पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/३९९

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CHAPTER VII--MEASUREMENT OF AREAS.

The rule for arriving at the minutely accurate values relating to a figure resembling (the longitudinal section of) the yava grain, and also to a figure having the outline of a bow :-

70${\tfrac {1}{2}}$.^[1] It should be known that the measure of the string (chord) multiplied by one-fourth of the measure of the arrow, and then multiplied by the square root of 10, gives rise to the (accurate) value of the area in the case of a figure having the outline of a bow as also in the case of a figure resembling the (longitudinal) section of a yava grain.

Examples in illustration thereof.

71${\tfrac {1}{2}}$. In the case of a figure resembling (the longitudinal section of the yava grain, the (maximum) length is 12 daṇdas. the two ends are needle points, and the breadth in the middle is 4 daṇdas. What is the area ?

72${\tfrac {1}{2}}$. In the case of a figure having the outline of a bow, the string is 24 in measure; and its arrow is taken to be 4 in measure. What may be the minutely accurate value of the area ?

The rule for arriving at the measure of the (bent) stick of the bow as well as of the arrow in the case of a figure having the outline of a bow :-

73${\tfrac {1}{2}}$.^[2] The square of the arrow measure is multiplied by 6. To this is added the square of the string measure. The square

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70${\tfrac {1}{2}}$.^  The figure resembling a bow is obviously the segment of a circle. The area of the segment as given here $=c\times {\tfrac {p}{4}}\times {\sqrt {10}}/$. This formula is not accurate. It seems to be based on the analogy of the rule for obtaining the area of a semi-circle, which area is evidently equal to the product of $\pi$, the diameter and one-fourth of the radius, i.e.,$\pi \times 2r\times {\tfrac {r}{4}}$

The figure resembling the longitudinal section of a good grain may be easily seen to be made up of two similar and equal segments of a circle applied to each other so as to have a common chord. It is evident that in this case the value of the arrow-line becomes doubled. Thus the same formula is made to hold good here also

73${\tfrac {1}{2}}$ & 74${\tfrac {1}{2}}$.^  Algebraically,

${\begin{aligned}&{\text{arc }}={\sqrt {6p^{2}+c^{2}}}\\&{\text{perpendicular }}={\tfrac {a^{2}-c^{2}}{\sqrt {6}}}\\&{\text{chord }}={\sqrt {a^{2}-6p^{2}}}\end{aligned}}$