7. A cone is a solid figure, having a circle for its base, and its top terminated in a point or vertex,

8. A sphere is a solid, bounded by one continued convex surface, every point of which is equally distant from a point within, called the centre. The sphere may be conceived to be formed by the revolution of a semicircle about its diameter, which remains fixed.

A hemisphere is half a sphere.

9. The segment of a pyramid, sphere, or any other solid, is a part cut off the top by a plane parallel to the base of that figure. 10. A frustum is the part that remains at the bottom after the segment is cut off.

11. The sector of a sphere is composed of a segment less than a hemisphere, and of a cone having the same base with the segment, and its vertex in the centre of the sphere.

12. The axis of a solid is a line drawn from the middle of one end to the middle of the opposite end; as between the opposite ends of a prism. The axis of a sphere is the same as a diameter, or a line passing through the centre, and terminating at the surface on both sides.

13. The height or altitude of a solid is a line drawn from its vertex, or top, perpendicular to its base.

MENSURATION OF SUPERFICIES AND SOLIDS.

PROBLEM I.

To find the area of a square or parallelogram.

RULE. Multiply the length by the breadth, and the product is the superficial contents.

1. What are the contents of a board 15 feet long and 2 feet wide ? Ans. 30 feet. 2. The State of Massachusetts is about 128 miles long and 48 miles wide. How many square miles does it contain? Ans. 6144 miles.

3. The largest of the Egyptian pyramids is square at its base, and measures 693 feet on a side. How much ground does it cover? Ans. 11 acres 4 poles.

4. What is the difference between a floor 40 feet square, and 2 others, each 20 feet square? Ans. 800 feet.

5. There is a square of 3600 yards area; what is the side of a square, and the breadth of a walk along each side of the square, and each end, which may take up just one half of the square? Ans. (42.42 yards, side of the square. 8.78+yards, breadth of the walk.

PROBLEM II.

To find the area of a rhombus or rhomboid.

RULE.- Multiply the length of the base by the perpendicular height. 6. The base of a rhombus being 12 feet, and its height 8 feet, required the area. Ans. 96 feet.

PROBLEM III.

To find the area of a triangle.

RULE. Multiply the base by half the perpendicular height; or, add the three sides together; then take half of that sum, and out of it subtract each side severally; multiply the half of the sum and these remainders together, and the square root of this product will be the area of the triangle.

7. What are the contents of a triangle, whose perpendicular height is 12 feet, and the base 18 feet? Ans. 108 feet.

8. There is a triangle, the longest side of which is 15.6 feet, the shortest side 9.2 feet, and the other side 10.4 feet. What are the contents ? Ans. 46.139+ feet.

PROBLEM IV.

Having the diameter of a circle given, to find the circumference. Multiply the diameter by 3.141592.

RULE. NOTE. The exact ratio of the diameter of a circle to its circumference has never yet been ascertained. Nor can a square, or any other rightlined figure, be found, that shall be exactly equal to a given circle. This is the famous problem, called the squaring the circle, which has exercised the abilities of the greatest mathematicians for ages, and has been the occasion of so many endless disputes. Van Ceulen, a Dutchman, was the first who ascertained this ratio to any great degree of exactness, which he extended to thirty-six places of decimals; and it was effected by

*

* This is said to have been thought so curious a performance, that the numbers were cut on his tombstone, in St. Peter's church-yard, at Leyden.

means of the continual bisection of an arc of a circle. This process was exceedingly troublesome and laborious; but since the invention of Fluxions and the Summation of Infinite Series, there have been several methods found for doing the same thing with less labor and trouble, and far more expedition. If, therefore, the diameter of a circle be 1 inch, the circumference will be 3.1415926535897932384626433832795028841971693

9937510582097494459230781640628620899862803482534211706798214808

65132823066470938446460955051822317253594081284802 inches nearly.

9. If the diameter of a circle is 144 feet, what is the circumference? Ans. 452.389248 feet. 10. If the diameter of the earth be 7964 miles, what is its circumference ? Ans. 25019.638688+ miles.

PROBLEM V.

Having the diameter of a circle given, to find the area.

RULE. Multiply half the diameter by half the circumference, and the product is the area; or, which is the same thing, multiply the square of the diameter by .785398, and the product is the area.

Demonstration. If we suppose a circle to be divided into an infinite number of triangles, by lines drawn from the centre of the circle to the circumference, we may find the contents of each triangle by multiplying its perpendicular height by half its base, but its perpendicular height is half the diameter of the circle, and half its base is half a certain portion of the circumference; and all the bases of all the triangles united form the whole circumference.

