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(4) How many acres are in a triangular field, whose base is 28, and perpendicular 20,5 chains?

(5) A triangular field 738 links long, and 583 in the perpendicular, brings in 12t. a-year. What is it let at per acre?

Case 2. When the three sides of a triangle are given, to find the area.

RULE.

3. From half the sum of the three sides subtract each side severally; multiply the half sum and the three remainders continually together, and the square root of the last product will be the area of the triangle, that is, a+b+c

2

=S= half the sum of the sides.

Then let s-a-e, and s-b=f, also ·s-c=g

the area.

last figure.

sefg

-Note, a=AC, b=AB, and e=BC. See the

EXAMPLES.

(6) Suppose I have a fish-pond of a triangular form, whose three sides measure 400, 348, and 312 yards; what quantity of ground does it contain?

PROBLEM IV.

To find the area of a trapezium.

RULE.

1. Divide it into triangles, according to the manner which you judge most convenient; then the sum of the area of those triangles, calculated by the last problem, will be the area of the trapezium; Or,.

2. Multiply the sum of the perpendiculars by half the diagonal, and the product will give the area; or multiply the sum of the perpendiculars by the diagonal, and half the product will be the area.

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Or, 3. Subtract severally each side from half the sum of the four sides, and the square root of the product of the four remainders will be the area required.

EXAMPLES.

(7) How many square yards of paving are there in a trapezium, ABCD, whose diagonal, BD, 45 feet, and the

perpendiculars, AE, equal to 17,25, CF, equal to 14 feet?

(8) Suppose the four sides of a trapezium are 15,60; 13,20; 10, and 26 chains: query, the area?

PROBLEM V.

To find the area of any regular polygon.

RULE I.

Let fall a perpendicular from the centre of the figure to one of its sides; then multiply together the perpendicular, the side of the figure, and the number of its sides, and half the product will be the area.

Here the number of sides is 5=N.

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Fig 6.

B

EXAMPLE.

(9) A piece of garden-box lies in the form of a regular pentagon, or figure of five equal sides (as above), each 48 feet and from the centre of the figure, C, to the middle of one of its sides, D, it measures 41,57 feet nearly the area of the figure will be the contents of these five triangles. Pray what is that?

RULE.

Multiply the square of the side of any regular figure by the multiplier standing opposite to its name in the following table; and the product will be the area.

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(10) What is the area of a hexagon, whose side is 30? (11) What is the area of an octagon, whose side is 24?

PROBLEM VI.

To find the diameter and circumference of a circle, the one from the other.

RULES.

1. Multiply the diameter by 3,1416, and the product will be the circumference. And therefore,

2. Divide the circumference by 3,1416, and the quotient will be the diameter.

3. See Sect. LIII. Case 3.

EXAMPLES.

(12) If the diameter of a circle be 7, what is the circumference?

(13) What is the diameter of a circle whose circumference is 22?

(14) What is the circumference of the earth, supposing it be perfectly round, and its diameter be 8000 miles?

PROBLEM VII..

To find the area of a circle.

RULES.

1. Multiply half the circumference by half the diameter, and the product will be the area.

Or,

2. Multiply the square of the diameter by ,7854, and the product will be the area. Or,

3. Multiply the square of the circumference by ,079574, and the product will be the area.

Or,

Or,

4. Multiply the square of the semi-diameter by 3,1416, and the product will be the area.

5. Multiply the circumference by the diameter, and a fourth part of the product will express the area.

**,7854, and 3,1416, are areas of circles whose dia meters are 1 and 2; and ,079577 is the area of a circle whose circumference is 1; likewise 452, and 1,273239, are squares of the diameters of circles, whose areas are 355 and 1; and 1,12837 is the diameter of a circle, whose area is equal to a square,

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whose side is 1. For if the diameter be unity, or 1, the circumference will be 3,1415926, then per rule I. 3,1415926

2

x=,79539816, or rather, ,7854, the area.

Also, if the diameter be 2, the circumference will be
6,2831853 2
X

6,2831853;

then per rule I.

2

3,1415926, or rather, 3,1416, the area,

=

2

Likewise

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Let C the circumference, and D the diameter, AB.

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(15) How many square feet are in a circle whose circum

ference is 6,2832?

(16) What is the area of a circle whose diameter is 12?

PROBLEM VIII.

To find the length of any arc of a circle, ABD.

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. Multiply together the radius, DC, the number of degrees in the given arc, and the number,01745329 (b), the

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