पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/३९७

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CHAPTER VII--MEASUREMENT OF AREAS.

62. In the case of a semicircular field of a diameter measuring 12, and of (another) field having a diameter of 36 in measure what is the circumference and what the area ?

The rule for arriving at the minutely accurate values relating to an elliptical figure:-

63.^[1] The square of the (shorter) diameter is multiplied 6 by and the square of twice the length (as measured by the longer diameter) is added square root sum ) the to this. (The of this gives measure of the circumference. This measure of the circumference multiplied by one-fourth of the (shorter) diameter gives the minutely accurate measure of the area of an elliptical figure.

An example in illustration thereof.

64. In the case of an elliptical figure, the length (as measured by the longer diameter) is 36, and the breadth (as measured by the shorter diameter) is 12. Tell me, after calculation. what the measure of the circumference is, and what the minutely accurate measure of the area.

The rule for arriving at the minutely accurate values in relation to a conchiform figure:-

65${\tfrac {1}{2}}$.^[2] The (maximum measure of the) breadth (of the figure), diminished by half (the measure of the breadth) of the mouth, and (then) multiplied by the square root of 10, gives rise to the measure of the perimeter. The square of half the (maximum)

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63.^  If a represents the measure of the longer diameter and b that of the shorter diameter of an ellipse, then, according to the rule given here the circumference is ${\sqrt {6b^{2}+4a^{2}}}$ and the area is ${\tfrac {1}{4}}b\times {\sqrt {6b^{2}+4a^{2}}}$. It may be noted that this stanza, as found in the MSS., omits to mention that the square root of the quantity is to be taken for arriving at the value of the circumference. The formula for the area given here is only an approximation, and seems to be based on the analogy of the area of a circle as represented by $\pi d\times {\tfrac {d}{4}}$ where d is the diameter and $\pi d$ is the circumference.

65${\tfrac {1}{2}}$.^  Algebraically, circumference $=(a-{\tfrac {1}{2}}m)\times {\sqrt {10}};$

${\text{ area }}=\left[\left\{(a-{\tfrac {1}{2}}m)\times {\tfrac {1}{2}}\right\}^{2}+\left({\tfrac {m}{4}}\right)^{2}\right]\times {\sqrt {10}}$; where a is the measure of the