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36
JAINA ANTIQUARY.
[Vol. I
“The diameter as diminished by the arrow being multiplied by four times the arrow becomes the square of the chord. Six times the square of the arrow being added by that becomes the square of the are.
"The square of the chord being added by four times the square of the arrow and then divided by four times the arrow gives as a rule the diameter of the circle. er of the circle.
. “A quarter of the product of the chord and arrow, and half the sum of the chord and arrow being multiplied respectively by ✓10 and the arrow give the gross and neat values of the area of the segment.
“The square of the chord being increased by the square of twice the arrow and divided by four times the arrow becomes the diameter. The difference of the squares of the chord and arrow is divided by six ; the square-root of the result is the arrow.
. “Subtract the square-root of the difference of the squares of the diameter and chord from the diameter. Know that half the remainder is the arrow..
“The square of the arc as divided by the twice the arrow is diminished by the arrow : half the remainder is the diameter. Diminish by the diameter the square-root of the square of the diameter as added by half the square of the arc: the remainder will be the arrow.
“The sum of the diameter and half the arrow being multiplied by four times the arrow gives the square of the arc. On subtracting six times the square of the arrow from that is obtained the square of the chord.”
If c=chord, h=arrow, a=aro and A'=area of a segment of a cirole of diameter d, we shall have tbe formulae.
(i) c2=4h (d–h), (ii) ao=6h2 +0%,
(iii) d = c++ th 2
4h
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