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RULE. Find the area of one side of the joist, or plank, by any of the preceding rules in superfi cies; multiply that area by the thickness in inches, the last product will be the superficial content of the joist, or plank..

Examples.

1. What is the superficial area, or board measure of a joist, that is 20 feet long 4 wide, and 3 thick? 20 ft. 0x0 ft. 46, ft. x3' thick equal 20 ft. Ans,

2. What is the area, or board measure of a plank that is 25 ft. long, and 16 in. wide, and 3 in. thick? Ans. 100 feet.

CASE IX.

To measure any irregular plane surface. RULE.-Divide the whole surface into trianglès and measure each triangle separately, as taught in case third, superficial measure.

Examples

1. What is the superficial content of a plat of ground, in form of the following figure; diagonal A B 50 ft.; perpendiculars, 24 and 12?

Ans. 900

12

24

1

CASE X.

B

To measure the surface of a circle. DEFINITION. Circles are round figures bounded every where by a circular line called the periphery, or arch, or sometimes the circumference; the line passing through the centre is called the diameter; half the diameter, or a line proceeding from the centre to the periphery is called the semidiameter, or radius.

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RULE 1.-AS 7, is to 22 so is the diameter to the circumference.

RULE 2 AS 113 is to 355 so is the diam. to the circum.

Examples.

1. What is the circumference of a circle, which has a diameter of, 14 rods?

By Rule 1. As 7: 22:14 to 44 rds. Ans. By Rule 2. As 113: 355 :: 14 to 4311 Ans.*

PROBLEM II.

The circumference given to find the diameter.

RULE 1.-As 22 is to 7 so is the cicum. to the diam.

RULE 2.-AS 355 is to 113 so is the cicum. to the diam.

RULE 3.-Annex two cyphers to the circum. and divide by 3.14 the quotient is the diameter nearly.

*The two methods do not exactly agree, the last method is nearest to the truth; an exact proportion between the diameter and circumference of a circle has not yet been discovered.

Examples.

1. What is the diameter of a circle whose cir cumference is 44 rods?

diam.

By Rule 1.-As 22: 7 44 14 Ans.
By Rule 2.-As 355 113: 44: 14 Ans.
By Rule 3-As 44.00÷3.141431 Ans.
PROBLEM III.

The diameter and circumference given to find the area. RULE. Multiply half the diameter by half the circumference, the product is the answer; or multiply the diameter and circumference together and the product is the area.

Examples.

1. What is the superficial content of a circle whose diameter is 14 rds. and circumference 44 rds.?

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1444154 square rods, Ans.

The same question done by rule second.
14X44-616154 sq. rds. Ans.

PROBLEM IV. 1.

The diameter given to find the area.

RULE. Multiply the square of the diameter by 7854 the product is the area.

Example.

1. What is the area of a circle whose diameter

is 14 rods?

rds.

14X14X 7854 153.9384, Ans.

PROBLEM V.

The circumference given to find the area.

RULE. Multiply the square of the circumfer ence by 07958, the product is the area.

Examples.

1. What is the area of a circle whose circumfer

ence is 44 rods?

44x44x-07958 154-06688, Ans.

PROBLEM VI.

The area of a circle given to find the diameter. RULE. Divide the area by 7854 and extract the square root of the quotient; the root is the diameter sought. (See application of square root, case 2.)

PROBLEM VII,

The area given to find the circumference.

RULE. Divide the area by 07958 and extract the square root of the quotient, and the root will be the circumference sought.

CASE XI.

To measure a sector of a circle.

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DEFINITION. The sector of a circle is a part of a circle bounded by an arc and two radii drawn to the extremities, see the figure AC B D. the arc; A D, or D B the radii.

DB the

ACB

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RULE. Find the length of the arc A C B; to do which say as 180°, is to the number of degrees in the arc, so is the radius multiplied by 3.1416, to the length of the arc: having found the length of the arc, multiply the radius by half the arc, the product is area required.

Examples.

1. What is the superficial content of the sector A C B D, the radius À D, or D B being 10 feet, and the arc A C B containing 135° ?

As 180 135

:

ft.

10x3.1416: 23:562 length of the arc.

ft.

then 10X23 56211781 Ans. RULE. Find the superficial content of a circle having the same radius; then say, as 360° (the degrees in a circle) are to the area of the whole circle; so are the number of degrees in the arc of the sector, to the superficial content of the same.

Examples.

1. What is the area of a sector of a circle whose are contains 45 degrees, and the radius of the same is 20 feet? Ans. 157 sq. ft.

NOTE. This method is exactly right, if a circle of 40 feet diameter contain tor of 45° with the same radius will be

CASE XII.

for it is evident that 1256 sq. ft. that a secof the circle,

To measure the segment of a circle.

A segment of a circle is a part of a circle bounded by part of the circle's periphery, and a chord connecting the two extremities of the periphery, ABC is a segment See the figure.

B

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NOTE. The quantity of the angle ADC is 820, instead "of 28°.

RULE.-Measure the sector ABCD by case 10, and also measure the triangle A D C, and sub

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