Imágenes de páginas
PDF
EPUB

tend at the centre, the ratio required will be the same as that of the angles above-named.

Thus, to find the ratio of the arc AB to the arc AC, in the same circle, subtending at the centre the angles AOB, AOC, respectively. Let the angles AOB, AOC be measured, and suppose them to be 45°35′, and 70776". These, when reduced to portions of 6′′ (that is, making 6" the unit), become respectively 27350 and 4271; therefore

B

O'

A

arc AB: arc AC-27350

4271

A rough approximation would be found from

C

27300

4200

[merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small]

mation would answer the

purpose, must depend upon the accuracy required in the particular work in hand.

A very useful mode of computing the magnitudes of straight lines, angles, circular arcs, &c. may be fitly introduced here, although it has not hitherto been noticed by writers on Mensuration. It consists in comparing the proposed magnitude with some unit of the same kind, by means of Continued Fractions. Thus

PROB. 15. Let it be required to measure the straight line CD in terms of the unit AB; that is, to find the CD numerical value of the fraction

required degree of accuracy.

AB approximately, to any

Open the compasses, until they exactly embrace AB.

[merged small][merged small][merged small][merged small][ocr errors][ocr errors]

Then, with the compasses thus fixed, step along CD, marking the intervals, each equal to AB; and suppose there are three such intervals in CD, with a remainder ED, less than AB.

Then

[blocks in formation]

Now set the compasses to ED, and step along AB; at intervals each equal to ED, and suppose ED to be contained twice in AB with remainder FB. Then

[merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small]

Similarly, let ED be divided by FB, and let FB go once with a remainder GD, so that

[blocks in formation]

Let this process be continued, till there either be no remainder, or the remainder be so small that it may be neglected. Suppose GD to be the last remainder; so

that GD goes twice exactly in FB, or

We have, then,

FB

= 2. GD

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small]
[blocks in formation]

that is, the area of a triangle, which has one angle of 30°, is equal to one-fourth of the product of the two sides containing that angle.

Ex. Let AC=24 yds., and AB=176 yds.; then

area of ABC:

=

24×17.6
4

=105'6 sq. yds.

[ocr errors][merged small][merged small][merged small][ocr errors]

PROB. 4. To find the length of a side of the square 'inscribed' in a given circle.

Let ABCD be the given circle, and O its centre: AC and BD two diameters at right angles to each other. Join AB, BC, CD, DA; then we have a square inscribed (155, B Part 11.). Now,

AB2=AO+BO-240-2 × (rad.); .. AB=rad.× √√2;

C

or the side of the inscribed square is equal to the radius multiplied by 2.

If rad.=1, the side of the square=√2.

PROB. 5. To find the length of a side of the equilateral triangle inscribed' in a given circle.

Let O be the centre of the given circle, and ABC an equilateral triangle inscribed in it. From O draw OD perpendicular to AB, and produce it to meet the circumference in E. Join AE, BE. Then it is easily shewn that AEBO is a lozenge, and that

OD=10E-rad.

E

B

D

[merged small][merged small][merged small][merged small][merged small][ocr errors]
[merged small][ocr errors][merged small][merged small]

If rad. 1, the side of the inscribed equilateral triangle=√3.

PROB. 6. To shew that the side of a square, together with the side of an equilateral triangle, both inscribed in the same circle, is equal to half the circumference of the circle, nearly.

By Prob. 4, the side of square

=

-rad. × √2=rad. ×1·414................

By Prob. 5, the side of triangle

=rad.×√3=rad.×1·732................;

.. sum of the two = rad. × (1·414+1·732),

=rad. × 3·146, nearly.

But half the circumference of the circle

=rad. x3.14159;

therefore the side of the square added to the side of the triangle does not differ from the semi-circumference of the circle by a quantity so great as '005, or

is, the 200th part of a unit.

5

, 1000

that

PROB. 7. To find the numerical value of the angle at the centre of a circle subtended by an arc equal to the radius.

By (240) we know, that the length of an arc subtending an angle of A° at the centre

A

[blocks in formation]

360

therefore, in this case,

rad.= 2 × rad. ×

A

to find A.

PART III.

4

[merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small]

PROB. 8. To construct a rectangle, or triangle, which shall be equal to a given circle.

Assuming that the area of the given circle is

[blocks in formation]

divide the given radius into 7 equal parts (168, Part 11.); then construct a rectangle, of which the base is 22 of such parts, and the height 7, that is, the radius. area of this rectangle

[blocks in formation]

The

[blocks in formation]

For a triangle of equal area, make the base the same, and the height equal to double the radius, that is, to the diameter.

PROB. 9. To measure the area of a circular ring.

Let it be required to find the area of the ring enclosed by the two concentric circles ABC, DEF. It is plain, that this area will be found by subtracting the area of the smaller circle from that of the greater. So that, if R represents the radius of the greater, and r the radius of the smaller, circle, we have, by (238),

D

F

E

B

« AnteriorContinuar »