(of this derived longish quadrilateral figure), gives rise to the measure of the top-side. The value of the perpendicular-side (of the derived longish quadrilateral figure) on being multiplied by two and increased by the value of the top-side (already arrived at, gives rise to the value of the base. The value of the base (of the derived longish quadrilateral figure) is (the same as that
The first thing we have to do is to construct a rectangle with the aid of the given bījas in accordance with the rule laid down in stanza 90; in this chapter. That rectangle comes to have 5 for the measure of its smaller side, 12 for the measure of its larger side, and 13 for the measure of its diagonal; and its area is 60 in value. Now the area given in the problem is to be multiplied by the square of the given optional number in the problem, so that we obtain . From this 63, we have to subtract 60, which is the measure of the area of the rectangle constructed on the basis of the given bījas: and this gives 3 as the remainder. Then the thing to be done is to construct a rectangle, the area whereof is equal to this 3, and one of the sides is equal to the longer side of the rectangle derived from the same bījas. Since this longer side is equal to 12 in value, the smaller side of the required rectangle has to be in value as shown in the figure here. Then the two triangles, into which the rectangle derived from the bījas, may be split up by its diagonal are added one on each side to this last rectangle, so that the sides measuring 12 in the case of these triangles coincide with the sides of the rectangle having 12 as their measure. The figure here exhibits the operation.
Thus in the end we get the quadrilateral figure having two equal sides, each of which measures 13, the value of the other two sides being and respectively. From this the values of the sides of the quadrilateral required in the problem may be obtained by dividing by the given optional number namely 3, the values of its sides represented by and .
