These two figures may be conceived to be formed from the others, by supposing their bases to be fixed, and the other sides to be moved on the lower angles into an oblique position. This, it is evident, would bring the base and opposite side nearer to each other, just in proportion to the degree of obliquity. It follows, therefore, that the space enclosed by the lines in that position, is less than when they were all perpendicular.

A rhombus, therefore, is less than a squa of equal sides, and a rhomboid than a rectangle.

Their area is only equal to that contained by a rectangle of a breadth, equal to the perpendicular distance from the base of the rhombus or rhomboid, to its opposite side, and of a length equal to the base of those figures respectively. The rectangle which measures each of these figures, is indicated in part, by the dotted lines; one of which encloses an area, not within the figure, just equal to the portion of it cut off by the other.

Fig. 5.

Therefore, to find the area of a rhombus or rhomboid : RULE.-Multiply the length by the perpendicular breadth. A TRIANGLE is a figure of three sides and three angles. If two of the sides are perpendicular to each other it is called a right-angled triangle. Thus, A, B, C, has a right-angle at B. By drawing lines parallel to the two sides adjacent to the right-angle, (as the dotted lines,) a rectangle is formed, and the hypotenuse bisects* it. The area of the triangle

is therefore, half that of a rectangle of equal sides. Consequently,

To find the area of a right-angled triangle:

RULE.-Multiply the base by the perpendicular, and take half the product; or, multiply one into half the other, and the result is the area.

If the triangle be not right-angled, that is, have not two

Cuts it into two equal parts

of its sides perpendicular to each other; as Fig. 6, we may determine the area, if we can get the perpendicular distance from the vertex to the base (C. D.)

RULE.-Multiply half that distance into the base.

Fig. 6.

D

The completion of the rectangle (shown by the dotted lines) makes the reason of the rule apparent.

In case the perpendicular distance from the base to the opposite angle, cannot be obtained, the following method may be adopted:

RULE.-Add the three sides together, and take half their sum; from that half sum subtract each of the sides separately; multiply the three remainders thus obtained, and the half sum, continually into each other, and the square root of the last product, will be the area of the triangle.

NOTE. The demonstration of this rule, would not be intelligible without a knowledge of geometry. Fig. 7.

When the vertex of a triangle is cut off by a line parallel to the base, the figure thus formed is called a trapezoid. Thus, Fig. 7, is a trapezoid. The rectangle indicated by the dotted lines, is the measure of its area. It includes a surface without the figure, precisely equal to the portion excluded within the figure. The

breadth of the rectangle is equal to half the sum of the two parallel sides.

RULE. Multiply half the sum of the two parallel sides, by the perpendicular distance between them.

All the figures we have considered, have been referred either to a square or to a rectangle. The area of a circle is obtained in a similar manner. If we draw a square about

a circle, so that each side shall touch its circunference, the diameter of the circle drawn from one of the points of contact, will meet the circle at the point of contact of the opposite side of the square.

Fig. 8.

Now, if we square the diameter of a circle, it is evident we shall have the area of a square, one side of which is equal to that diameter; or, in other words, of a square, which if drawn about the circle, (in the manner of the diagram,) would touch it in four points. But this would exceed the area of the circle, by the spaces included between its circumference, and each of the four angles of the square. It has, therefore, been an interesting problem with mathematicians, to ascertain the

ratio between the area of a circle, and of a square drawn about it; and it is demonstrated, that the area of the circle is to that of a square, as the decimal .7854 is to unity.* If, therefore, the square contain 1 yard, a circle of that diameter would contain 7854 ten thousandths of a yard. If the square contain 4 yards, (its side being 2 yards,) then a circle of 2 yards in diameter, would contain .7854X 4 3.1416 yards.

RULE.-Multiply the square of the diameter by the decimal .7854, and the product is the area.

When the circumference of a circle is known, the diameter may be found by the following proportion; as 22:7: so is the circumference to the diameter; or, as 3.1416:1::, &c.

* Nearly.

The converse of this statement will give the circumference, when the diameter is known.

DEFINITION.

A cylinder is a round body of uniform diameter, its two ends being parallel, and its axis perpendicular to them.

To find the area of a cylinder :

RULE.-Multiply the perimeter by the length.

To find the area of a sphere or globe:

RULE.-Multiply the circumference and diameter totgether; or, multiply the square of the circumference by

3.1416

QUESTIONS UNDER THE FOREGOING RULES.

1. What is the difference in quantity, between two pieces of land, one of which is a square with sides 40 rods in length, and the other a rhombus, with equal sides, but of a breadth from the base to the opposite side of only 34 rods? Ans. The square contains 240 rods the most. 2. There is a square floor, 18 feet on a side; what number of square feet does it contain; and how many yards of carpeting, 1 yard wide, will cover it?

Ans.

S 324 feet. 36 yards. 3. There is a barn 50 feet by 36: the body is 20 feet in height to the eaves. How many feet of boards will make the siding of the main body; and if the boards were all 15 inches in breadth, and 10 feet in length, what number of boards will it require? 3440 feet.

Ans. {275.+ boards.

4. On a base of 120 rods in length, a surveyor wished to lay off a rectangular lot of land, to contain 60 acres, what distance in rods must he run out from his base line?

Ans. 80 rods. 5. How many square yards in a triangle, whose base is 48 feet, and perpendicular height 25 feet?

Ans. 67 yards. 6. The three sides of a triangular field are severally 60, 30, and 40 rods; how many acres does it contain?

Ans. 6 acres 32 rods.

7. How many acres in a triangular piece of land, whose sides are 30, 40, and 50 rods? Ans. 32 acres. 8. What is the difference between the area of a square, whose sides are 64 feet, and a rhombus of the same dimensions, whose perpendicular height is 52 feet?

Ans. 768 feet.

9. A man bargaining for a piece of land of a rectangular shape, the length of which was 198 rods, and its breadth 150 rods, agreed to give $ 32 an acre, provided the owner would deduct from the quantity, the contents of a circular pond in it, which was admitted to be 100 rods in circumference. What was deducted for the pond, and what did the form come to? 795 rods deducted. $5781 the cost of the farm.

Ans.

10. If the forward wheels of a coach are 4 feet, and the hind ones 5 feet in diameter; how many more times will the former revolve than the latter, in going a mile ?

11. If the diameter of a circular what is the area of its bottom?

Ans. 84 times. cistern be 6 feet; Ans. 28.2 feet.

12. What quantity of land in a field of the form of a trapezoid, the parallel sides of which are severally 60 rods and 40 rods, and their distance asunder 50 rods?

Ans. 15 acres. 13. How many square feet in a board 2 feet wide at the larger, and 1 foot 8 inches at the smaller end, and 14 feet long? Ans. 25 feet.

14. If you wish to plant 600 apple trees in a rectangular form, and are limited in one direction to 20 trees; how many trees must you have in the other direction to complete your orchard? Ans. 30 trees. 15. What is the quantity of surface in a solid cylinder, 20 feet long and 2 feet in diameter; the area of the two ends included? Ans. 131.9 feet.

NOTE. The perimeter is the measure round; that is, the circumference. The area of the ends is to be calculated separately, and then added.

16. What is the convex area of a 12 inch globe; and suppose it to be of the uniform thickness of of an inch; what is the concave area of the same globe?