Again, if multiplying half the circumference by half the diameter give the area of a circle, it is evident that the area will be obtained by multiplying one fourth the circumference by the whole diameter; and, as the circumference of a circle, whose diameter is 1, is 3.141592, therefore by multiplying 1 by one fourth of 3.141592 we shall obtain the area of a circle whose diameter is 1. Thus, 3.141592 ÷4 = .785398. And as circles are to each other as the squares of their diameters (see page 246), therefore, if we wish to obtain the area of a circle whose diameter is 20 feet, we make the following statement.

As 12 foot: 202 feet: : .785398: 314.1592 feet, Ans. And this process is equivalent to multiplying the square of the diameter of the given circle by .785398. Q. E. D.

11. If the diameter of a circle be 761 feet, what is the area? Ans. 454840.475158 feet. 12. There is a circular island, three miles in diameter; how many acres does it contain ? Ans. 4523.89+ acres.

PROBLEM VI.

Having the diameter of a circle given; to find the side of an

equal square.

RULE. Multiply the diameter by .886227, and the product is the side of an equal square.

Demonstration. - We have seen in Problem V. that the area of a circle, whose diameter is 1, is .785398163397; if, therefore, we extract the square root of this number, we shall obtain the side of a square of a circle whose diameter is 1. Thus, ✓.785398163397: = .886227. And since, as we have before stated, the diameters of circles are to each other as the sides of their similar inscribed figures, therefore, as 1, the diameter of the given circle, is to the diameter of the required circle, so is .886227, the side of a square equal to the given circle, to the side of a square equal to the required circle. If, therefore, the diameter of a circle were 20 feet, and it was required to find the side of a square that would contain that quantity, we should make the following statement:

As 1 foot: 20 feet :: .886227 : 17.72454 feet, Ans.

We see, from this process, that multiplying the diameter of the required circle by .886227 gives the side of an equal square. Q. E. D.

13. I have a round field, 50 rods in diameter; what is the side of a square field, that shall contain the same area?

PROBLEM VII.

Ans. 44.31135+ rods.

Having the diameter of a circle given, to find the side of a square inscribed.

RULE.-Multiply the diameter by .707106, and the product is the side of a square inscribed.

Demonstration.

B

Let ABCD represent A a square inscribed in a circle whose diameter is 1. DB is the diameter of the circle, and it is also the diagonal of the square. As DAB is a right-angled triangle, the squares of DA and A B are equal to the square of BD; but AD is equal to AB, therefore the square of DA is equal to half the square of B D. The square of B D is 1, therefore the square of DA is .5; and the square root of .5 is .5 = .707106 AD, the side of the square, whose diameter is 1. Q. E. D. There

D

C

fore, to find the side of a square inscribed in any circle, we say, as 1 is to the diameter of any required circle, so is .707106 to the side of a square inscribed in the required circle.

14. I have a piece of timber 30 inches in diameter; how large a square stick can be hewn from it?

Ans. 21.21+ inches square. 15. Required the side of a square, that may be inscribed in a circle 80 feet in diameter. Ans. 56.56848+ feet.

PROBLEM VIII.

In a given circle to describe a hexagon and an equilateral triangle, and to find the length of one of the sides of the inscribed triangle.

RULE.-Multiply the diameter by .8660254 and the product is the side of an inscribed equilateral triangle.

Demonstration. Let ABCDEF be the given circle, and G the centre. From the point B in the circumference apply the radius BG six times to the circumference, and join B C, CD, DE, EF, FA, and AB, and the figure ABCDEF, thus formed, is an equilateral inscribed hexagon.

Join the alternate angles A E, EC, and CA, and the figure AE C, thus

formed, is an equilateral triangle inscribed. It is equilateral because the three sides subtend the equal arches of the circumference.

ABCG is a rhombus, and the diagonal B G is equal to either side of the rhombus. If, therefore, the diameter of the circle AD is 1, the semidiameter AG or BG will be .5; and BH, which is half of BG, will be .25. AHB is a right-angled triangle, and therefore A H is equal to the square root of the difference of the squares of AB and BH. Thus AH= A B2B H2 = √.52 — .252.25 -.0625.1875 .4330127. Now if AH be .4330127, AC, which is twice AH, will be .8660254. But AC is the side of the equilateral triangle inscribed; and as we have before shown, page 246, that all similar figures are in proportion to their homologous sides, it will therefore follow, that as the diameter of the given circle, which is 1, is to the side of its inscribed equilateral triangle