3. If a piece of land lie in the form of a right-angled triangle, its base being 37 rods, and the perpendicular line being 24 rods, how many acres are in it?

Ans. 2,8617+ acres. 4. If the base of a triangular field be 7 chains and 50 links, and the perpendicular 4 chains and 25 links, how much does it contain? Ans. 1ac. 2roo. 15rods.

Joists and Planks are measured by the following

RULE. Find the area of one side of the joist or plank, by one of the preceding rules; then multiply it by the thickness in inches; and the last product will be the superficial content.

EXAMPLES.

1. What is the area, or superficial content, or board measure, of a joist, 20 feet long, 4 inches wide, and 3 inches thick?

20×4

12

-x3=20 feet. Ans.

2. If a plank be 32 feet long, 17in. wide, and 3in. thick, what is the board measure of it? Ans. 136 feet.

NOTE.-There are some numbers, the sum of whose squares makes a perfect square; such are 3 and 4, the sum of whose squares is 25, the square root of which is 5; consequently, when one leg of a right-angled triangle is 3, and the other 4. the hy. potenuse must be 5. And if 3, 4, and 5, be multiplied by any other numbers each by the same, the products will be sides of true right-angled triangles. Multiplying them by 2, gives 6. 8, and 10-by 3, gives 9, 12 and 15-by 4, gives 12, 16 and 20, &c.; all which are sides of right-angled triangles. Hence architects, in setting off the corners of buildings, commonly measure 6 feet on one side, and 8 feet on the other; then, laying a 10 feet pole across from those two points, it makes the corner a true right-angle. RULE 2.-To find the area of any triangle when the three sides only are given.

RULE. From half the sum of the three sides subtract each side severally; multiply these three remainders and the said half sum continually together; then the square root of the last product will be the area of the triangle.

EXAMPLE.

Suppose I have a triangular fish-pond, whose three sides measure 400, 348, and 312yds.; what quantity of ground does it oover? Ans. 10 acres, 3 roods, 8+ rods.

S

CASE 4. To measure irregular surfaces.

RULE.-Divide the figure or plane into triangles, by drawing diagonal lines from one angle to another; then measure all the triangles, by either of the rules in Case 3; and the sum of their several areas will be the area of the given figure.

EXAMPLE.

If a piece of ground be divided into two triangles by a diagonal line drawn through it, measuring 30 rods, and two perpendiculars be let fall, one measuring 8 rods, and the other 14 rods; how many acres does it contain? Ans. 2 acres.

CASE 5. To measure a circle.

Definition. A circle is a figure bounded by a curve or circular line, every part of which is equally distant from the middle or centre, The curve line is called the periphery or circumference; a line drawn, from one side to the other, through the centre, is called the diameter; and a line drawn, from the center to the circumference, is called the semidiameter, (half diameter,) or radius.

PROBLEM I.

The diameter given to find the circumference. RULE. AS 7 are to 22, so is the given diameter to the circumference; or, more exactly, as 113 are to 855, so is the diameter to the circumference, &c.

EXAMPLES.

1. What is the circumference of a wheel whose diameter is 4 feet?

As 7:22:4: 12,57+ feet the circumference. Ans. 2. What is the circumference of a circle whose diamcter is 35 rods? As 7:22:35:110 rods. Ans.

NOTE. To find the diameter, when the circumference is given, reverse the foregoing rule, and say, as 22 are to 7, so is the given circumference to the required diameter; or, as 355 are to 113, so is the circumference to the diameter. 3. What is the diameter of a circle whose circumference is 1,10 rods?

As 22 7: 110 : 35 rods the diam. Ans.

:

PROBLEM II.

To find the area of a circle.

RULE.-Multiply half the diameter by half the circumference, and the product is the area; or, if the diameter alone is given, multiply the square of the diameter by ,785398 or, which is near enough, by ,7854-and the product will be the area.

EXAMPLES.

1. Required the area, or superficial content of a circle whose diameter is 12 rods, and circumference 37,7 rods. 18,85-half the circumference.

6 half the diameter.

113,10 area in square rods. Ans.

