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CHAPTER VII--MEASUREMENT OF AREAS.

The rule for arriving at the (longish quadrilateral) figure associated with a diagonal having a numerical value optionally determined:-

122${\tfrac {1}{2}}$.^[1] Each of the various figures that are derived with the aid of the, given (bījas) is written down; and by means (of the measure) of its diagonal the (measure of the) given diagonal is divided. The perpendicular-side, the base, and the diagonal (of this figure) as multiplied by the quotient (here) obtained, give rise to the perpendicular-side, the base and the diagonal (of the required figure).

An example in illustration thereof.

123${\tfrac {1}{2}}$-124${\tfrac {1}{2}}$.O mathematician, quickly bring out with the aid of the given (bījas) the (value of the) perpendicular-sides and the bases of the four longish quadrilateral figures that have respectively 1 and 2, 2 and 3, 4 and 7, and 1 and 8, for their bījas, and are also characterised by different bases. And, (in the problem) here, the diagonal is (in value) 65. Give out (the measures of) what may be the (required) geometrical figures (in that case).

The rule for arriving at the numerical values of the base and the perpendicular side of that derived longish quadrilateral figure, the numerical measures of the perimeter as also of the diagonal whereof are known:-

125${\tfrac {1}{2}}$.^[2] Multiply the square of the diagonal by two; (from the resulting product), subtract the square of half the perimeter; (then) get at the square root (of the resulting difference). If (this square root be thereafter) utilized in the performance of the

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122${\tfrac {1}{2}}$.^  The rule is based on the principle that the sides of a right angled triangle vary as the hypotenuse, although for the same measure of the hypotenuse there may be different sets of values for the sides.

125${\tfrac {1}{2}}$.^  If a and b represent the sides of a rectangle, then ${\sqrt {a^{2}+b^{2}}}$ is the measure of the diagonal, and $2a+2b$ is the measure of the perimeter. It can be seen easily that

${\Bigg \{}{\tfrac {2a+2b}{2}}+{\sqrt {2\left({\sqrt {a^{2}+b^{2}}}\right)^{2}-\left({\tfrac {2a+2b}{2}}\right)^{2}}}{\Bigg \}}\div 2=a$; and

${\Bigg \{}{\tfrac {2a+2b}{2}}-{\sqrt {2\left({\sqrt {a^{2}+b^{2}}}\right)^{2}-\left({\tfrac {2a+2b}{2}}\right)^{2}}}{\Bigg \}}\div 2=b$.

These two formulas represent algebraically the method described in the rule here.