678. A Circle is a figure bounded by one uniform curved line.

679. The Circumference of a circle is the curved line bounding it.

680. The Diameter of a circle is a straight line passing through the center, and terminating in the circumference.

681. A Semi-Circle is one half of a circle.

682. A Prism is a solid whose bases or ends are any similar, equal, and parallel plane figures, and whose sides are parallelograms.

683. A Parallelopiped is a solid bounded by six parallelograms, the opposite ones of which are parallel and equal to each other. Or, it is a prism whose base is a parallelogram.

684. A Cube is a solid bounded by six The cube is sometimes called a

equal squares. Right Prism.

685. A Sphere or Globe is a solid bounded by a single curved surface, which in every part is equally distant from a point within called its center.

686. The Diameter of a sphere is a straight line passing through its center, and terminating at its surface.

687. A Hemisphere is one half of a globe or sphere..

688. Similar Figures and Similar Solids are such as have their like dimensions proportional.

PROBLEM I.

689. To find either side of a right-angled triangle, the other two sides being given.

Let us take any right-angled triangle, as ABC, and form the square, A E D C, on the hypotenuse. Now take a portion, A B C, of this square, and move it as on a hinge at A, until the points B and C

are brought to the positions of H
and E, respectively. Take also
another portion, D F C, and move
it as on a hinge at D, until the
points F and C are brought to the
positions of G and E, respectively.
Then the figure formed by the
parts thus moved and the remain-
ing part will be composed of two
new squares, one on AB, the
base of the triangle, and one on G
DF, which is equal to the per-
pendicular of the triangle. Hence,

D

The square of the hypotenuse of a right-angled triangle is equal to the sum of the squares of the other two sides.

From this property we derive the following

RULE. I. To find the hypotenuse; ·Add the squares of the two sides, and extract the square root of the sum.

II. To find either of the shorter sides; Subtract the square of the given side from the square of the hypotenuse, and extract the square root of the remainder.

EXAMPLES FOR PRACTICE.

1. The top of a tower standing 22 feet from the shore of a river, is 75 feet above the water, and 256 feet in a straight line from the opposite shore; required the width of the river.

Ans. 222.76 ft.

2. Two ships set sail from the same port, and one sails due east 50 leagues, the other due north 84 leagues; how far are they apart?

3. A ladder 50 ft. long will reach a window 30 ft. from the ground on one side of the street, and without moving the foot, will reach a window 40 ft. high on the other side; what is the breadth of the street?

4. What is the distance through a cubical block, measured from one corner to the opposite diagonal corner, the side of the cube being 6 feet?

Ans. 10.39 ft

PROBLEM II.

690. To find the side of a square equal in area to a given rectangle.

NOTE. This case, arithmetically considered, requires us to find a mean proportional between two given numbers.

The product of the sides of the rectangle will be the area which the square is to contain; hence

RULE. Multiply the sides of the rectangle together, and extract the square root of the product.

EXAMPLES FOR PRACTICE.

1. There is a field whose length is 208 rods, and whose breadth is 13 rods; what is the length of the side of a square lot containing an equal area? Ans. 52 rods.

2. If it cost $312 to inclose a farm 216 rods long and 24 rods wide, how much less will it cost to inclose a square farm of equal area with the same kind of fence?

3. What is the mean proportional between 12 and 588?

Ans. 84. 4. A and B traded together. A put in $540 for 480 days, and received of the gain; and the number of dollars which B put in was equal to the number of days it was employed in trade. What was B's capital? Ans. $720.

PROBLEM III.

691. To find the two sides of a rectangle, the area and the ratio of the sides being given.

NOTE. This case, arithmetically considered, requires us to find two numbers whose product and ratio are given

If we multiply together the terms of the given ratio, the product will be the area of a rectangle similar in form to the rectangle whose sides are required. Now we perceive, by the accompanying figures, that multiplying both sides of any rectangle by 2, 3, 4, etc., multiplies the area by the 34* 2 A

squares of these numbers, or 4, 9, 16, etc. If, therefore, we divide the given area by the rectangle of the terms proportional to the required sides, the quotient will be the square of that number which must be multiplied into these proportional terms to produce the required sides.

Hence the following

RULE. I. Divide the given area by the product of the terms proportional to the sides, and extract the square root of the quotient. II. Multiply the root thus obtained by each proportional term; the products will be the corresponding sides.

EXAMPLES FOR PRACTICE.

1. The sides of a rectangle containing 432 square feet are as 4 to 3; required the length and breadth.

Ans. Length, 24 feet; breadth, 18 feet. 2. Separate 23 into two factors which shall be to each other as 2 to 3. Ans. 3.91578+; 5.87367 +. 3. It is required to lay out 283 A. 2 R. 27 P. of land in the form of a rectangle whose length shall be 3 times the width; what will be the dimensions?

NOTE. The proportional terms are 3: 1.

Ans. 369 rods; 123 rods.

PROBLEM IV.

692. To find the radius, diameter, or circumference of a circle, the ratio of its area to a known circle being given.

All examples of this class relating to circles, may be solved by means of the following property: —

The areas of two circles are to each other as the squares of their radii, diameters, or circumferences.

NOTE. This property of the circle is only a particular case of a more general principle, viz. That the areas of similar figures are to each other as the squares of their like dimensions. This principle is rigidly demonstrated in Geometry, but cannot be easily proved here.

EXAMPLES FOR PRACTICE.

1. The radius of a circle containing 28.2744 sq. ft., is 6 ft. what is the radius of a circle containing 175.7150 sq. ft.?

;

225, square of radius re

2. If it cost $75 to inclose a circular pond containing a certain area, how much will it cost at the same rate to inclose another, containing 5 times the area of the first? Ans. $167.70.

3. If a cistern 6 feet in diameter hold 80 barrels of water, what must be the diameter of a cistern of the same depth to hold 1200 barrels ?

4. If a pipe 1.5 in. in diameter will fill a cistern in 5 h., what must be the diameter of a pipe that will fill the same cistern in 55 min. 6 sec.? Ans. 3.5 in.

PROBLEM V.

693. To find the side of a cube, the solid contents being given.

NOTE. This case, arithmetically considered, requires us to separate a number into three equal factors.

The solid contents of a cube are found by cubing the length of one side; hence,

RULE. Extract the cube root of the given contents.

EXAMPLES FOR PRACTICE.

1. What must be the length of the side of a cubical bin that shall contain the same quantity as one that is 24 ft. long, 18 ft. wide, and 4 ft. deep? Ans. 12 ft.

2. What must be the length of the side of a cubical bin that will contain 150 bushels?

3. What must be the depth of a cubical cistern that will hold 200 bbl. of water?

4. How many sq. ft. in the surface of a cube whose solidity is 79507 cu. ft.?

Ans. 11094.