2. What is the superficial content of a circular garden whose diameter is 11 rods?

By the second method,

11x11=121. ,7854×121=95,0334 rods. Ans. 3. What will be the cost of a circular platform to the curb of a round well, at 10 cents per square foot; if the diameter of the well be 42 inches, and the breadth of the platform be 14 inches? Ans. $1,87 cts. +

PROBLEM III.

To find the area of a circle when the circumference alone is given. RULE.-Multiply the square of the circumference by ,079577 or, which is near enough, by ,07958—and the product will be the area.

NOTE.-,785398 is the area of a circle whose diameter is 1, and ,079577 is the area of a circle whose circumference is 1.

EXAMPLE.

rods?

What is the area of a circle whose circumference is 30 30x30x,07958=71,62200 rods. Ans. PROBLEM IV.

The area of a circle given to find the diameter. RULE. Divide the area by ,7854-and the square root of the quotient is the required diameter.

EXAMPLES.

1. Required the diameter of a circle that will contain within its circumference the quantity of an acre of land. acre=4840sq. yds. Then 484-88,5+yds. Ans. 2. In the midst of a meadow abounding with feed,

For two acres, to tether my horse, I've agreed; How long must the rope be, that, feeding all round, He may n't graze less or more than the two acres of ground? Ans. 55 yards. 3. A, B, and C, join to buy a grindstone, 36 inches in diameter, which cost $33, and towards which A paid $11, B, $13, and C, 833cts. The waste hole for the spindle was 5 inches square. To what diameter ought the stone to be worn, when B and C begin severally to work with it, allowing for the hole, and A first grinding down his share, next B, and then C?

Ans. 29,324+inch. diameter where B begins to grind; and 19,013+in. diam. C begins. NOTE. Twice the square of the side of a square, will be the square of the diameter of its circumscribing circle. PROBLEM V.

The area of a circle given to find the circumference. RULE. Divide the given area by ,07958-and the square root of the quotient is the required circumference.

EXAMPLES.

1. The expense of turfing a round plot, at 4 pence per square yard, was £2 9s. 93d. ,8; what was its circumference? Ans. 130+ feet.

2. How many feet of boards will fence a round garden, containing just two acres, the fence five feet high; and what will be the expense at 63 mills per square foot ? Ans. 5231+ ft. boards; and cost $32,69cts. 61m.+

CASE 6. To measure a sector of a circie.

Definition. A sector is a part of a circle, contained between an arch line and two radii, or semidiameters of the circle.

RULE. Find the length of the arch by saying, as 180 degrees are to the number of degrees in the arch, so is the radius, multiplied by 3,1416, to the length of the arch, which length, divided by 2, and multiplied by the radius, will become the required area.

EXAMPLE.

What is the area of the sector of a circle whose radius is 25 feet, its arch containing 125° ?*

As 180° 125° :: 25×3,1416: 54,5416+ft. length of arch. 54,5416

RULE 2.--Find the area of a circle having the same radius; then say, as 360 degrees, [the number of degrees into which all circles are divided,] are to the area of the said circle, so is the number of degrees in the arch of the sector, to the area required.

EXAMPLE.

Required the area of a sector of a circle whose arch contains 65 degrees, and radius 35 feet. Ans. 695,5 sq.ft.

CASE 7.--To find the area of a segment of a circle. Definition.--A segment of a circle is any part of a circle cut off by a right line drawn across the circle, which does not pass through the centre, and is always greater or less than a semicircle.

RULE. Find the area of the sector having the same arch as the segment, by Case 6; find also the area of the triangle formed by the chord of the segment and the radii of the sector, by Case 3; subtract the area of the latter from that of the former, and the remainder will be the area of the segment, when the segment is less than a semicircle; but the sum of the two areas is the answer, when it is greater.

EXAMPLE.

What is the area of the segment, whose arch contains 55°; its chord 12,5 rods; the perpendicular of its triangle 16 rods; and its semidiameter 17,2 rods?

First, find the area of a circle whose diameter is 34,4 rods. As 7 : 22:: 34,4 : 108,1+rods circumference.

Then, as 360°: 929,66 :: 55°: 142+ area of the sector.

As we have not been able to obtain engravings to represent any of the figures in the preceding or subsequent Examples, the Teacher will, we trust, be so good as to draw them, for the Pupil, on paper or a slate.