If for π we use 3.1416, in the case before supposed, the circumference thus obtained will be 314-16 inches: whereas the more correct value is 314-1592... inches; therefore the error is less than 001 inches, that is, less than one-thousandth part of an inch.

If for the value of π we use

355

, 113

that is, 3.1415929,

it is plain on comparing this with the value given before up to 20 places of decimals, that the difference is less than 000001; and therefore, in the example before given, the error in the circumference would be less than one-millionth part of an inch.

And so on, by using a sufficient number of decimal places in the value of, we can approach to the true value of the circumference, as near as we please, expressed in the same linear unit as the given radius or diameter.

238. To measure the area of a given circle.

We have seen (227) that the area of any regular polygon, as ABCDEF, is equal to the perimeter x half the perpendicular from the centre of the circumscribing circle upon one of the sides. So also, if the polygon had been described about the circle, as abcdef, the area would be equal to the perimeter of the polygon x half the perpendicular upon one of its sides, (which is OP, the radius of the circle).

Now the area of the circle is evidently greater than

the former polygon, and less than the latter; but if the number of the sides be indefinitely increased in each case, and therefore the length of each indefinitely diminished, the perimeter of each polygon approaches to the circumference of the circle, and the area of the circle is the ultimate value of the area of either polygon, when the number of the sides is indefinitely great.

Hence, putting the circumference of the circle for the perimeter of the polygons, and the radius for the perpendicular, we have

area of circle = circumference × & radius,

=x twice radius radius, (237),

X

=πx (radius).

COR. Since does not admit of being found in a terminating, or a recurring, decimal (237), therefore, the area of a circle does not admit of being converted into a rectangle without error; and therefore, cannot be exactly measured by square units. Hence arose the impossible problem of squaring the circle, as it is called, which means finding a square numerically equal to a proposed circle.

239. That the area of a circle is equal to half circumf.× rad. may be thus exhibited to the eye.

Divide each half of the circle into any the same number of equal sectors, and draw the chord at the base of each sector, forming as many isosceles triangles as there are sectors. Take a straight line AB, and place these triangles on it in juxta

position, having their bases coinciding with AB, one-half above and the other half below AB, as shewn by the dark triangles in the diagram. Then it is plain, that as the number of sectors is increased, AB approaches nearer and nearer to the semi-circumference of the circle, and the altitude of each triangle to the radius of the circle.

Through A and B draw CAD and EBF at right angles to AB, and through the vertices of the triangles CE and DF parallel to AB. Then it is easily seen, that

the sum of the dark triangles is half the rectangle CDEF, whatever the number of them may be. But when that number is increased indefinitely, the aggregate area of the triangles is the area of the circle, whilst AB is the semi-circumference, and AC the radius ;

.. area of circle = ABXCD,

=ABXAC= circumf. x rad.

240. To measure a given circular arc.

1st. Suppose the centre of the circle given; and draw the radii from it to both extremities of the arc. Then, with the Protractor, or by some other means, measure the angle contained by these radii, and let it be expressed by A°; measure also the radius, if it be not already known. Then, since (84), in the same circle, any two arcs are proportional to the angles which they subtend at the centre, the proposed arc whole circumference: A° : 360°,

or, arc: 2 × rad. ; A°: 360°;

Ex. Let the arc subtend at the centre an angle of 36°, and let the radius be 7 feet, then

4 feet.

2nd. Suppose the given arc simply traced on a plane surface, but the centre of the circle not known. The centre may readily be found by (147, Part 11.); then proceed as before.

3rd. Suppose the centre of the circle to be inaccessible, as in the case of a vertical section of a railwaybridge.

D

Measure the chord of the whole arc, and the perpendicular distance of the highest point of the arc from that chord: lay down on paper these two lines in proper proportion according to these measurements, as AB, and CD in the annexed diagram; that is, if the chord of the given arc be 30 feet, suppose, and the greatest height of the arc be 10 feet, make AB equal to 1 in., and

A

CD (drawn from the middle point between A and B, at right angles to AB), equal to half an inch. Then, by (134, Part II.) construct the circle whose circumference shall pass through the three points A, B, D (there is only one such circle, see Cor. 1, p. 118); find the centre of this circle (50, Part 1.); and proceed as in the 1st case, remembering that in the result 1 inch represents 20 feet.

COR. It has been shewn, that the length of any circular arc which subtends an angle of A° at the centre

A is equal to 2 × rad. x ; therefore the length of the arc 360

which subtends an angle of 1o

Hence, since in the same circle, or in equal circles, arcs are proportional to the angles which they subtend at the centre, an arc, which subtends an angle of Ao, will be equal to Ax arc of 1o,

=Ax 01746 × rad. nearly.

Ex. Let the arc subtend at the centre an angle of 36°, and let the radius be 7 feet, then

the arc=36×7×01746=4.3999 ft.=4.4 ft. nearly.

The RULE in this case is-Multiply the number of degrees which the arc subtends at the centre by the radius, and the product by 01746; the result will be the length of the arc expressed in the same unit as the radius.

N.B. If the arc subtends an angle of degrees and minutes, or degrees, minutes, and seconds, the whole must be converted into a decimal, with a degree for the unit, before the Rule is applied.

Ex. Suppose an arc subtends an angle at the centre of 10°36', and the radius is a mile; then, since

10°36' 1038 10.6°,

.. length of arc-10'6×1× 01746 miles,

=185076 miles,

= 325 yards nearly.

241. To measure a given sector of a circle.

Let the arc of the given sector subtend an angle of A° at the centre; and suppose the whole circle divided into 360 equal sectors; then each of these sectors will have an arc subtending an angle of 1o at the centre; and it is plain, that

given sector: whole circle :: A° : 360°;

Hence, if the number of degrees which the arc subtends at the centre, and the radius, be given, we use the formula

A

area of sector =π × (rad.)3×