It would, however, be very desirable to have another sub stitute in terms of the upper and lower diameter of the solid, in place of the diameter of a great circle ; for this purpose, we shall here give the following rules : Given any two parallel chords in a circle, and their distance, to ,find the distance of the greater chord from the centre.
To the square of the distance between the chords add the square of half the lesser chord. The difference between this sum and the square of half the greater chord, divided by twice the distance of the chords, gives the distance from the greater chord to the centre.
Example.—Suppose the greater chord, c D, is 4S feet, and the lesser, A B, 30 feet, and their distance, E a, 13 feet ; what is the distance, E F, from the centre to the greater chord, c n ? 13 30 = 15 48 = 24 13 2 15 2 24 39 75 9613 15 4S [chord. — [chord.
169 225 square of the less 576 square of greater 169 square of distance. 394 394 26) (7 dist. required.
- 152 Given the chord of a circle, and its distance from the centre, to find the radius of the circle.
To the square of the half chord, add the square of the distance from the centre, and the square root of the sum will be the radius required.
Example.—Given the chord o D, 48 feet, and its distance E r from the centre, 7 feet, the radius of the circle is required.
7 X 7 = 49 2 24 96 48 576 49 625 (25 the radius required.
4 45) 225 c),-)r --a Suppose, then, that we would wish to obtain the area o. the curved surface of a dome, contained between two parallel planes, the greater of which is 30 feet diameter, and the lesser 20 feet diameter, and the distance between them being 5 feet.
10 x 10 = 100 15 x 15 = 225 5 x 5 = 25 125 125 1,0) 10,0 the difference.
10. feet, the distance of the greater chord from the centre.
.2796 Say then, that the diameter is 36 feet, omitting the very small fractional part .02.
Then 36 X 3.1410 = 113.976 the circumference of a great circle.
Therefore 113.976 X 5 = 569.SSO feet, the superficial content of the dome.
Pi !lure 2.—In order to show the truth of the above rule, for finding the above diameter of a circle, from the two chords, and the distance between them being given ; Let y = F I, the height of the lesser segment ; F B, the distance between the chords ; x = B D, the distance from the nearest chord A a, to the centre 1; c = A a, half the greater chord A o; c = E F, half the lesser chord ; Then, by the property of the circle, we have