2, 3.

(7) A sphere is equal to a cone, whose height is equal to the radius, and whose base is equal to four great circles of the sphere.

(8) The contents of cubes, spheres, similar parallelopipeds, cylinders and cones, are as the cubes of their homologous lineal dimensions.

THE COMPUTATIONS OF ARTIFICERS.

199. DEF. Artificers generally take the dimensions of their work in yards, feet, inches, parts, &c.: and it is usual to reduce the yards to feet, so that the different denominations are all connected by the same number 12, or decrease in a twelvefold ratio, from the place of feet towards the right hand. For the sake of uniformity, the denominations after feet are termed primes, seconds, thirds, &c., distinguished respectively by accents &c., placed a little to the right, contiguous to the figures to which they belong: thus, 20 feet, 8 inches, 5 parts, &c., is written 20. 8'. 5′′. &c.