# पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/४३३

एतत् पृष्ठम् अपरिष्कृतम् अस्ति
237
CHAPTER VII--MEASUREMENT OF AREAS.

root (of the quotient resulting thus), the value of the diameter happens to result. In relation to a regular circular figure, the measure of the area and the circumference are to be made out as explained before.

An example in illustration thereof

164${\displaystyle {\tfrac {1}{2}}}$. In the case of a regular circular figure, the accurate measure of the area has been pointed out to be 5. Calculate quickly and tell me what the diameter of this (circle) may be.

On knowing the approximate measure as well as the accurate measure of an area, tho rule for arriving at a quadrilateral figure with two equal side, as well as at a quadrilateral figure with three equal sides, having those same approximate and accurate measures (as such measures of their areas):-

165${\displaystyle {\tfrac {1}{2}}}$. In the case of (the quadrilateral with) two equal sides, the square root of the difference between the squares of the (approximate and accurate) measures of the area is to be obtained. On adding (this square root) to the optionally chosen quantity and on subtracting (the same square root from the same optionally chosen quantity), the base and the top-side are so obtained as to have to be divided by the square root of the optional quantity. The approximate measure of the area, gives rise to the value of the sides so as to have to be divided by the square root of the optional quantity.

165${\displaystyle {\tfrac {1}{2}}}$. If R represents the approximate area of a quadrilateral with two equal sides, and the accurate value thereof, and p is the optionally chosen number, then

${\displaystyle base={\tfrac {{\sqrt {R^{2}-r^{2}}}+p}{\sqrt {p}}};{\text{ top-side }}={\tfrac {p-{\sqrt {R^{2}-r^{2}}}}{\sqrt {p}}};}$ and each of the
${\displaystyle equalsides={\tfrac {R}{\sqrt {p}}}.}$. If a,b,c and d be the measures of the sides of the quadrilateral with two equal series, then it may be seen that